Answer:100m west
Explanation:
List Five examples from daily life in which you see periodic motion caused by a pendulum
(Marking Brainliest)
Answer:
by a rocking chair, a bouncing ball, a vibrating tuning fork, a swing in motion, the Earth in its orbit around the Sun, and a water wave.
Explanation:
Is a parked car potential or kinetic ?
Answer:
Potential energy is the energy that is stored in an object. ... When you park your car at the top of a hill, your car has potential energy because the gravity is pulling your car to move downward; if your car's parking brake fails, your vehicle may roll down the hill because of the force of gravity.
A 1kg cannon ball.is fired horizontally with an initial velocity of 5m/s. If the cannon was atop a wall 20m above the ground, what is the total
change in KE?
Answer:
Ek = 196.2 [J]
Explanation:
The question concerns the KE kinetic energy.
That is, we must find the kinetic energy at the moment the cannon is fired and the kinetic energy of when the ball hits the ground after having fallen 20 meters.
At the moment when the ball is fired it is 20 meters above ground level. If the ground level is taken as the reference level of potential energy, where it is equal to zero, in this way when the ball is at the highest (20 meters) you have the maximum potential energy.
In this way, the energy in the initial state is equal to the sum of the kinetic energy plus the potential energy. As the energy is conserved this same energy will be present when the ball hits the ground, where the potential energy is zero and will have only kinetic energy.
[tex]E_{1}=E_{2}\\E_{k1}+E_{p1}=E_{k2}\\\frac{1}{2} *m*v^{2} +m*g*h=E_{k2}\\E_{k2}=0.5*1*(5)^{2} +1*9.81*20\\E_{k2}=208.7[J][/tex]
The kinetic energy in the initial state can be easily calculated by means of the following equation.
[tex]E_{k1}=\frac{1}{2} *m*v^{2}\\E_{k1}=0.5*1*(5)^{2}\\E_{k1}=12.5 [J][/tex]
Therefore the change in KE
[tex]E_{k} = 208.7 - 12.5\\E_{k} = 196.2 [J][/tex]
Ball 1 (1.5 kg) moves to the right at 2 m/s and ball 2
(2.5 kg) moves to the left at 1.5 m/s. The balls stick together after collision. What is the speed and direction of ball 2 after the collision?
Answer:
0.1875 m/s leftward
Explanation:
Taking rightwards as positive
We are given:
Ball 1:
Mass (m1) = 1.5 kg
velocity (u1) = 2 m/s
Ball 2:
Mass (m2) = 2.5 kg
velocity (u2) = -1.5 m/s [negative because it is in the opposite direction]
Speed and Direction of Ball 2:
We are told that the balls stick together after the collision
We can say that the balls have the same velocity since they are sticking together
So, Final velocity of Ball 1 (v1) = Final velocity of Ball 2 (v2) = V m/s
According to the law of conservation of momentum
m1u1 + m2u2 = m1v1 + m2v2
replacing the variables
1.5(2) + (2.5)(-1.5) = V (1.5 + 2.5) [v1 = v2 = V]
3 + (-3.75) = 4V
-0.75 = 4V
V = -0.75/4 [dividing both sides by 4]
V = -0.1875 m/s
Hence, the balls will move at a velocity of 0.1875 m/s in the Leftward direction
Suppose a popular FM radio station broadcasts radio waves with a frequency of 96. MHz. Calculate the wavelength of these radio waves. Be sure your answer has the correct number of significant digits.
Answer:
3.125 meters.
Explanation:
(3.0*10^8)/(96*10^6)
= 3.125 meters.
Hope this helped!
An ideal gas expands quasi-statically and isothermally from a state with pressurepand volumeVto a state with volume 4V. How much heat is added to the expanding gas?
Answer:
Q = PV(In 4)
Explanation:
We are told that the volume expands from V to a state with volume 4V.
Thus, initial volume is V and Final volume is 4V.
We want to find How much heat is added to the expanding gas.
For an isothermal process, the work done is calculated from;
W = nRT(In(V_f/V_i))
Where;
V_f is final volume
V_i is initial volume
Thus;
W = nRT(In(4V/V))
W = nRT(In 4)
Now, from ideal gas equation, we know that;
PV = nRT
Thus;
W = PV(In 4)
Now from first law of thermodynamics, we know that internal energy is zero and thus; Q = W
Where Q is quantity of heat
Thus;
Q = PV(In 4)
what causes the coriolis effect
A. Earths orbit around the sun.
B. Wind currents.
C. Earths rotation around its axis
D. Uneven solar heating of earth
A projector lens projects an image from a 6.35 cm wide LCD screen onto a
screen 3.25 m wide. If the focal length of the projector lens is 13.8 cm, the screen
must be how far from the projector
Answer:
For any given projector, the width of the image (W) relative to the throw distance (D) is know as the throw ratio D/W or distance over width. So for example, the most common projector throw ratio is 2.0. This means that for each foot of image width, the projector needs to be 2 feet away or D/W = 2/1 = 2.0.
An explanation of the relationships among particular phenomena.
Answer:
Theory
Explanation:
Theory is a term that is used often in academic work or scientific research to explain certain things or conditions established on universal principles or laws.
It is used to describe the "why and how" or the reason behind the occurrence of a situation.
Hence, it is correct to conclude that THEORY is "an explanation of the relationships among particular phenomena."
Answer:
E) Theory
Explanation:
Edge 2020
Brainliest?
An airtight box, having a lid of area 80.0 cm^2, is partially evacuated. Atmospheric pressure is 1.01 Times 10^5 Pa. A force of 108 lb is required to pull the lid off the box. The pressure in the box was:_________.
Answer:
5×10^4Pa
Explanation:
Given force of 108 lb is required to pull the lid off the box,
To convert "Ib"to Newton ,we use conversation rate below
1 pounds = 4.4482216282509 newtons
Then 108 lb=x Newton
Cross multiply we have
X= 480.41Newton
The force that is needed to open the lid is F and pressure P.
We know that Pressure= Force/Area
Area is given as 80.0 cm^2, we can convert to m^2 for unit consistency since 1cm^2= 0.001m^2 then
80.0 cm^2 = 80×10^-4m^2
Substitute to the equation of the pressure we have
P= 480.41Newton/(80×10^4m^2)
P=6×10^4 Pa
The pressure in the box will be difference between the initial pressure and final pressure
=( 1.01 ×10^5 Pa)-(6×10^4 Pa)
= 50100Pa
= 5×10^4Pa
Therefore, The pressure in the box was
5×10^4Pa
a current of 200 mA through a conductor converts 40 joules of electrical energy into heat in 30 second
s determine the p
otential drop across the conductor
Answer:
V = 6.65 [volt]
Explanation:
First, we must calculate the power by means of the following equation, where the voltage is related to the energy produced or consumed in a given time.
[tex]P=E/t\\P = 40/30\\P = 1.33[s][/tex]
Using the power we can calculate the voltage, by means of the following equation that relates the voltage to the current.
[tex]P=V*I[/tex]
where:
V = voltage [Volts]
I = current = 200 [mA] = 0.2 [A]
[tex]V = 1.33/0.2\\V = 6.65 [volt][/tex]
A good baseball pitcher can throw a baseball toward home plate at 97 mi/h with a spin of 1540 rev/min. How many revolutions does the baseball make on its way to home plate
Answer:
10778292789403987593790
Explanation:
I am a Cow'
SI unit differ from one country to another . true or false
Answer:
false ..........false
Answer:
FALSE
Explanation:
Aluminum wire with a diameter of 0.8650 mm is wound onto a spool. The wire is insulated, but you have access to both ends. The resistivity of aluminum at 20.0 °C is 2.65 x 10^-8 Ω-m. You measure the resistance of the wire at that temperature, and it is 2.48 Ω. What is the length of the wire?
a. 8.10 x 10^4 m
b. 22.0 m
c. 5.68 m
d. 0.111 m
e. 55.0 m
Answer:
e. 55.0 m
Explanation:
Given;
diameter of the aluminum wire, d = 0.865 mm
radius of the wire, r = d/2 = 0.4325 mm = 0.4325 x 10⁻³ m
resistivity of the wire, ρ = 2.65 x 10⁻⁸ Ω-m
resistance of the wire, R = 2.48 Ω
The resistance of a wire is given by;
[tex]R = \frac{\rho \ L}{A} \\\\[/tex]
where;
L is length of the wire
A is area of the wire = πr² = π(0.4325 x 10⁻³ )² = 5.877 x 10⁻⁷ m²
Substitute the givens and solve for L,
[tex]L = \frac{RA}{\rho} \\\\L = \frac{(2.48)(5.877*10^{-7})}{2.65*10^{-8}}\\\\L = 55.0 \ m[/tex]
Therefore, the length of the wire is 55.0 m
A sealed cubical container 28.0 cm on a side contains three times Avogadro's number of molecules at a temperature of 24.0°C. Find the force exerted by the gas on one of the walls of the container.
Answer:
3.32 atm
__________________________________________________________
We are given:
side of the cubical container = 28 cm
number of molecules in the container = 3 * Nₐ
[where Nₐ is the Avogadro's number]
Temperature = 24°C OR 297 K
We need to find the pressure exerted by the gas on the walls of the container
__________________________________________________________
Some Calculations:
Volume of the container
we are given the side of the cubical container = 28 cm
Volume of the cubical container = side³
Volume = 28³
Volume = 21952 cm³
We know that 1 cm³ = 1 mL
So,
Volume = 21952 mL
We also know that 1 L = 1000 mL
Volume = 21.952 L
Number of moles of Gas
We know that:
number of moles = number of molecules / Avogadro's number
number of moles = 3 * Nₐ / Nₐ [number of molecules = 3 * Nₐ]
number of moles = 3 moles
__________________________________________________________
Pressure Exerted by the Gas:
Using the ideal gas equation:
PV = nRT
Since the volume is in L, and Temperature is in K. R is equal to
0.082 L atm /mol K and the pressure will be in atm
P(21.952) = 3*(0.082)*(297)
P = 3.32 atm
Hence, the gas will exert a pressure of 3.32 atm on the walls of the container
The earth's radius is 6.37×106m; it rotates once every 24 hours.What is the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator? (Hint: what is the radius of the circle in which the point moves?) Express your answer to two significant figures and include the appropriate units.
Answer:
v = 120 m/s
Explanation:
We are given;
earth's radius; r = 6.37 × 10^(6) m
Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s
Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.
The angle will be;
θ = ¾ × 90
θ = 67.5
¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.
The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:
v = r(cos θ) × ω
v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)
v = 117.22 m/s
Approximation to 2 sig. figures gives;
v = 120 m/s
The x and y coordinates of a particle at any time t are x = 5t - 3t2 and y = 5t respectively, where x and y are in meter and t in second. The speed of the particle at t = 1 second is
Answer:
[tex]v=\sqrt{26}~m/s[/tex]
Explanation:
Parametric Equation of the Velocity
Given the position of the particle at any time t as
[tex]r(t) = (x(t),y(t))[/tex]
The instantaneous velocity is the first derivative of the position:
[tex]v(t)=(v_x(t),v_y(t))=(x'(t),y'(t))[/tex]
The speed can be calculated as the magnitude of the velocity:
[tex]v=\sqrt{v_x^2+v_y^2}[/tex]
We are given the coordinates of the position of a particle as:
[tex]x=5t-3t^2[/tex]
[tex]y=5t[/tex]
The coordinates of the velocity are:
[tex]v_x(t)=(5t-3t^2)'=5-6t[/tex]
[tex]v_y(t)=(5t)'=5[/tex]
Evaluating at t=1 s:
[tex]v_x(1)=5-6(1)=-1[/tex]
[tex]v_y(1)=5[/tex]
The velocity is:
[tex]v=\sqrt{(-1)^2+5^2}[/tex]
[tex]v=\sqrt{1+25}[/tex]
[tex]\mathbf{v=\sqrt{26}~m/s}[/tex]
A freshly caught catfish is placed on a spring scale, and it oscillates up and down with a period of 0.19 s. If the spring constant of the scale is 2330 N/m, what is the mass of the catfish?
Answer:
The mass of the catfish is 2.13 kg
Explanation:
Period of oscillation, T = 0.19 s
spring constant, k = 2330 N/m
The period of oscillation of the spring is given by;
[tex]T = 2\pi \sqrt{\frac{m}{k} }\\\\\frac{T}{2\pi} = \sqrt{\frac{m}{k} }\\\\\frac{T^2}{4\pi^2} = \frac{m}{k}\\\\m = \frac{kT^2}{4\pi^2}[/tex]
where;
m is mass of the catfish
substitute the given values and solve for m;
[tex]m = \frac{kT^2}{4\pi^2} \\\\m = \frac{(2330)(0.19)^2}{4\pi^2} \\\\m = 2.13 \ kg[/tex]
Therefore, the mass of the catfish is 2.13 kg
A car is accelerated at a constant rate from 15 m/s to 25 m/s. It takes the car 6 s to reach its final speed. What is the car’s acceleration?
Answer:
1.67 m/s²Explanation:
The car’s acceleration can be found by using the formula
[tex]a = \frac{v - u}{t} \\ [/tex]
where
v is the final velocity
u is the initial velocity
t is the time taken
a is the acceleration
From the question we have
[tex]a = \frac{25 - 15}{6} = \frac{10}{6} = \frac{5}{3} \\ = 1.666666...[/tex]
We have the final answer as
1.67 m/s²Hope this helps you
Two blocks with different masses are dropped, hitting the ground with the same velocity. Which of the following is true?
They have same change in velocity but different changes in kinetic energy
The lighter object started at a smaller height.
The heavier object started at a smaller height
They started at the same height
They have same change in kinetic energy but different changes in velocity
Answer: • They have same change in velocity but different changes in kinetic energy
•They started at the same height.
Explanation:
First and foremost, we need to note that both balls have thesame acceleration due to gravity and due to this, even though they've different masses, they'll fall at same speed.
Also, since kinetic energy that's, the energy relating to motion of a mass, us dependent on mass and speed, their kinetic energy will be different.
Therefore, based in the explanation, the correct options are:
• They have same change in velocity but different changes in kinetic energy
•They started at the same height.
If 10 calories of energy are added to 2 grams of ice at -30° C, calculate the final temperature of the ice. (Notice that the specific heat of ice is different from that of water.) Assume the specific heat of ice is 0.5
-30° C
40° C
-20° C
30° C
Answer:
-20°C
Explanation:
The specific heat capacity of ice using the cgs system is 0.5cal/g°C
The enthalpy change is calculated as follows
ΔH=MC∅ where M represents mass C represents specific heat and ∅ represents the temperature change.
10cal = 2g×0.5cal/g°C×∅
∅=10cal/(2g×0.5cal/g°C)
∅=10°C
Final temperature= -30°C+ 10°C= -20°C
Answer:
-20 degrees Celsius
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A ball is thrown vertically upward with an initial velocity of 23 m/s. What are its position and velocity after 2 s?
Answer:
The position of the ball after 2 s is 26.4 mThe velocity of the ball after 2 s is 3.4 m/sExplanation:
Given;
initial velocity of the ball, u = 23 m/s
time of motion, t = 2 s
The position of the ball after 2 s is given by;
h = ut - ¹/₂gt²
h = (23 x 2) - ¹/₂ x 9.8 x 2²
h = 46 - 19.6
h = 26.4 m
The velocity of the ball after 2 s is given by;
v² = u² + 2(-g)h
v² = u² - 2gh
v² = 23² - (2 x 9.8 x 26.4)
v² = 529 - 517.44
v² = 11.56
v = √11.56
v = 3.4 m/s
Which phrase desenbes an irregular galaxy ?
has a round shape
contains many young stars
has arms that extend from the center
Is larger than other types of galaxies
Answer:
contains many young stars
Explanation:
Irregular galaxies have no definite shape, which means that the first option is incorrect. They are definitely not round.
However, they contain many young stars because the degree of star formation is fast. They also contain old stars. Thus, the second choice is correct.
The "spiral galaxy" is the type of galaxy that has arms that extend from the center. These arms look "spiral," which influenced its name. This makes the last choice incorrect.
They are actually smaller than the other types of galaxies. This makes them prone to collisions. This makes the last choice incorrect.
Answer:
Contains many young stars
Explanation:
At high noon, the sun is almost directly above (about 2.0 degrees from the vertical) and a tall redwood tree casts a shadow that is 10m long. How tall is the redwood tree?
290m
The light always travels in a straight line.
At high noon, the ray from the sun is 2° from the vertical axis.
tan θ = (Opposite side)/(Adjacent side)
On applying above trigonometric formula, we get,
tan 2 = 10/h
0.035 = 10/h
∴ h = 10/0.035 = 290 m
An object in FREE-FALL on the MOON would experience which of the following
FORCES?
O a. Weight
O b. Normal
O c. Air Resistance
d. a and c
O e. None of these
Answer:
e. none of these
Explanation:
An object in FREE-FALL on the MOON would experience only acceleration
Given F1: a force of magnitude 6 N at an angle of 30°
F2: a force of magnitude 8 N at an angle of 50°C
a. Find F1+ F2 analytically (using equations instead of graphing) and write it in the form Fr1i + Fr2 j
b. Find the magnitude FR and θ_resultant
Answer:
13.8 N
[tex]41.44^{\circ}[/tex]
Explanation:
[tex]F_1=6\ \text{N}[/tex]
[tex]F_2=8\ \text{N}[/tex]
[tex]F_1\cos\theta_1\hat{i}+F_1\sin\theta_1\hat{j}\\ =6\cos30^{\circ}+6\sin30^{\circ}\hat{j}\\ =5.2\hat{i}+3\hat{j}[/tex]
[tex]F_2\cos\theta_2\hat{i}+F_2\sin\theta_2\hat{j}\\ =8\cos50^{\circ}+8\sin50^{\circ}\hat{j}\\ =5.14\hat{i}+6.13\hat{j}[/tex]
[tex]F_R=F_1+F_2=10.34\hat{i}+9.13\hat{j}[/tex]
[tex]|F_R|=\sqrt{10.34^2+9.13^2}=13.8\ \text{N}[/tex]
The magnitude of the resultant is 13.8 N
Direction is given by
[tex]\tan^{-1}=\dfrac{y}{x}=\tan^{-1}\dfrac{9.13}{10.34}=41.44^{\circ}[/tex]
The angle of the resultant is [tex]41.44^{\circ}[/tex]
The CEO, ellen misk, left her martian office but accidentally left a cylindricall can of coke (3.1 inches in diameter, 5.42 inches in height) on her desk. If the can exerts a pressure of 510 Pascals, what is the specific gravity of the can?
Answer:
Specific Gravity = 0.378
Explanation:
First, we will find the force exerted by the can on the table. This force will be equal to the weight of the can:
Pressure = Force/Area = Weight/Area
Weight = Pressure*Area
where,
Area = πdiameter²/4 = π[(3.1 in)(0.0254 m/1 in)]²/4 = 4.8 x 10⁻³ m²
Weight = (510 N/m²)(4.8 x 10⁻³ m²)
Weight = 2.48 N
Now, the weight is given as:
Weight = mg
2.48 N = m(9.8 m/s²)
m = (2.48 N)/(9.8 m/s²)
m = 0.25 kg
Now, we calculate volume of can:
Volume = (Area)(Height) = (4.8 x 10⁻³ m²)(5.42 in)(0.0254 m/1 in)
Volume = 6.6 x 10⁻⁴ m³
Hence, the density of can will be:
Density of Can = m/Volume = 0.25 kg/6.6 x 10⁻⁴ m³
Density of Can = 378.32 kg/m³
So, the specific gravity of Can will be:
Specific Gravity = Density of Can/Density of Water
Specific Gravity = (378.32 kg/m³)/(1000 kg/m³)
Specific Gravity = 0.378
The mass of a paper-clip is 0.50 g and the density of its material is 8.0g/cm'. The total volume of
a number of clips is 20 cm.
How many paper-clips are there?
Answer:
320 paper clips
Explanation:
mass = volume × density = 20cm³ × 8g/cm³ = 160g
mass of 1 paper clip = 0.50g
mass of x paper clips = 160g
x = 160/0.50 = 320
If you start at a speed of 4m/s and slow down to 2m/s in 4s what is your
acceleration?
Answer:
penis
Explanation:
A cheetah can maintain a maximum constant velocity of 34.2 m/s for 8.70 s. What is
the displacement the cheetah covered at that velocity?
Answer:
297.54mExplanation:
step one:
given data
velocity v=34.2m/s
time t= 8.7s
Step two
Required is the distance the cheetah has covered on the condition
we know that speed= distance/time
make distance subject of formula we have
distance= velocity *time
distance= 34.2*8.7
distance = 297.54m
Therefore the displacement the cheetah covered at that velocity
is 297.54m