How high does a rocket have to go above the earth's surface to be subject to a gravitational field from the earth that is 50.0 percent of its value at the earth's surface?
A) 2.650 km
B) 3,190 km
C) 9.020 km
D) 12.700 km
Answer:
A) 2.650 km
Explanation:
The relationship between acceleration of gravity and gravitational constant is:
[tex]g = \frac{Gm}{R^2}[/tex] ---- (1)
Where
[tex]R = 6,400 km[/tex] -- Radius of the earth.
From the question, we understand that the gravitational field of the rocket is 50% of its original value.
This means that:
[tex]g_{rocket} = 50\% * g[/tex]
[tex]g_{rocket} = 0.50 * g[/tex]
[tex]g_{rocket} = 0.5g[/tex]
For the rocket, we have:
[tex]g_{rocket} = \frac{Gm}{r^2}[/tex]
Where r represent the distance between the rocket and the center of the earth.
Substitute 0.5g for g rocket
[tex]0.5g = \frac{Gm}{r^2}[/tex] --- (2)
Divide (1) by (2)
[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}/\frac{Gm}{r^2}[/tex]
[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}*\frac{r^2}{Gm}[/tex]
[tex]\frac{1}{0.5} = \frac{1}{R^2}*\frac{r^2}{1}[/tex]
[tex]2 = \frac{r^2}{R^2}[/tex]
Take square root of both sides
[tex]\sqrt 2 = \frac{r}{R}[/tex]
Make r the subject
[tex]r = R * \sqrt 2[/tex]
Substitute [tex]R = 6,400 km[/tex]
[tex]r = 6400km * \sqrt 2[/tex]
[tex]r = 6400km * 1.414[/tex]
[tex]r = 9 049.6\ km[/tex]
The distance (d) from the earth surface is calculated as thus;
[tex]d = r - R[/tex]
[tex]d = 9049.6\ km - 6400\ km[/tex]
[tex]d = 2649.6\ km[/tex]
[tex]d = 2650\ km[/tex] --- approximated
How long does it take a plane, traveling at a constant speed of 123 m/s, to fly once around a circle whose radius is 4330 m?
Answer:
3.7 minExplanation:
Step one:
given data
speed = 123m/s
radius of circle= 4330m
Step two:
We need to find the circumference of the circle, it represents the distance traveled
C=2πr
C= 2*3.142*4330
C= 27209.72m
Step three:
We know that velocity= distance/time
time= distance/velocity
time= 27209.72/123
time=221.2 seconds
in minute = 221.2/60
time= 3.7 min
7. DRAW A PICTURE TO SHOW WORK.
Brandon buys a new Seadoo. He goes 12
km north from the beach. He jumps
wakes for 6 km to the east. Then chases
a boat 12 km south. He then turns and
goes 3 km to the West. What distance
did he cover? What was his
displacement?
Students are using a mallet to hit a drum. The diagram below shows this action.
Drum
Drum
When the mallet
hits the drum,
the drum skin
begins to vibrate.
particles of air
compressed
together
particles of air
compressed
together
Mallet
particles of air
moving apart wave of
again
compressed
particles moving
outward
Which statement BEST describes the phenomenon illustrated in the model?
Sound waves are being transmitted through the air in the drum.
Sound waves are being transferred from the mallet to the drum.
O Sound waves are being absorbed by the drum.
Sound waves are being reflected by the drum.
Answer:
A
Explanation:
i think
[2.21] Please help me find a) and b)
Answer:
A. 28.42 m/s
B. 41.21 m.
Explanation:
A. Determination of the initial velocity of the ball:
Time (t) to reach the maximum height = 2.9 s
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going against gravity)
Initial velocity (u) =?
Thus, we can obtain the initial velocity of the ball as follow:
v = u + gt
0 = u + (–9.8 × 2.9)
0 = u – 28.42
Collect like terms
u = 0 + 28.42
u = 28.42 m/s
Therefore, the initial velocity of the ball is 28.42 m/s.
B. Determination of the maximum height reached.
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going against gravity)
Initial velocity (u) = 28.42 m/s.
Maximum height (h) =?
Thus, we can obtain the maximum height reached by the ball as follow:
v² = u² + 2gh
0² = 28.42² + (2 × –9.8 × h)
0 = 807.6964 + (–19.6h)
0 = 807.6964 – 19.6h
Collect like terms
0 – 807.6964 = – 19.6h
– 807.6964 = – 19.6h
Divide both side by – 19.6
h = –807.6964 / –19.6
h = 41.21 m
Therefore, the maximum height reached by the ball is 41.21 m
find the vector parallel to the resultant of the vector A=i +4j-2k and B=3i-5j+k
Answer:
2008
Explanation:
2000+3+5======2008
Answer:
[tex]8\hat i-2\hat j-2\hat k[/tex]
Explanation:
Vectors in 3D
Given a vector
[tex]\vec P = P_x\hat i+P_y\hat j+P_z\hat k[/tex]
A vector [tex]\vec Q[/tex] parallel to [tex]\vec P[/tex] is:
[tex]\vec Q = k.\vec P[/tex]
Where k is any constant different from zero.
We are given the vectors:
[tex]\vec A = \hat i+4\hat j-2\hat k[/tex]
[tex]\vec B = 3\hat i-5\hat j+\hat k[/tex]
It's not specified what the 'resultant' is about, we'll assume it's the result of the sum of both vectors, thus:
[tex]\vec A +\vec B = \hat i+4\hat j-2\hat k + 3\hat i-5\hat j+\hat k[/tex]
Adding each component separately:
[tex]\vec A +\vec B = 4\hat i-\hat j-\hat k[/tex]
To find a vector parallel to the sum, we select k=2:
[tex]2(\vec A +\vec B )= 8\hat i-2\hat j-2\hat k[/tex]
Thus one vector parallel to the resultant of both vectors is:
[tex]\mathbf{8\hat i-2\hat j-2\hat k}[/tex]
_____ are group of tissue working together to perform a certain job.
Answer:
organ
Explanation:
Answer:
an organ
Explanation:
cell -> tissue -> organ -> organ system -> organism
How to calculate net radiation
Answer:
(1) R n = ( 1 − α ) R si − L ↑ + L ↓ where Rn is the net radiation (W m−2), Rsi is the solar radiation (W m−2), α is the soil surface albedo (α = 0–1
Explanation:
Light travels from a
laser across an 8 m
room where it reflects
off a mirror. Uniform or non uniform velocity
Answer:
Non-uniform velocity as the laser light beam has got reflected by the mirror
And as the light got reflected there is a change in velocity making it non-uniform velocity
A force Ě = F, î + Fy h acts on a particle
that undergoes a displacement of g = s, it
Sy Ĵ where F, = 10 N, F, = -1 N, 8z = 4 m,
and sy=1 m.
Find the work done by the force on the
particle.
Answer in units of J.
Answer:
Pt 1: [tex]W=39J[/tex]
Pt 2: θ = [tex]19.7[/tex]
Explanation:
Part 1:
[tex]W=FS[/tex]
[tex]W=F_{_xs_x}+F{_ys_y}[/tex]
imagine the "i"and the "j" with the hat
[tex]W=(10i-1j)(4i+3j)[/tex]
[tex]W=(10*4)-(1*1)[/tex]
[tex]W=40-1[/tex]
[tex]W=39J[/tex]
Part 2:
[tex]|F|=\sqrt{10^2+(-1)^2}[/tex]
[tex]|F|=\sqrt{101}[/tex]
[tex]|S|=\sqrt{4^2+1^2}[/tex]
[tex]|S|=\sqrt{17}[/tex]
[tex]W=|F||S|cos[/tex]θ
[tex]39=(\sqrt{101})(\sqrt{17} )cos[/tex]θ
θ = [tex]19.7[/tex]
Hope it helps
Which equation is correct according to Ohm’s law? Which equation is correct according to Ohm’s law? A.) V = IR B.) I = R/V C.) R = I/V D.) R = IV
Answer:
[tex]V = IR[/tex]
Explanation:
Required
Which equation represents ohm's law?
Literally, ohm's law implies that current (I) is directly proportional to voltage (V) and inversely proportional to resistance (R).
Mathematically, this can be represented as:
[tex]I\ \alpha\ \frac{V}{R}[/tex]
Convert the expression to an equation
[tex]I\ =\ \frac{V}{R}[/tex]
Multiply both sides by R to make V the subject
[tex]I\ * R\ =\ \frac{V}{R} * R[/tex]
[tex]I\ * R\ =V[/tex]
Reorder
[tex]V = I\ * R[/tex]
[tex]V = IR[/tex]
Option (a) is correct; Others are not
Answer:
V=ir
Explanation:
Ohm's law deals with the relation between voltage and current in an ideal conductor. It states that: Potential difference across a conductor is proportional to the current that pass through it. It is expressed as V=IR. The correct answers from the choices are:
v = ir
When weather predictions are incorrect what is the most likely cause
A: measurements of the initial conditions may have been very in accurate
B: small differences in models can lead to large differences in complex systems
C: The person predicting the weather may have had a bias
D: The elevation of different landforms I have been significantly in accurate
Answer:small differences in models can lead to large differences in complex systems
Explanation: this is the most accurate phrase
A horse has a momentum of 1200 kg·m/s. If the horse has a mass of 313 kg, what is the speed of the horse?
Answer:
3.83 m/sExplanation:
The speed of the horse can be found by using the formula
[tex]v = \frac{p}{m} \\ [/tex]
p is the momentum
m is the mass
From the question we have
[tex]v = \frac{1200}{313} \\ = 3.83386..[/tex]
We have the final answer as
3.83 m/sHope this helps you
Bobbie is on a skateboard riding it down a hill. As he approaches the hill his velocity is 26 m/s. While he travels down the hill he realizes his acceleration is 6 m/s2. At the bottom of the hill he finds out that his velocity is 102 m/s. What time will be required to create this change in velocity?
Answer:
Bobbie will need 12.7 seconds.
Explanation:
Constant Acceleration Motion
It's a type of motion in which the velocity of an object changes uniformly in time. If the speed changes from vo to vf in a time t, then the acceleration is:
[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]
The question states Bobbie is riding on a skateboard down a hill, starting from vo=26 m/s with an acceleration of a=6\ m/s^2 until his speed is 102 m/s. We are required to find the time needed to create that change of velocity.
Solving for t:
[tex]\displaystyle t=\frac{v_f-v_o}{a}[/tex]
[tex]\displaystyle t=\frac{102-26}{6}=\frac{76}{6}[/tex]
t = 12.7 s
Bobbie will need 12.7 seconds.
What causes friction between two solids?
Answer:
Friction is when 2 solids move against each other. The cause of friction is adhesion, and surface roughness. Surface roughness is when a surface is rough enough that is causes friction against another surface. Adhesion is when 2 surfaces collide because of thier molecular force.
A vertical tube one meter long is open at the top. It is filled with 50 cm of water. If the velocity of sound is 344 m/s, what will the fundamental resonant frequency be (in Hz)?
Answer:
The fundamental resonance frequency is 172 Hz.
Explanation:
Given;
velocity of sound, v = 344 m/s
total length of tube, Lt = 1 m = 100 cm
height of water, hw = 50 cm
length of air column, L = Lt - hw = 100 cm - 50 cm = 50 cm
For a tube open at the top (closed pipe), the fundamental wavelength is given as;
Node to anti-node (N ---- A) : L = λ / 4
λ = 4L
λ = 4 (50 cm)
λ = 200 cm = 2 m
The fundamental resonance frequency is given by;
[tex]f_0 = \frac{v}{\lambda}\\\\f_0 = \frac{344}{2}\\\\f_0 = 172 \ Hz\\\\[/tex]
Therefore, the fundamental resonance frequency is 172 Hz.
What is the name for family labeled #4 (Yellow)?
#3
#5
#2
#
341 sud-
lasa 1
17:55
Alkaline Earth Metals
Metalloids
Transition Metals
Alkali Metals
Answer:
transition metals im sorry if this was too late
Object A has a mass of 2 kg; object B’s mass is 10 kg. If object B is at rest and object A runs into it at low speed, object B:
A. Will move off at high speed.
B. Cannot move off with high speed.
C. Will loose speed in the collision.
D. Will remain at rest.
Answer:
Cannot move off with high speed
Explanation:
Just did it
In order to prevent injury in a car crash, it is recommended that you _______.
A) Increase the time of the collision.
B) Increase the change in momentum of the collision.
C) Increase the force in the collision.
D) Increase the initial velocity of the collision.
Increase of momentum of the collision will have the car in a unsettling position, creating an unsettling spot for it. Moreover increasing this would most likely take worse effect, it is better than increasing the time, which will only create the car faster. Let me show you what I mean...
A) Creates the car go faster, creating an even worse tragedy. B) save this for later...C) Increasing force will only result in worse damage.And finally, D) Increasing the velocity is basically increasing speed, once again, making things worse.So overall in this piece, the answer may very well end up being B. I sincerely hope this helped you by whatever means possible. It's logic that helps in real life situations, so take this as a little lesson- I guess :3A disk with a rotational inertia of 2.0 kg m2 and a radius of 1.6 m is free to rotate about a frictionless axis perpendicular to the disk's face and passing through its center. A force of 5.0 N is applied tangentially to the rim of the disk. What is the angular acceleration of the disk as a result of this applied force?
a) 4.0 rad/s2
b) 1.0 rad/s2
c) 2.0 rad/s2
d) 0.40 rad/s2
e) 0.80 rad/s2
Answer:
a) 4.0 rad/s2
Explanation:
For rigid bodies, Newton's 2nd law becomes :τ = I * α (1)
where τ is the net external torque applied, I is the rotational inertia
of the body with respect to the axis of rotation, and α is the angular
acceleration caused by the torque.
At the same time, we can apply the definition of torque to the left side of (1), as follows:[tex]\tau = F*r*sin \theta (2)[/tex]
where τ = external net torque applied by Fnet, r is the distance
between the axis of rotation and the line of Fnet, and θ is the
angle between both vectors.
In this particular case, as Fnet is applied tangentially to the disk, Fnet
and r are perpendicular each other.
Since left sides of (1) and (2) are equal each other, right sides are equal too, so we can solve for the angular acceleration as follows:[tex]\alpha = \frac{F*r}{I} = \frac{5.0N*1.6m}{2.0 kg*m2} = 4.0 rad/s2 (3)[/tex]
Mike rides his horse with a constant speed of 20 km/h. How far can he travel in 4 hours?
Answer:
Mike can travel 80 Km in 4 hours
At the top of Mt. Everest (8850 m) a climber and their equipment (80 kg) has how much
potential energy?
Does anyone knows this for physics?
Which of the following is the recommended amount of fats per meal for male clients
Answer:
44 grams- 55 grams through the whole day. Probably about 14.6 grams per meal.
Explanation:
Answer:
2 thumbs (ISSA Guide)
Explanation:
A merry-go-round on a playground consists of a horizontal solid disk with a weight of 805 N and a radius of 1.58 m. A child applies a force 49.5 N tangentially to the edge of the disk to start it from rest. What is the kinetic energy of the merry-go-round disk (in J) after 2.95 s?
Answer:
The value is [tex]KE = 259.6 \ J[/tex]
Explanation:
From the question we are told that
The weight of the horizontal solid disk is [tex]W = 805 \ N[/tex]
The radius of the horizontal solid disk is [tex]r = 1.58 \ m[/tex]
The force applied by the child is [tex]F = 49.5 \ N[/tex]
The time considered is [tex]t = 2.95 \ s[/tex]
Generally the mass of the horizontal solid disk is mathematically represented as
[tex]m_h = \frac{W}{ g}[/tex]
=> [tex]m_h = \frac{805}{ 9.8 }[/tex]
=> [tex]m_h = 82.14 \ N[/tex]
Generally the moment of inertia of the horizontal solid disk is mathematically represented as
[tex]I = \frac{1}{2} * m * r^ 2[/tex]
=> [tex]I = \frac{1}{2} * 82.14 * 1.58^ 2[/tex]
=> [tex]I = 102.5 \ kg \cdot m^2[/tex]
Generally the net torque experienced by the horizontal solid disk is mathematically represented as
[tex]T = I * \alpha = F * r[/tex]
=> [tex]\alpha = \frac{ F * r }{ I }[/tex]
=> [tex]\alpha = \frac{ 49.5 * 1.58 }{ 102.53 }[/tex]
=> [tex]\alpha = 0.7628[/tex]
Gnerally from kinematic equation we have that
[tex]w = w_o + \alpha t[/tex]
Here [tex]w_o[/tex] is the initial angular velocity velocity of the horizontal solid disk which is [tex]w_o = 0\ rad/s[/tex]
So
[tex]w = 0 + 0.7628 * 2.95[/tex]
=> [tex]w = 2.2503 \ rad/s[/tex]
Generally the kinetic energy is mathematically represented as
[tex]KE = \frac{1}{2} * I * w^2[/tex]
=> [tex]KE = \frac{1}{2} * 102.53 * 2.2503 ^2[/tex]
=> [tex]KE = 259.6 \ J[/tex]
name the force that help us to walk
Ask For
6.
A horizontal 100 N force is applied to a 50 kg classmate resting on a level tile
floor. The coefficient of kinetic friction is 0.15.
a. Draw a force diagram to represent this situation.
b. What is the acceleration of the classmate?
c. Suppose the classmate was resting on a carpet where the coefficient of static
friction is 0.25. Is the horizontal 100 N force sufficient to cause the classmate to
accelerate? Draw a force diagram, and then explain why or why not.
Two ice-skaters are skating in circles on a frozen pond. Maria is making large circles with a radius of 12 m and skating 4.5 m/s. Her friend, Samantha, is making smaller circles with a radius of 6 m but is not skating as quickly, going only 3.8 m/s. ii. How could each skater increase her centripetal acceleration without changing the size of her path? Explain your reasoning.
Answer:
v_maria > 4.5 m/s
v_samantha > 3.8 m/s
Explanation:
Formula for centripetal acceleration is;
a_c = v²/r
Where;
v is speed
r is radius
For Maria;
r = 12 m
v = 4.5 m/s
Thus;
a_c = 4.5²/12
a_c = 1.6875 m/s²
For Samantha;
r = 6
v = 3.8 m/s
Thus;
a_c = 3.8²/6
a_c = 2.41 m/s²
We want to find how could each skater increase her centripetal acceleration without changing the size of her path.
From the centripetal acceleration formula, since the size of path can't be changed it means the radius can't be changed and so the only thing that can now increase the centripetal acceleration is when the speed increases.
Thus;
For Maria, she has to move with a faster speed. Thus: v_maria > 4.5 m/s
For Samantha, she also has to move with a faster speed. Thus; v_samantha > 3.8 m/s
Question 25
The drawings below show four objects that are acted upon by different forces The direction of the forces are represented by the arrows, and the units of the forces are in newtons (N) Based on Newton's law of
inertia, which object is moving at a steady speed, assuming that all four objects are moving on a frictionless surface?
Answer:
The one with equal forces on both sides, D is your answer
Explanation:
A, 5 and 10 No
B 13 and 8 no
C 5 and 8 no
D 6 and 6 YES
Object 4 have steady speed. Object 4 have same force on both the sides of object. Same force is applied on both the sides.
What is steady speed?Uniform or constant velocity describes a body's motion when there is no acceleration. A body's velocity is its rate of movement in a specific direction.
When running steadily, your exhaustion should be caused by the length of the run, not by your speed. A steady-state run pace, according to the McMillan Running website, is halfway between your current 30k and half-marathon pace. This speed should allow you to run for 25 to 75 minutes.
Therefore, Object 4 have steady speed. Object 4 have same force on both the sides of object. Same force is applied on both the sides.
To learn more about velocity, refer to the link:
https://brainly.com/question/18084516
#SPJ2
A student mixes .075 kg of an unknown substance at 96.5°C with .075 kg of water at 25.0°C. If the final temperature of the system is 31.15°C, what is the specific heat capacity of the substance?
Answer:
The specific heat of the substance is 393.939 joules per kilogram-degree Celsius.
Explanation:
We notice that the student is mixing a substance with a high temperature and another substance with a low temperature, where the first release heat to the latter one until thermal equilibrium is reached. By the First Law of Thermodynamics and assuming that the entire system has no energy interactions with the surroundings, we get the following model:
[tex]\Delta U_{x}+\Delta U_{w} = 0[/tex] (1)
Where [tex]\Delta U_{x}[/tex] and [tex]\Delta U_{w}[/tex] are the changes in internal energy for the unknown substance and water, measured in joules.
By definition of internal energy, we expand the equation above now:
[tex]m_{x}\cdot c_{x}\cdot (T_{o,x}-T_{f,x})+m_{w}\cdot c_{w}\cdot (T_{o,w}-T_{f,w}) = 0[/tex] (2)
Where:
[tex]m_{x}[/tex], [tex]m_{w}[/tex] - Masses of the unknown substance and water, measured in kilograms.
[tex]c_{x}[/tex], [tex]c_{w}[/tex] - Specific heats of the unknown substance and water, measured in joules per kilogram-degree Celsius.
[tex]T_{o,x}[/tex], [tex]T_{f,x}[/tex] - Initial and final temperatures of the unknown substance, measured in degrees Celsius.
[tex]T_{o,w}[/tex], [tex]T_{f,w}[/tex] - Initial and final temperatures of water, measured in degrees Celsius.
Then, we clear the specific heat of the unknown substance:
[tex]c_{x} = \frac{m_{w}\cdot c_{w}\cdot (T_{f,w}-T_{o,w})}{m_{x}\cdot (T_{o,x}-T_{f,x})}[/tex]
If we know that [tex]m_{w} = m_{x} = 0.075\,kg[/tex], [tex]c_{w} = 4186\,\frac{J}{kg\cdot^{\circ}C}[/tex], [tex]T_{f,w} = T_{f,x} = 31.15\,^{\circ}C[/tex], [tex]T_{o,x} = 96.5\,^{\circ}C[/tex] and [tex]T_{o,w} = 25\,^{\circ}C[/tex], then the heat capacity of the unknown substance is:
[tex]c_{x} = \frac{(0.075\,kg)\cdot \left(4186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (31.15\,^{\circ}C-25\,^{\circ}C)}{(0.075\,kg)\cdot (96.5\,^{\circ}C-31.15^{\circ}C)}[/tex]
[tex]c_{x} = 393.939\,\frac{J}{kg\cdot ^{\circ}C}[/tex]
The specific heat of the substance is 393.939 joules per kilogram-degree Celsius.