Answers
From the given list, let's select the statement that best defines a conductor.
A conductor can be defined as any material that allows the flow of electric charge.
Examples of conductors are:
• Copper
,• Aluminium
,• Silver...
Therefore, the best statement that defines a conductor is that a conductor is a material that allows easy movement of charge.
• Option B is wrong because most conductors are made of metal.
,• Option C is wrong because it is an ,insulator ,hinders the passage of electricity.
,• Option D is wrong because a glass is an insulator not a conductor.
ANSWER:
A. allows easy movement of charge
Related Questions
Part of a light ray striking an interface between air and water is refracted, and part is reflected, as shown. The index of refraction of air is 1.00 and the index of refraction of water is 1.33. The frequency of the light ray is 7.85 x 10^16 Hz.(a) If angle 1 measures 40°, find the value of angle 2.(b) If angle 1 measures 40°, find the value of angle 3.(c) Calculate the speed of the light ray in the water.(d) Calculate the wavelength of the light ray in the water.(e) What is the largest value of angle 1, that will result in a refracted ray?
Answers
We will use Snell's law, which states:
[tex]n_1\sin \theta_1=n_2\sin \theta_2[/tex]Where n1 and n2 are the refraction indexes and their respective angles are "theta1" and "theta2".
For part A we replace the values:
[tex]1\sin 40=1.33\sin \theta_2[/tex]Now we solve for "theta2" first by dividing both sides by 1.33:
[tex]\frac{\sin40}{1.33}=\sin \theta_2[/tex]Now we use the inverse function for sine:
[tex]\arcsin (\frac{\sin 40}{1.33})=\theta_2[/tex]Solving the operation:
[tex]28.9=\theta_2[/tex]For part B, since "theta1" and "theta3" are angles of reflection, according to the reflection law, these angles are equal, therefore:
[tex]\theta_3=\theta_1=40[/tex]For part C. The index of refraction is defined as:
[tex]n=\frac{c}{v}[/tex]Where "c" is the speed of light in a vacuum and "v" is the speed of light in the medium. Replacing the values:
[tex]1.33=\frac{3\times10^8\text{ m/s}}{v}[/tex]Now we solve for "v":
[tex]v=\frac{3\times10^8\text{ m/s}}{1.33}[/tex]Solving the operation:
[tex]v=2.26\times10^8\text{ m/s}[/tex]For part d. We will use the following formula:
[tex]\lambda=\frac{v}{f}[/tex]Where "v" is the speed and "f" is the frequency. Replacing we get:
[tex]\lambda=\frac{2.26\times10^8\text{ m/s}}{7.85\times10^{16}s^{-1}}[/tex]Solving the operations:
[tex]\lambda=0.288\times10^{-8}m[/tex]For part e. The largest value of the angle of incidence that will result in refraction is 90 degrees.
Hey! I really need help with this question please :)
Answers
Answer: B
Explanation:
The formula for calculating the efficiency of a heat engine is expressed as
Efficiency = useful work done/Heat energy supplied x 100
From the information given,
Heat energy supplied = 500
useful work done = 50
Efficiency = 50/500 x 100
Efficiency = 10%
A sample of unknown material weight 900N In air and and 400N when submerged in an alcohol solution with a density of 0.7 g/cm³.What is the density of the material ?
Answers
1.26 g/cm³
ExplanationStep 1
given
[tex]\begin{gathered} F_{g(air)}=900\text{ N} \\ F_{g(alchodol)}=400\text{ N} \\ \rho_{alcohol}=0.7\text{ }\frac{g}{cm^3} \end{gathered}[/tex]unknown; the density of the material, so
[tex]\begin{gathered} F_B=F_{g(air)}-F_{g(alcohol)} \\ F_B=900\text{ N-400 N=500 N} \end{gathered}[/tex]so, the proportion is
the ratio of the force equals the ratio of the density ,so
[tex]\begin{gathered} \frac{F_{g(air)}}{F_B}=\frac{\rho_{material}}{\rho_{alcholol}} \\ replace \\ \frac{900\text{ N}}{500\text{ N}}=\frac{\rho_{material}}{0.7\text{ }\frac{g}{cm^3}} \\ mutliply\text{ both sides by 0.7}\frac{g}{cm^3} \\ \frac{900\text{N}}{500\text{N}}*0.7\text{ }\frac{g}{cm^3}=\frac{\rho_{mater\imaginaryI al}}{0.7\text{\frac{g}{cm^{3}}}}*0.7\frac{g}{cm^3} \\ 1.26\frac{g}{cm^3}=\text{ density of the material } \\ \end{gathered}[/tex]so, the density of the material is
1.26 g/cm³
I hope this helps you
If an astronaut weighs 148 N on the Moon and 893 N on Earth, then what is his mass on Earth?
_____ kg
Answers
The mass on Earth is 91.1 kg
The weight on the Moon is 148 N
The weight on the Earth is 893 N
We need to apply the concept of force.
Weight=massxacceleartion
893=mass of man x acceleration of gravity on earth
893= mass x 9.8
mass = 91.1 kg
Therefore, the mass on earth is 91.1 kg.
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Answer:
the mass on Earth is 91.1 kg
Explanation:
A worker pushes horizontally on a large crate with a force of 245 N, and the crate is moved 3.5 m. How much work was done? answer in : ___J
Answers
The amount of work done by a force can be written as the following:
[tex]W=F.\Delta x[/tex]For our case, we can replace our values and we'll get:
[tex]W=245*3.5=857.5J[/tex]Thus, the amount of work done is 857.5J
Multiple part question Here are the needed details:Five rotations took 5.15 seconds 1 rotation took 1.07s Distance from shoulder to elbow is 29 cm distance from shoulder to middle of the hand is 57cm.Questions:2. A how far in degrees did the hand travel during the five rotations?B. How far in radians did the hand travel during the five rotations?C. How far in meters did the hand travel during the five rotations?3. A. What was the average angular speed (degrees/s and rad/s) of the hand?B. What was the average linear speed (m/s) of the hand?C. Are the answers to a and b the same or different? Explain.4. A. What was the average angular acceleration (degrees/s squared and rad/s squared) of the hand. How do you know?B. What was the average centripetal acceleration (m/s squared) of the hand?C. Are the answers to a and b the same or different. Explain.5. A. How far (degrees and rad) did the elbow travel during the five rotations?B. How far (m) did the elbow travel during the five rotations?C. How do these compare to the hand? Why are they the same and or/different?6. A. What was the average angular speed (degrees/s and rad/s) of the elbow?B. What was the average linear speed (m/s) of the elbow?C. How do these compare to the hand? Why are they the same and or/different?7. A. What was the average angular acceleration (degrees/s squared and rad/s squared) of the elbow?B. What was the average centripetal acceleration (m/s squared) of the elbow?C. How do these compare to the hand? Why are they the same and or/ different?
Answers
Given:
Time taken for 5 rotations = 5.15 seconds
Time for 1 rotation = 1.07 seconds
Distance from shoulder to elbow = 29 cm
Distance from shoulder to the middle of the hand = 57 cm
Let's use the information above to answer the following questions.
Question 2:
Let's determine how far in degrees the hand travelled during the five rotations.
In one full rotation, we have 360 degrees.
Thus, 5 full rotations = 5 * 360 = 1800 degrees
Therefore, in 5 full rotations, the hand travelled 1800 degrees.
B. In radians, we have:
180 degrees = π rad
[tex]1800\degree=\frac{\pi}{180}\ast1800=10\pi\text{ radians}[/tex]C. To find the distance in meters, we have:
Distance from elbow to shoulder = 29 cm = 0.29 meters
[tex]2\pi\ast5\ast0.29=9.11\text{meters}[/tex]Therefore, the hand travelled 9.11 meters during the five rotations.
Question 3:
A. To find the average angular speed, apply the formula:
[tex]\begin{gathered} w=\frac{10\pi}{t}\text{ (rad/s)} \\ \\ w=\frac{1800}{t}\text{ (degre}es\text{/s)} \end{gathered}[/tex]Where t = 5.15 seconds
Thus, we have:
[tex]\begin{gathered} w=\frac{10\pi}{5.15}=6.1\text{ rad/s} \\ \\ w=\frac{1800}{5.15}=349.5\text{ degre}es\text{/s} \end{gathered}[/tex]B. Average linear speed of the hand.
To find the average linear speed of the hand, we have:
[tex]v=\frac{10\pi r}{t}=\frac{10\pi}{5.15}\ast\frac{1}{2}=3.05\text{ m/s}[/tex]C. The average angular speed and average linear speed are the same
A 72.5 kg student sits at a desk 1.25 m away from a 80.0 kg student. What is the magnitude of the gravitational force between the two students?
Answers
Given:
The mass of one student is,
[tex]m_1=72.5\text{ kg}[/tex]The mass of the other student is,
[tex]m_2=80.0\text{ kg}[/tex]The distance between them is,
[tex]d=1.25\text{ m}[/tex]The gravitational force between them is,
[tex]F=G\frac{m_1m_2}{d^2}[/tex]Here the gravitational constant is,
[tex]G=6.6\times10^{-11}\text{ }\frac{N.m^2}{\operatorname{kg}}[/tex]Substituting the values we get,
[tex]\begin{gathered} F=\frac{(6.6\times10^{-11})\times72.5\times80.0}{(1.25)^2} \\ =2.5\times10^{-7}\text{N} \end{gathered}[/tex]Hence the second option is correct.
Block A in (Figure 1) has mass 0.900 kg , and block B has mass 3.00 kg . The blocks are forced together, compressing a spring S between them; then the system is released from rest on a level, frictionless surface. The spring, which has negligible mass, is not fastened to either block and drops to the surface after it has expanded. Block B acquires a speed of 1.35 m/s .
Part A: What is the final speed of block A?
Part B: How much potential energy was stored in the compressed spring?
Answers
(a) The final speed of block A is determined as 4.5 m/s.
(b) The potential energy that was stored in the compressed spring is 11.85 J.
What is the final speed of block A?
The final speed of block A is determined by applying the principle of conservation of linear momentum as follows;
Pa = Pb
where;
Pa is the momentum of block APb is the momentum of block Bmv (block A) = mv (block B)
(0.9 kg)(v) = (3 kg)(1.35 m/s)
0.9v = 4.05
v = 4.05/0.9
v = 4.5 m/s
The potential energy stored in the compressed spring is calculated as follows;
Apply the principle of conservation of energy.
U = K.E
where;
K.E is the kinetic energy of the blocksU = ¹/₂mv² (A) + ¹/₂mv² (B)
U = ¹/₂(0.9)(4.5²) + ¹/₂(3)(1.35²)
U = 11.85 J
Thus, the potential energy that was stored in the compressed spring is determined by applying the principle of conservation of energy.
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Which statement best describes Earth's oceans?A. The waters of Earth's five major oceans rarely mix with oneanother.B. More than 95% of the water in the hydrosphere is found in oceans.C. The oceans contain almost all of Earth's freshwaterD. Oceans cover about 25% of Earth's surface.
Answers
From the given list, let's select the best statement that describes Earth's oceans.
The oceans of the earth are the principal component of the Earth's hydrosphere. The major oceans are the pacific ocean, atlantic ocean, indian ocean, souther ocean and the Arctic ocean.
The ocean covers more than 70% of the surface of the Earth and 97% of the Earth's water.
More than 95% of the Earth's water are found in the oceans.
The oceans contain only about 3% of the Earth's freshwater.
The waters of the Earth's five major oceans are connected with one another.
Therefore, the best statement which best describes the Earth's oceans is:
More than 95% of the water in the hydrosphere is found in oceans.
ANSWER:
B. More than 95% of the water in the hydrosphere is found in oceans.
On which of the following does the speed of a falling object depend?a.) v ∝ mb.)v ∝ mc.)v ∝ Δh
Answers
Given:
The falling object
To find:
The dependence on the speed of the falling object
Explanation:
For an object falling freely, the total mechanical energy remains always constant. So, we can write, that the decrease in potential energy will be equal to the increase in the kinetic energy that is
[tex]\begin{gathered} \frac{1}{2}mv^2=mgh \\ v^2=2gh \end{gathered}[/tex]Hence, the speed of the falling object does not depend on the mass it depends on the height difference.
Car A rear ends Car B, which has twice the mass of A, on an icy road at a speed low enough so that the collision is essentially elastic. Car B is stopped at a light when it is struck. Car A has mass m and speed v before the collision. After the collision.. A)each car has half the impulse of before B)each car has double the impulse of before C)each car have the same impulse D)each car has an impulse in ratio to its mass.
Answers
We are given the following information.
The collision is elastic.
Car B has twice the mass of car A.
Recall that in an elastic collision, the momentum and the kinetic energy are conserved.
Impulse is basically the change in momentum.
Since car B has 2 times the mass of car A, the momentum of car B will be 4 times the momentum of car A.
This means that the impulse of car B will be greater than the impulse of car A.
Option D says that each car has an impulse in ratio to its mass meaning that a car with a larger mass will have a larger impulse and vice versa.
Therefore, we can conclude that after the collision, each car has an impulse in ratio to its mass.
What is absolute zero?0k0c273-100
Answers
The absolute zero is measured in Kelvin scale
Thus absolute zero is 0K.
order the colors of the discuses to show the size of the force applied to throw each discus
Answers
Since the acceleration will be the same for all, the order would be the following [From larger to smaller force]:
Blue
Green
Orange
Red
Purple
Analyze the collision of a baseball with a bat. During which portion of the collision does the baseball’s velocity reach zero?1) before the collision2) during the collision3) one second after the collision4) one-hundredth of a second after the collision
Answers
ANSWER:
2) during the collision
STEP-BY-STEP EXPLANATION:
We have that when a baseball ball collides with a bat, its velocity changes from positive to negative, that is, the ball when hit with an opposing force changes its direction exactly opposite to the initial one.
Therefore, during the collision, at some point in time, the velocity of the ball reaches zero and then finally changes its direction associated with an increase in the velocity of the ball.
How much potential energy due to gravity would a person have if they were standing on top of a building that is 36.2 m high? Assume that they have a mass of 79.2 kg.
Answers
Given:
The mass of thee person is
[tex]m=79.2\text{ kg}[/tex]The height at which person standing is
[tex]h=36.2\text{ m}[/tex]Required: calculate the potential energy of the person
Explanation:
when anybody of mass m is at a distance h from the earth's surface then it has potential energy that is given as
[tex]P.E=mgh[/tex]where g is the acceleration due to gravity whose value is
[tex]9.8\text{ m/s}^2[/tex]Plugging all the values in the above relation, we get;
[tex]\begin{gathered} P.E=79.2\text{ kg}\times9.8\text{ m/s}^2\times36.2\text{ m} \\ P.E=28096.992\text{ J} \end{gathered}[/tex]Thus, the potential energy is
[tex]28096.992[/tex]carts, bricks, and bands
5. What acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with one brick?
a. About 0.25 m/s2
b. About 0.50 m/s2
c. About 0.75 m/s2
d. About 1.00 m/s2
Answers
B. The acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with one brick is 0.5 m/s².
What is the applied force on an object?The force applied on object is obtained by multiplying the mass and acceleration of the object.
According to Newton's second law of motion, the force exerted on an object is directly proportional to the product of mass and acceleration of the object.
Also, the applied force is directly proportional to the change in the momentum of the object.
Mathematically, the force acting on object is given as;
F = ma
a = F/m
where;
a is the acceleration of the objectm is the mass of the objectF is the applied forceFrom the trials, the acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with one brick is 0.5 m/s².
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Jeff tosses a can of soda pop to Karen, who is standing on her 3rd floor balcony a distance of 8.5m above Jeff’s hand. Jeff gives the can an initial velocity of 16m/s, fast enough so that the can goes up past Karen, who catches the can on its way down. Calculate the velocity of the can the instant before Karen grabs the can. How long after Jeff tosses the can does Karen have to prepare to catch it?
Answers
ANSWER
9.51 m/s
EXPLANATION
We know that Jeff is 8.5m below Karen. He tosses the can up with initial velocity u = 16m/s and it passes where Karen is, so the maximum height of the can is 8.5m plus some more meters x. Then Karen catches the can in its way down, so when she does the can goes this distance x.
Let's find this distance. The height of an object thrown up with initial velocity u is:
[tex]y=ut-\frac{1}{2}gt^2[/tex]We know u = 16m/s but we don't know the time. This we can find from the final velocity of the can:
[tex]v=u-gt[/tex]At its maximum height the velocity is zero:
[tex]0=u-gt[/tex]Solving for t:
[tex]t=\frac{u}{g}[/tex]If we assume g = 9.8m/s²:
[tex]t=\frac{16m/s}{9.8m/s^2}=1.63s[/tex]We know that the can was in the air for 1.63 seconds until it reached its maximum height. The maximum height is:
[tex]y=16m/s\cdot1.63s-\frac{1}{2}\cdot9.8m/s^2\cdot1.63^2s^2[/tex][tex]y=26.08m-13.02m=13.06m[/tex]This is the maximum height of the can. The extra distance the can travelled above Karen is:
[tex]x=13.06m-8.5m=4.56m[/tex]In the can's way down, the initial velocity is 0, because it starts falling after stopping in its way up. The acceleration is still the acceleration of gravity and the height it falls is x. We can find the time it took to reach Karen's hand after it started falling:
[tex]y=\frac{1}{2}gt^2[/tex]Note that in this case we use the acceleration of gravity positive because it is in the same direction of the can's motion. Solving for t:
[tex]t=\sqrt[]{\frac{2y}{g}}[/tex][tex]t=\sqrt[]{\frac{2\cdot4.56m}{9.8m/s^2}}=\sqrt[]{0.93s^2}=0.97s[/tex]Knowing that the can was in the air for another 0.97 seconds after starting falling until it reached Karen's hand, we can find its velocity at that instant:
[tex]v=u+gt[/tex]Remember that in this case u = 0:
[tex]v=gt=9.8m/s^2\cdot0.97s=9.51m/s[/tex]The velocity of the can the instant before Karen grabs it is 9.51 m/s
An electric motor is used to do the 2.30 x104 J of work needed to lift an engine out of a car. If the motor draws a current of 3.2 A for 30 s, calculate the potential difference across the motor.
Answers
Given data
*The given energy is U = 2.30 x 10^4 J
*The given current is I = 3.2 A
*The given time is t = 30 s
The formula for the charge is given as
[tex]q=It[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} q=(3.2)(30) \\ =96\text{ C} \end{gathered}[/tex]The formula for the potential difference across the motor is given as
[tex]U=qV[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} 2.30\times10^4=(96)V \\ V=239.58\text{ V} \\ \approx240\text{ V} \end{gathered}[/tex]Hence, the potential difference across the motor is V = 240 V
How much power is created when you perform 55 Joule of work with a time 20 sec?
Answers
Answer:
2.75 watts
Explanation:
The power is equal to the work divided by time, so
P = W/t
Then, replacing W = 55 J and t = 20 sec, we get:
P = 55 J / 20 s = 2.75 Watts
Therefore, the power created is 2.75 watts
A 60.0 kg skier with an initial speed of 14 m/s coasts up a 2.50 m high rise as shown in the figure.
Find her final speed right at the top, in meters per second, given that the coefficient of friction between her skis and the snow is 0.38?
Answers
The final speed of the skier at the top mountain is determined as 9.27 m/s.
What is the change in the energy of the skier?
The change in the energy of the skier due to frictional force is calculated as follows;
ΔP.E = Pi + Ef
where;
Pi is the initial potential at the topEf is the energy lost to frictionThe distance of the plane travelled is calculated as;
sin35 = 2.5/L
L = 2.5 / sin35
L = 4.36 m
ΔP.E = mghi - μmgcosθ(L)
where;
m is the masshi is the initial heightg is acceleration due to gravityμ is coefficient of frictionΔP.E = (60 x 9.8 x 2.5) - (0.38)(60)(9.8) cos(35) x (4.36)
ΔP.E = 671.98 1 J
The final speed of the skier at the top of the plane;
P.E = K.E
P.E = ¹/₂mv²
v² = 2P.E /m
v = √(2P.E /m)
v = √(2 x 671.98) / 60)
v = 4.73 m/s
Total speed = -4.73 m/s + 14 m/s = 9.27 m/s
Thus, due to frictional force opposing the upward motion of the skier, the final speed at the top will be smaller than the initial speed.
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Giving a test to a group of students, the grades and gender are summarized belowGrades vs. Gender ABCMale10316Female465If one student was chosen at random, find the probability that the student got a B.
Answers
Answer:
20.45%
Explanation:
The probability that the student got a B is
[tex]\frac{total\#of\text{ students that got B}}{\text{total students }}\times100\%[/tex]Now, how many students are there in total?
The answer is
[tex]10+3+16+4+6+5=44\; \text{students}[/tex]How many students got a B?
The answer is
[tex]3+6=9\; \text{students}[/tex]therefore, the probability that the student has got a B is
[tex]\frac{9\text{ students }}{44\text{ students }}\times100\%=20.45\%[/tex]Hence, the probability that a student has got a B is 20.45%
A sound wave of wavelength 1.66m at a temperature of 23 C is produced for 2.5 seconds. How far does this wave travel? How many complete waves are emitted in this time interval?
Answers
To find how far it travels, we just have to multiply
[tex]1.66\times2.5=4.15[/tex]It travels 4.15 meters.
During this interval, there are emitted 2.5 waves.
You observe waves on the beach and measure that a wave hits the beach every 5 seconds. What is the period and the frequency of the waves ?
Answers
Answer:
Period = 5 seconds
Frequency = 0.2 Hz
Explanation:
The period is the time per cycle. So, if a wave hits the beach every 5 seconds, the period will be 5 seconds.
Additionally, the frequency is the inverse of the period, so the frequency of the waves can be calculated as:
[tex]\text{frequency = }\frac{1}{Period}=\frac{1}{5}=0.2\text{ Hz}[/tex]So, the answers are:
Period = 5 seconds
Frequency = 0.2 Hz
An object is dropped from a height of 65 m above ground level. A) determine the final speed in m/s, at which the object hits the ground c) determine the distance in meters, traveled during the last second of motion before hitting the ground.
Answers
Given:
height = 65 m
Given that the object is in free fall, let's solve for the following:
• (a). determine the final speed in m/s.
To find the final velocity, apply the kinematics equation:
[tex]v^2=u^2-2ax[/tex]Where:
v is the final velocity
u is the initial velocity = 0
a is the acceleration due to gravity = 9.8 m/s²
x is the displacement = 65 m
Thus, we have:
[tex]\begin{gathered} v^2=0^2-2(-9.8)(65) \\ \\ v^2=-(-1274) \\ \\ v^2=1274 \\ \\ \text{ Take the square root of both sides:} \\ \sqrt{v^2}=\sqrt{1274} \\ \\ v=35.69\text{ m/s} \end{gathered}[/tex]Therefore the final speed will be -35.69 m/s.
• (c). The distance traveled during the last second of motion before hitting the ground.
To find the distance, apply the formula:
[tex]H=ut+\frac{1}{2}at^2[/tex]Where:
H is the height.
u is the initial velocity = 0 m/s
t is the time
a is acceleration due to gravity.
Let's rewrite the formula to find the time traveled.
[tex]\begin{gathered} H=0t+\frac{1}{2}at^2 \\ \\ H=\frac{1}{2}at^2 \\ \\ t=\sqrt{\frac{2H}{a}} \end{gathered}[/tex]Thus, we have:
[tex]\begin{gathered} t=\sqrt{\frac{2*65}{9.8}} \\ \\ t=\sqrt{\frac{130}{9.8}} \\ \\ t=\sqrt{13.26} \\ \\ t=3.64\text{ s} \end{gathered}[/tex]Therefore, the time is 3.64 seconds.
Now, to find the distance traveled during the last second of motion, apply the formula:
[tex]s=\frac{1}{2}a(t_2^2-t_1^2)[/tex]Where:
t2 = 3.64 seconds
t1 = 3.64 seconds - 1 second = 2.64 seconds
Thus, we have:
[tex]\begin{gathered} s=\frac{1}{2}(9.8)((3.64)^2-(2.64)^2) \\ \\ s=4.9(13.2496-6.9696) \\ \\ s=4.9(6.28) \\ \\ s=30.77 \end{gathered}[/tex]Therefore, the distance in meters, traveled during the last second of motion before hitting the ground is 30.77 meters.
ANSWER:
(A). -35.69 m/s
(C). 30.77 m
Timmy walks 5 m North, 3m West, and finally 1 m South. What is his displacement from his starting point?
Answers
Timmy walks 5 m North, 3m West, and finally 1 m South then his displacement from the starting point would be 5 meters in the northwest direction.
What is displacement?Displacement describes this shift in location and it is calculated with the help of the initial and the final position of the object.
As given in the problem If Timmy walks 5 m North, 3m West, and finally 1 m South ,
The resultant displacement of the Timmy = √(4² + 3²)
= 5 meters
Thus, the resultant displacement of the Timmy would be 5 meters
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Two ropes support a 15.0 kg load between them. One rope points NW at an angle of 15.0 degrees to the horizontal. Second rope points NE at an angle of 20.0 degrees to the vertical. Determine the magnitude of the tension force in each of the ropes.
Answers
The tension in the first rope is 38.05 N and the tension in the second rope is 138.1 N.
What is the weight of the load?
The weight of the load due to the force of gravity is calculated as follows;
W = mg
where;
m is mass of the loadg is acceleration due to gravityW = 15 kg x 9.8 m/s²
W = 147 N
Since the weight of the load is acting downwards, the tension in each rope is calculated as follows;
The tension in the first rope, T1 = W sinθ
where;
θ is the angle of inclination above the horizontalT1 = 147 sin(15)
T1 = 38.05 N
The tension in the first rope, T2 = W sinθ
where;
θ is the angle of inclination above the horizontal = 90 - 20 = 70⁰T2 = 147 x sin(70)
T2 = 138.1 N
Thus, the tension in each rope is determined by calculating the vertical component of force in each rope.
Learn more about tension in ropes here: https://brainly.com/question/187404
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Based on the circuit voltage and the wattage consumption,determine the approximate ampere rating of the followingappliances. Remember amps = watts divided by voltage.a = w÷ VRound to the nearest whole amp.1. AC Compressor on a 240 volt line and using 5,000 watts, amps =_____2. baseboard heater on a 120 volt line and using 1,200 watts, amps =_____3. vacuum cleaner on a 120 volt line and using 500 watts, amps =______4. blender on a 115 volt line and using 300 watts, amps5. toaster on a 120 volt line using 1,100 watts, amps =_____
Answers
Given:
1.
The voltage of AC compressor is V = 250 V
The power of the AC compressor is P = 5000 W
2.
The voltage of the baseboard heater is V = 120 V
The power of the baseboard heater is P = 1200 W
3.
The voltage of the vacuum cleaner is V = 120 V
The power of the vacuum cleaner is P = 500 W
4.
The voltage of the blender is V = 115 V
The power of the blender is P = 300 W
5.
The voltage of the toaster is 120 V
The power of the toaster is P = 1100 W
Required:
1. The approximate ampere rating of the AC compressor.
2. The approximate ampere rating of the baseboard heater.
3. The approximate ampere rating of the vacuum cleaner.
4. The approximate ampere rating of the blender.
5. The approximate ampere rating of the toaster.
Explanation:
1. The approximate ampere rating of the AC compressor can be calculated as
[tex]\begin{gathered} I\text{ = }\frac{P}{V} \\ =\frac{5000}{240} \\ =20.833\text{ A} \\ \approx21\text{ A} \end{gathered}[/tex]2. The approximate ampere rating of the baseboard heater can be calculated as
[tex]\begin{gathered} I=\frac{1200}{120} \\ =\text{ 10 A} \end{gathered}[/tex]3. The approximate ampere rating of the vacuum cleaner can be calculated as
[tex]\begin{gathered} I\text{ = }\frac{500}{120} \\ =4.2\text{ A} \\ \approx4\text{ A} \end{gathered}[/tex]4. The approximate ampere rating of the blender can be calculated as
[tex]\begin{gathered} I\text{ =}\frac{300}{115} \\ =2.6\text{ A} \\ \approx3\text{ A} \end{gathered}[/tex]5. The approximate ampere rating of the toaster can be calculated as
[tex]\begin{gathered} I\text{ =}\frac{1100}{120} \\ =9.2\text{ A} \\ \approx\text{ 9 A} \end{gathered}[/tex]Final Answer:
1. The approximate ampere rating of the AC compressor is 21 A.
2. The approximate ampere rating of the baseboard heater is 10 A.
3. The approximate ampere rating of the vacuum cleaner is 4 A.
4. The approximate ampere rating of the blender is 3 A.
5. The approximate ampere rating of the toaster is 9 A.
How long does it take to stop a 1000 kg object moving at 20 m/s with a force 5000N? 2500 N, 1000 N, 500 N, 400 N, 200 N, 100 N
Answers
ANSWER
[tex]\begin{gathered} 5000N\Rightarrow4s \\ 2500N=8s \end{gathered}[/tex]EXPLANATION
To find the time taken to stop the object, we first have to find the acceleration of the object.
Since the force is working to stop the object (slow down the object), it means that the object is decelerating (slowing down).
To find the acceleration (deceleration), apply Newton's second law of motion:
[tex]F=ma[/tex]where F = force; m = mass; a = acceleration
Therefore, for a force of 5000N, we have that:
[tex]\begin{gathered} 5000=1000\cdot a \\ \Rightarrow\frac{5000}{1000}=a \\ \Rightarrow a=5m\/s^2 \end{gathered}[/tex]Now, we can apply Newton's equation of motion to find the time taken to stop the object:
[tex]v=u-at[/tex]where v = final velocity = 0 m/s; u = initial velocity = 20 m/s; t = time taken
Note: the negative sign indicates deceleration
Hence, the time taken for a force of 5000N to stop the object is:
[tex]\begin{gathered} 0=20-5\cdot t \\ \Rightarrow5t=20 \\ \Rightarrow t=\frac{20}{5} \\ t=4s \end{gathered}[/tex]For a force of 2500N, the deceleration is:
[tex]\begin{gathered} 2500=1000\cdot a \\ a=\frac{2500}{1000} \\ a=2.5m\/s^2 \end{gathered}[/tex]Hence, the time taken for a force of 2500N to stop the object is:
[tex]\begin{gathered} 0=20-2.5\cdot t \\ \Rightarrow2.5t=20 \\ t=\frac{20}{2.5} \\ t=8s \end{gathered}[/tex]It’s an assessment Jane multiplied 825x22 and got 3,300. Flynn multiples the same numbers and got 18,150 Which student is correct?
Answers
In order to determine which student is correct, multiply the given numbers, as follow:
8 2 5
x 2 2
1 6 5 0
1 6 5 0
1 8 1 5 0
As you can notice, the result of the multiplcation is 18,150, hence, Flynn is right.
The heating element of an iron operates at 110 V with a current of 11 A.(a) What is the resistance of the iron? Ω(b) What is the power dissipated by the iron? W
Answers
(a)
In order to calculate the resistance, we can use Ohm's Law:
[tex]\begin{gathered} R=\frac{V}{I}\\ \\ R=\frac{110}{11}\\ \\ R=10\text{ \Omega} \end{gathered}[/tex](b)
To calculate the power, we can use the formula below:
[tex]\begin{gathered} P=I\cdot V\\ \\ P=11\cdot110\\ \\ P=1210\text{ W} \end{gathered}[/tex]