During their physics field trip to the amusement park, Leslie and Maria took a ride on the Whirligig. The Whirligig ride consists of long swings which spin in a circle at relatively high speeds. As part of their lab, Leslie and Maria estimate that the riders travel through a circle with a radius of 6.5 m and make one circle every 5.8 seconds. Determine the frequency, the period, the speed and acceleration of the riders on the Whirligig
Answer:
a) frequency = 0.1724 Hz
b) Period = 5.8 sec
c) speed = 7.04 m/s
d) acceleration = 7.62 m/s²
Explanation:
Given that;
radius = 6.5m
time period = 5.8 sec every circle
a) the frequency
frequency is the number of rotation in unit time
frequency = 1 / time period = 1/5.8
frequency = 0.1724 Hz
b) the period
period is time taken in one rotation
period = total time / rotation = 5.8 / 1
Period = 5.8 sec
c) the speed
speed = distance/time = circumference/time period = 2πr / t = (2π×6.5) / 5.8
speed = 7.04 m/s
d) acceleration
To find the acceleration we take the linear velocity squared divided by the radius of the circle.
so
acceleration = v² / r = (7.04)² / 6.5 = 49.5616 / 6.5
acceleration = 7.62 m/s²
A ball player catches a ball 3.10 s after throwing it vertically upward. a) with what speed did he throw it? b) what height did it reach?
Answer:
speed=30.41, height=94.27
Explanation:
since the motion is vertical upward means the motion is against gravitational force and the initial velocity of the ball was zero.
9. What is the speed of a 25 kg object that has 2500 J of kinetic energy?
Answer:
We conclude that the speed of a 25 kg object that has 2500 J of kinetic energy is 14.14 m/s.
Explanation:
Given
Mass m = 25 kg Kinetic Energy (K.E) = 2500 JTo determine
The speed v = ?
We know that a body can possess energy due to its movement — Kinetic Energy.
The speed with which a body moves can be determined using the formula
[tex]K.E=\frac{1}{2}mv^2[/tex]
[tex]v=\sqrt{\frac{2\times \:K.E}{m}}\:\:\:[/tex]
substituting m = 25 and K.E = 2500 in the formula
[tex]v=\sqrt{\frac{2\times 2500}{25}}[/tex]
[tex]v=\sqrt{\frac{5000}{25}}[/tex]
[tex]v=\sqrt{200}[/tex]
[tex]v=14.14[/tex] m/s
Therefore, we conclude that the speed of a 25 kg object that has 2500 J of kinetic energy is 14.14 m/s.
In a game of ice hockey, you use a hockey stick to hit a puck of mass 0.16 kg that slides on essentially frictionless ice. During the hit you exert a constant horizontal force on the puck that gives it an acceleration of 75 m/s2 for a fraction of a second. During the hit, what is the magnitude of the horizontal force that you exert on the puck
Answer:
12N
Explanation:
given- mass, acelation
Fnet=ma= .16kg*75m/s2
Fnet=12 N only force no friction given.
A cart was allowed to roll down a ramp from rest. It reached the bottom of the ramp with a speed of 25 cm/s. A student could conclude that the cart
Answer:
A student could conclude that the cart gained momentum as it rolled down.
Explanation:
If the cart was allowed to roll down the ramp from rest, it means that the initial velocity of the cart is zero (0). Also, if the final velocity of the cart became 25 cm/s, it means that the velocity of the cart increased from 0 to 25 cm/s.
Therefore, A student could conclude that the cart gained momentum as it rolled down.
Cheetahs can accelerate to a speed of 20.0 m/s in 2.50 s and can continue to accelerate to reach a top speed of 30.0 m/s. Assume the acceleration is constant until the top speed is reached and is zero thereafter. Let the +x-direction point in the direction the cheetah runs. Express the cheetah’s top speed (vtop) in miles per hour (mi/h).
Starting from a crouched position, how much time (t-accel) does it take a cheetah to reach its top speed, and what distance (d) does it travel in that time?
Answer:
The cheetah’s initial velocity is 0 m/s. Let’s use the following equation to determine the acceleration.
Explanation:
Acceleration = (v/t) = (20/2.5) = 8m/sec^2.
1) Time to 29m/sec. = (v/a) = (29/8) = 3.625 secs. (3.6)
Distance covered to top speed = (v^2/2a) = 52.5625 metres (52.6).
2) Distance to run at top speed = (120 - 52.5625) = 67.4375 metres.
Time = (d/v) = (67.4375/29) = 2.325 secs.
Total time to warthog = (2.325 + 3.625) = 5.95 secs.
Cheetah's top speed will be "67.12 mi/h". A complete solution is provided below.
Given that:
Speed,
20.0 m/sTime,
2.50 sAcceleration to top speed,
30.0 m/sNow,
→ Acceleration (a) of cheetah will be:
= [tex]\frac{20}{2.5}[/tex]
= [tex]8 \ m/s^2[/tex]
→ Top speed ([tex]V_{top}[/tex]) of cheetah
= [tex]\frac{30}{1609}\times 3600[/tex]
= [tex]67.12 \ mi/h[/tex]
→ The time required (t) to reach the top speed will be:
= [tex]\frac{30}{8}[/tex]
= [tex]3.75 \ sec[/tex]
and,
→ The distance travelled (d) by cheetah will be:
= [tex]\frac{1}{2}\times 8\times (3.75)^2[/tex]
= [tex]\frac{1}{2}\times 8\times 14.0625[/tex]
= [tex]56.25[/tex]
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For each scenario below, choose the best graph.
(a) Maria bikes from home to work.
A line graph that plots either speed or distance against time.
Really, the type of graph depends on what quantity you want the graph to show.
A north magnetic pole is facing another north magnetic pole with a distance xx. If the distance between the poles becomes 12x12x, what happens to the magnitude of the field energy between them
Answer:
Explanation:
When two north magnetic poles are placed close to each other , they will repel each other . If distance between them is increased due to mutual repulsion from x to 12 x , work is done by the magnetic field . This results in decrease of magnetic field energy .
Hence , when the distance between the poles becomes 12x from x , the magnitude of the field energy between them decreases .
A car traveling 85 km/h is 250 m behind a truck
traveling 73 km/h.
Time needed = t = 20.83 s
Further explanationGiven
car speed = 85 km/h
truck speed = 73 km/h
Required
the time it takes for the car to reach the truck
Solution
When the car reaches the truck, the distance between them will be the same
x car - 250 m = x truck
General formula for distance (d) :
d = v.t
So the equation becomes :
85t-250 = 73t
12t=250
t = 20.83 s
Predictions about the future based on the position of planets is an example of
- physics
- biology
- earth science
- pseudoscience
Answer:
astronomi
Explanation:
sciemce ya ya
Using your knowledge on personal care products, how does sunscreen
lotion protect your skin from the damaging effect of ultraviolet rays?
Explain why of x-rays and gamma rays are commonly used in
radiotherapy.
Answer:
Ultraviolet rays from sun are very harmful from skin and can cause sunburn and skin diseases especially ultraviolet B rays. A sunscreen lotion act as a protection barrier on the skin that restrict the direct contact of UV rays with skin and filter the harmful rays to enter the skin.
Radiotherapy is a medical therapy use to treat cancer. Radiotherapy commonly uses x-rays and gamma rays because they are high-energy particles or waves that kills or destroys the cancer cells.
Sunscreen lotion is able to filter this damaging ultraviolet radiation and prevent it from damaging the skin.
The sun reaches us from outer space brings ultraviolet rays to us. Ultraviolet rays are known to have some damaging effects on the skin. One way to protect our skin from this damaging ultraviolet rays is to use sunscreen lotion which is able to filter this damaging ultraviolet radiation and prevent it from damaging the skin.
X-rays and gamma rays are used in radiotherapy because they are light energy rays which are able to penetrate and destroy malignant cells in the body.
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if |v|= 11, |w|= 23, |v-w|= 30, find |v+w|
Answer:
[tex]|v+w|=20[/tex]
Explanation:
We are given that
|v|=11
|w|=23
|v-w|=30
We have to find the value of |v+w|
|a-b|^2=(a+b)\cdot (a+b)=a^2+b^2-2|a||b|cos\theta
Using the formula
[tex](30)^2=(11)^2+(23)^2-2(11)(23)cos\theta[/tex]
[tex]900=121+529-506cos\theta[/tex]
[tex]900-121-529=-506cos\theta[/tex]
[tex]250=-506cos\theta[/tex]
[tex]cos\theta=-\frac{250}{506}[/tex]
[tex]|a+b|^2=|a|^2+|b|^2+2a\cdot bcos\theta[/tex]
Using the formula
[tex]|v+w|^2=(11)^2+(23)^2+2(11)(23)\times (-\frac{250}{506})[/tex]
[tex]|v+w|^2=400[/tex]
[tex]|v+w|^2=(20)^2[/tex]
[tex]|v+w|=20[/tex]
Imagine that I have a ping-pong ball and a bowling ball resting on the floor of our classroom. I go up to the bowling ball and give it a push so that it starts rolling. If I give a push of the exact same strength to the ping-pong ball, how will the resulting motion of the ping-pong ball be different
Answer:
the speed and aceleration of the ping pong ball is greater than that of the bowling ball.
Explanation:
We can analyze this exercise from several points of view, if we use Newton's second law
Bowling ball
F = M a₁
pingpongg ball
F = m a₂
as the forces the same
M a₁ = m a₂
a₂ = [tex]\frac{M}{m}[/tex] a₁
Since the mass of the bowling ball is much greater than the ping pong ball,
a₂ »a₁
so the acceleration of the ping pong ball is much greater than the acceleration of the bowling ball.
If we use the relationship of momentum and momentum, assuming that the time for the two cases is the same and that both start from rest
Bowling ball
I = F t = Δp
I = M (v₁ - v₀)
Ping pong ball
I = F t = Δp
I = m (v₂ -v₀)
the impulse itself
M v₁ = m v₂
v₂ = [tex]\frac{M }{ m}[/tex] v₁
so we conclude that the speed of the ping pong ball is much greater than the speed of the bowling ball.
In conclusion the speed and aceleration of the ping pong ball is greater than that of the bowling ball.
Calculate the change in momentum of a ball has 35 kg mass and it changed its velocity from 3m/s to 6 m/s
please hellllpppppp
Answer:
105 kg.m/s
Explanation:
Formula for change in momentum is given as;
ΔP = m(v - u)
We are given;
Mass; m = 35 kg
Initial velocity; u = 3 m/s
Final velocity; v = 6 m/s
Thus;
ΔP = 35(6 - 3)
ΔP = 35 × 3
ΔP = 105 kg.m/s
Glucose is the main source of energy.
True or false
Answer:
True :p
Explanation:
Carbohydrates are foods that get converted into glucose.
Which of the following statements is FALSE regarding the law of conservation of energy?
The total amount of energy remains constant.
Energy can neither be created nor be destroyed.
Energy can be transformed from one form to another.
The total amount of energy changes upon an energy transformation.
Answer:
D
Explanation:
because it doesn't agree with the law of definite proportion
and energy can niether be created nor destroyed but can be transformed from one form to another
The false statement will be, The total amount of energy changes upon an energy transformation.
What is conservation of energy?The conservation of energy according to the first law of thermodynamics is that energy neither be created nor be destroyed, but can be transformed from one form to another form.
Now we can take many examples like if we are running a bicycle our efforts are converted to the rotation of the wheel and the rotation energy is converted to the linear energy of our motion.
But the necessary point is that from the second law of thermodynamics it states that the energy can not be completely converted to the other form but some part of it is losses in the environment.
Thus the false statement will be, The total amount of energy changes upon an energy transformation.
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How does energy keep the world going? How is energy conserved?
Answer:
The most notable way that reducing energy helps the environment is by decreasing power plant emissions.
Answer:
Explanation:
The conservation of energy law states that energy cant be created or destroyed, only transferred. Since energy cant be created or destroyed, the amount of energy in the universe is completely constant.
Energy keeps the world "going" because we use it every single day in our life. Everything in the universe uses some sort of energy. We digest food for energy, we use electrical energy for electronics, we convert electrical energy into heat energy for heaters and such, we use solar panels to power our houses. There are countless examples of how energy keeps the world "going."
Look at the equation. What detail is missing? 3 m/s2= (33 m/s - X)/30 S
Answer:
The starting velocity.
Explanation:
We must understand that this equation comes from the following equation of kinematics.
[tex]v_{f}=v_{o}+a*t[/tex]
where:
Vf = final velocity = 33 [m/s]
Vo = starting velocity [m/s]
a = acceleration = 3 [m/s²]
t = time = 30 [s]
So, these values can be assembly in the following way:
[tex]v_{f}=v_{o}+a*t\\a*t=v_{f}-v_{o}\\3=\frac{33-v_{o}}{30}[/tex]
Marquette King, formerly of the Denver Broncos, is practicingkicking off using a kicking holder with the ball on the ground.For one of the kicks the ball reaches a height of 90.6 m andlands on the ground 53 yds (48.5 m) away. Find the magnitudeof the initial velocity given by his kick to the ball. Treat airresistance as negligible. Hint: Even though the horizontal andvertical motions are independent, there is a quantity that iscommon to both of them.
Answer:
Explanation:
Maximum height reached = 90.6 m . Range = 48.5 m
. Let u be the initial velocity at angle α .
Horizontal range is covered by horizontal component of u .
Vertical height is achieved by vertical component
v² = u² sin² α - 2gh , t is time taken to attain maximum height .
0 = u² sin² α - 2 x 90.6 x 9.8
u² sin² α = 2 x 887.88 -------( 1 )
Range R = u² sin2α / g
48.5 = 2 u² sinα . cos α / 9.8
u² sinα . cos α = 237.65 ----------------------------- ( 2 )
( 1 ) / ( 2 )
Tan α = 2 x 887.88 / 237.65 = 7.47
α = 82⁰
u² sin² α = 2 x 887.88
u² sin² 82 = 2 x 887.88
u² = 1812
u = 42.56 m /s
A particle with a charge of -4.3 μC and a mass of 4.4 x 10-6 kg is released from rest at point A and accelerates toward point B, arriving there with a speed of 80 m/s. The only force acting on the particle is the electric force. What is the potential difference VB - VA between A and B? If VB is greater than VA, then give the answer as a positive number. If VB is less than VA, then give the answer as a negative number.
Answer:
ΔV = - 3274 V
Explanation:
For this exercise we can use conservation of energy
starting point.
Em₀ = U = q ΔV
final point
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
q ΔV = ½ m v²
ΔV = [tex]\frac{m \ v^2 }{q}[/tex]
let's calculate
ΔV = [tex]\frac{4.4 \ 10^{-6} \ 80^2 }{ 2 \ 4.3 10^{-6} }[/tex]
ΔV = 3274.4 1 V
since the charge q is negative, the potential at point B must be less than the potential at point A, so the answers
ΔV = - 3274 V
Is it possible for the earth to be sucked into a black hole? If so, what would happen?
Answer:
si es posible
Explanation:
ya no existiria la tierra
if possible
the earth would no longer exist.
Answer:
Yes,
Explanation:
We would all die immediatly, its like being stretched and crushed
What is a watt a unit of?
light
time
distance
power
Answer:
power...............m....
Answer:
The dimension of power is energy divided by time. In the International System of Units (SI), the unit of power is the watt (W), which is equal to one joule per second.
SI unit: watt (W)
In SI base units: kg⋅m2⋅s−3
Explanation:
An object is traveling at a velocity of 100 rn/s to the left for about 5 seconds. How far has the object traveled? (What is the distance?)
Answer:
500 rn
Explanation:
The surface area of a postage
stamp is 0.00600 m^2, and the air
exerts 1.00 atm of pressure on it.
How much force does it exert on
the stamp?
(Hint: The standard unit for
pressure is Pa.)
(Unit = N)
Answer:
Force = 607.95 Newton
Explanation:
Given the following data;
Area = 0.00600 m^2
Pressure = 1 atm to Pascal = 101325 Pa
To find the force;
Pressure = Force/area
Force = pressure * area
Substituting into the equation, we have;
Force = 101325 * 0.00600
Force = 607.95 Newton.
Therefore, the amount of force exerted by the air on the stamp is 607.95 Newton.
A negatively-charged rod is brought close to (but does not touch) two neutral spheres that are in contact with each other but insulated from the ground. If the two spheres are then separated, what kind of charge will be on the spheres
Answer:
One sphere obtains a positive charge, while the other obtains a negative charge.
Explanation:
When the negatively-charged rod is brought close to the two neutral spheres in contact with each other, the electrons in both spheres are repelled away from the rod and thus those electrons move into the second sphere farther away from the rod.
The first sphere is now short of electrons.
When the spheres are separated, the first sphere which is short of electrons now carries a net positive charge, while the other sphere which has excess electrons now carries a net negative charge.
A stunt man jumps from a burning building and comes to
rest at ground level. Consider his impulse during landing if
he lands on a soft mat, on a concrete surface, or in a safety
net. Which of the following is true?
a. His impulse is greatest if he lands on a soft mat.
b. His impulse is greatest if he lands on a concrete
surface.
c. His impulse is greatest if he lands in a safety net.
d. The impulse is the same in all three cases.
e. There is zero impulse in all three cases because he
comes to rest.
Answer:
I think it's E it just sounds right to me tbh
Explanation:
A pendulum is placed on a distant planet. The length is one meter, and the measured period is 1.4 seconds, what is the acceleration of gravity on that planet?
Answer:
[tex]a=20.14\ m/s^2[/tex]
Explanation:
The time period of the simple pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
l is the length of the pendulum
g is the acceleration due to gravity
We have,
T = 1.4 s, l = 1 m
So,
[tex]T^2=\dfrac{4\pi^2 l}{g}\\\\g= \dfrac{4\pi^2 l}{T^2}\\\\g= \dfrac{4\pi^2 \times 1}{(1.4)^2}\\\\g=20.14\ m/s^2[/tex]
So, the acceleration due to gravity of that planet is [tex]20.14\ m/s^2[/tex].
Someone help me with this PE question?
Answer:
Your abdominals
Explanation:
According to Coulomb's Law, if the distance between two charged particles is doubled, the electric force will be _________. *
Answer: reduced by 1/4
Explanation:
The force will be reduced by 1/4. Try plugging in 2r, then squaring it. You will get 4r^2, which is essentially dividing the force by 4
Which statement about personal strength is TRUE?
1. It is inborn and cannot be developed with training or practice.
2. It is something you can acquire with hard work and lots of practice.
3. It is strictly a physical quality, like beauty, or physical ability, like the ability to swim fast.
4. It is such a natural part of a person that they may not even recognize it.