At the top of the building, the speed of the ball is zero.
The speed increases steadily as the ball falls.
When it hits the street, its speed is 39.6 m/s.
....................................
Answer:
yeah
Explanation:
Which of the following explains why NASA mandated that space food could NOT crumble or come apart?
Crumbs may cause electronics to malfunction.
There is no way to clean crumbs, so all food must be contained.
Floating crumbs may get into astronauts’ eyes.
Crumbs increase the risk of bacteria development.
Answer:
C.
Floating crumbs may get into astronauts’ eyes.
UV light has ______ wavelengths than visible light waves, and its peak energy is at a wavelength of ______.
Explanation:
Shorter wavelength since UV light has more energy than visible light
365 nm - What is the peak energy wavelength of UV light? It allows both infrared daylight and ultraviolet night-time communications by being transparent between 320 nm and 400 nm and also the longer infrared and just-barely-visible red wavelengths. Its maximum UV transmission is at 365 nm
If the mass of the Earth is doubled, how will the moon’s orbit be affected?
What Would Happen to the Orbit of the Moon if it Were Twice as Massive? ... The orbit would not change.
HOPE IT HELPS:)
PLS FOLLOW:)
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The energy transferred per unit electric charge in a circuit is .
A current
B charge
C power
D potential difference
Answer:
(D)
E = V Q energy of charge Q thus V = E / Q
A car accelerates from 20m/s to 30m/s in 10 sec. Find the cars acceleration using v=u+at
Explanation:
we derive the formula of acceleration using the formula so v - u /t so
30-20/10
so the acceleration is 1 m/s square
Please find attached photograph for your answer.
Hope it helps.
Do comment if you have any query.
A large solar farm has 21 700 solar panels and generates 5.0 MW of power.
1.0 MW = 1.0 × 106 W
Calculate the average power each panel produces.
Answer:
answer is
Explanation:
average power each panel produces = 21700/ 5×10^6
4340 ×10^-6
that's mean 4340 micro Watt
Which of the followng is Newton's Second Law?
a. An object in motion will stay in motion unless acted upon by an unbalanced outside force
b .Force = mass times acceleration
c.When one object puts a force on the another object, the second object puts a force back on the first, equal in magnitude and opposite in direction
Answer:
A
Explanation:
The table shows the diameters of the planets in our solar system. Assume that a basketball whose diameter is 25
centimeters represents the planet Jupiter. Find the scale (ratio) between Jupiter and the basketball. Then use this ratio to
find the scaled diameter of the other planets. Enter these numbers into the table. (Saturn has been done for you.) Finally,
choose a real-world spherical or nearly spherical object that matches the scaled diameter of each planet.
Saturn is shown as an example. To find what Saturn's size would be after adjusting it to scale, follow these steps. Let the
scaled diameter of Saturn be d centimeters. If Jupiter's diameter, 142,984 kilometers, is scaled to 25 centimeters, Saturn's
diameter of 120,536 kilometers will be scaled to about 21 centimeters:
120,536
d
120,536x25
142,984
d ~ 21 cm
Use this process to complete the table.
142,984
25
Answer:
Mercury:
.85
pea
Venus:
2.1
gumball
Earth:
2.2
gumball
Mars:
1.2
marble
Uranus:
9
grapefruit
Neptune:
8.6
softball
Explanation:
I have no clue if I'm right but hopefully, I am
[tex] \huge \mathcal{Question}[/tex]
Can someone Help me with this question ?
(I need proper Explanation if possible)
Thanks for Answering ~
Answer:
c. Force applied is 75 N
EDITED
tan 37 = a/g
a = 3g/4
a in equation
t× 3/5 = 2 × 3g/4
t = 5× 10/2
t = 25n
f = ma
f = ( mt + mb )a
F = 10 × 3× 10/4
F = 75 n
• Correct option is (C).
We know that the angle made by string is θ=
[tex]tan {}^{ - 1} ( \frac{a}{g} )[/tex]
where a is the acceleration of the frame in which the string is hanging in our case it is a trolley.
[tex]putting \: θ=37° \: or \: tan \: 37° = \frac{3}{4} [/tex]
[tex]we \: get \: a \: \frac{3g}{4} = \frac{30}{4} = 7.5 \: m/s^2[/tex]
[tex] \boxed{Force \: F=total \: mass \: of \: the \: system \: on \: which \: force \: is \: acting ×acceleration \: of \: the \: system=(8+2)×7.5=75 }[/tex]
[tex] \boxed{ so \: force \: is \: 75 \:N. }[/tex]
What is the weight, on Earth, of a book with a mass of 1. 5 kg? 1. 5 N 6. 5 N 11. 3 N 14. 7 N.
Answer:
The weight, on Earth, of a book with a mass of 1.5kg is 14.7N.
Explanation:
Hi there!
Weight of an object = mass (kg) × acceleration due to gravity (N/kg)
Plug in the given values using the g constant g = 9.8 N/kg:
W = 1.5 × 9.8 = 14.7 N
A ball is falling toward the ground. Which of the following statements are false? (a) The force that the ball exerts on Earth is equal in magnitude to the force that Earth exerts on the ball. (b) The ball undergoes the same acceleration as Earth. (c) The magnitude of the force the Earth exerts on the ball is greater than the mag- nitude of the force the ball exerts on the Earth.
Hi there!
(A) The force that the ball exerts on Earth is equal in magnitude to the force that Earth exerts on the ball.
Gravitational force is an example of an action-reaction pair:
[tex]F_{1,2} = -F_{2,1}[/tex]
Gravitational force is given by the equation:
[tex]F_g = \frac{Gm_1m_2}{r^2}[/tex]
Plugging in the masses of the Earth and ball, gravitational constant, and radius will result in the SAME GRAVITATIONAL FORCE exerted by both objects on each other.
please help me thank you
Answer:
Refer to the attachment
Use the graph of velocity versus time for an object to answer the question.
Which statement fairly compares segment 2 and segment 3?
(1 point)
O These represent equal periods of time, but the force during segment 2 is different than the force during segment 3.
These represent different periods of time, but the force acting on the object is the same during each period of time.
O These represent different periods of time, and the force during segment 2 is different than the force during segment 3.
O These represent equal periods of time, and the force acting on the object is the same during each period of time.
The statement which fairly compares segment 2 and segment 3 is These represent equal periods of time, but the force during segment 2 is different than the force during segment 3.
Since segment 2 starts at t = 60 s and ends at t = 150 s, the time interval is Δt = 150 - 60 = 90 s.
Also, segment 3 starts at t = 150 s and ends at t = 240 s, the time interval is Δt = 240 - 150 = 90 s.
So, their time periods are the equal.
We notice that segment 2 is less steep than segment 3 this implies that the acceleration in each segment is different, since the acceleration is the slope of the graph.
Since force is determined by acceleration, this implies that the force on segment 2 is different form the force acting in segment 3.
So, we have equal time periods but different forces.
So, the statement which fairly compares segment 2 and segment 3 is These represent equal periods of time, but the force during segment 2 is different than the force during segment 3.
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Answer:
Segments 2 and 3 have equal periods of time but the force during segment 2 is different than the force during segment 3 !! :)
Explanation:
a comet or asteroid impact on the earth is a frequently suggested cause for many of the mass extinctions. for which of the major mass extinctions does the most evidence exist that this was the main trigger for the event?
Answer:
Cretacious-Tertiary
Explanation:
I hope this helps!
2. Summarize how to use a spring scale to measure a pull and how to use it to apply a push with
a specific force.
If the spring constant is known, the extension or compression of the spring can be used to calculate the applied force (pull or push) on the spring.
According to Hook's law, the force applied to an elastic material is directly proportional to the extension of the material.
The force applied;
F = kx
where;
k is the springThe applied force can be in form of push or pull. A push force can result in compression of the spring while a pull force can result in extension of the spring.
If the spring constant is known, the extension or compression of the spring can be used to calculate the applied force (pull or push) on the spring.
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A child is stationary on a swing.(a)The child is given a push by his brother to start him swinging.His brother applies a steady force of 84 N over a distance of 0.25 m.(i) Calculate the work done by this force.(2)..............................................................................................................................................(ii) State how much energy is transferred by this force.(1)..............................................................................................................................................(iii) After several more pushes, the child has a kinetic energy of 71 J.The mass of the child is 27 kg.Show that the velocity of the child at this point is about 2.3 m/s.(2)(iv) Which one of these quantities changes in both size and direction while he is swinging?Put a cross ( ) in the box next to your answer.(1)Ahis gravitational potential energyBhis momentumCthe force of gravity acting on himDhis kinetic energy
Answer:
Work done = Fs
Explanation:
(i) W = Fxs
= 84N x0.25m
= 21 Joules
(ii) Ek = 1/2mv^2
71 = 1/2×27×v^2
71×2/27 = v^2
v = 23m/s
The work done by this force is 21 Joule.
Energy transferred by the force is 21 Joule.
The boy's momentum changes periodically, that is, in both size and direction.
What is force?An external force is an agent that has the ability to change the resting or moving condition of a body. It has a direction and a magnitude. So, it is a vector quantity. Newton is the SI unit of force (N).
(i) Given parameter:
Steady force applied by his brother, F = 84 N.
Displacement, d = 0.25 m.
The work done by this force, W = F.d = 84×0.25 = 21 Joule.
(ii) Amount of work done is transferred by this force as an energy. So, amount of energy transferred by the force is 21 Joule.
(iii) Given parameter:
Mass of the child, m= 27 kg.
Kinetic energy, E = 71 J.
Let, velocity of the child = v, then kinetic energy is,
E = 1/2 × mv²
⇒ v = √(2E/m)
= √(2×71/27)
= 2.3 m/s.
Hence, the velocity of the boy is 2.3 m/s. (proved).
(iv) During swing, his velocity changes periodically. That's why, his momentum changes periodically, that is, in both size and direction.
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What would the kinetic energy of a 20kg person running at a velocity of 2.5m/s?
Answer:
62.5 JExplanation:
The kinetic energy of an object can be found by using the formula
[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]
m is the mass in kg
v is the velocity in m/s
From the question
m = 20 kg
v = 2.5 m/s
From the question we have
[tex]k = \frac{1}{2} \times 20 \times {2.5}^{2} \\ = 10 \times 6.25 \\ = 62.5\: \: \: \: \: \: \: \: \: \: [/tex]
We have the final answer as
62.5 JHope this helps you
What are Newtons Laws famous for
Answer:
Why are Newton's laws of motion important? Newton's laws of motion are important because they are the foundation of classical mechanics, one of the main branches of physics. Mechanics is the study of how objects move or do not move when forces act upon them.
explique ¿por que un objeto que tiene energía es capaz de realizar un trabajo?
pls help me do these physic questions
Answer:
1 since it is meatl and sharp
2 since it is sharp and iron
a child has the kinetic energy of 71 j the mass of the child is 27kg show that the velocity of the child at this point is 2.3m/s
Explanation:
kinetic energy of child = 1/2 m v^2
71 j. = 1/2×27kg ×v^2
71×2/27= v^2
142/27 = v^2
5.3. =v^2
v= √5.3
v= 2.3 m/s
What is a lever?....
Answer:
A lever is a simple machine in the subject "S.T.E.M"
Explanation:
Levers use the following forces in order to be used: push or pull.
examples are crow bars,shovels, brooms and ect.
Which of these is a unique use of beta particles?
A. CT scanners
B. Tracing chemicals in the environment C. Killing bacteria
D. Smoke detectors
Answer:
The answer is A, CT scanners.
The kinetic energy of an object can sometimes be greater than a potential energy a originally possessed, true or false?
Answer:
true
Explanation:
Energy stored in the nuclei of atoms can be used to generate electricity. ... Most of the energy of food is converted to heat.
The mass of the hammer is 0.454 kg. Calculate the weight of the hammer.
Explanation:
weight =0.454 × 9.8=4.4492N
Answer:
4.45 n sorry if I am wrong
if you exert a force on an object, but the object remains stationary as you apply the force, then how much work have you done on the object?
Answer:
no work done
Explanation:
unless you moved the object
When we apply a force on an object, but the object remains at rest when the force is applied, then we have zero work done on the object because the object does not move.
What is Work done?Work done by a force is defined as the product of the displacement and the component of the force exerted by the object in the direction of displacement. Work done is expressed as:
Work done= Force* Displacement
When an object is at rest, there is no external force on it except gravity acting on it, in which case work will be zero because the force acting on it is perpendicular. Thus displacement is zero and if displacement is zero then work done will be zero.
Suppose that displacement is zero then,
W=f×0
Work done is zero
Thus, the work done is zero.
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251mL=25100L
True or false and why or why not?
Answer:
False
Explanation:
To find how many liters 251mL is you would need to divide 251 by 100 which is 0.251.
A box of mass 6.0 kg is accelerated from rest
across a floor at a rate of 6.0 m/s² for 9.0 s.
Find the network done on the box
[tex]\text{Given that },\\\text{Mass, m = 6 kg}\\\\\text{Acceleration, a = 6 } \text{m} \text{s}^{-2}\\\\\text{Time, t=9 sec}\\\\\text{Displacement, }s=ut + \dfrac 12 at^2 = \dfrac 12 at^2 = \dfrac 12 \times 6 \times 9^2 = 243 ~m}\\\\\ \text{Work,} ~W = Fs = mas = 6 \times 6 \times 243 = 8748 ~ J}[/tex]
Hi there!
We know that:
[tex]W = \Delta KE = \frac{1}{2}m(\Delta v)^2[/tex]
We can begin by solving for the final velocity using the equation:
[tex]v_f = v_i + at[/tex]
The box starts from rest, so:
[tex]v_f = at\\\\v_f = 6(9) = 54 m/s[/tex]
Now, we can find the kinetic energy:
[tex]KE = \frac{1}{2}mv^2\\\\KE = \frac{1}{2}(6)(54^2) = 8748 J[/tex]
Thus:
[tex]W = \Delta KE\\\boxed{W = 8748 J}[/tex]
If two stars are in a binary system with a combined mass of 5.5 solar masses and an orbital period of 12 years, what is the average distance between the two stars
The average distance between the two stars is 792 light years
Let the mass of the first star be [tex]m_1[/tex]
Let the mass of the second star be [tex]m_2[/tex]
The combined mass of the two stars, [tex]m_1+m_2=5.5[/tex] solar masses
The orbital period of the stars, P = 12 years
Average distance between the two stars, D = ?
The average distance between the two stars can be calculated using Kepler's equation
[tex]D=(m_1+m_2)P^2[/tex]
Substitute [tex]m_1+m_2=5.5[/tex] and P = 12 into the formula [tex]D=(m_1+m_2)P^2[/tex]
[tex]D=5.5(12^2)[/tex]
D = 5.5(144)
D = 792 light years
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