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Ksivusya [100]
2 years ago
9

Think about the process of pulling down the spring and letting it go. For this event, list each form of energy associated with t

he spring and the weight. (Hint: Use the Energy Graph.)
Physics
1 answer:
Afina-wow [57]2 years ago
6 0

Answer:

  1. gravitational 2.force of gravity

Explanation:

1.it ocours when an object is thrown into the sky. 2.iy ocurs when an object is falling or being pulled from the sky

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True or False: Warm air has greater air pressure than cool air 21 points
alukav5142 [94]

Answer:

False

Explanation:

Warmer air is less dense than cold, which is why warm air tends to rise and cold air sinks. Being acted on by gravity, colder, denser air weighs more and exerts greater pressure per unit area.

8 0
3 years ago
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A 65-cm segment of conducting wire carries a current of 0.35 A. The wire is placed in a uniform magnetic field that has a magnit
Artyom0805 [142]

Answer: The angle between the wire segment and the magnetic field 66.42°

Explanation:

Please see the attachment below

8 0
3 years ago
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An elevator ascends at a constant speed of 4 m/s, how much time is required for the elevator in order to travel 120 m upwards?
vlabodo [156]

We simply use the formula,

velocity =\frac{distance}{time}

Given, velocity = 4 m per s and distance = 120 m.

Substituting these values, we get

4\ m/s =\frac{120\ m}{time}\\\\\ time=\frac{120\ m}{4\ m/s} = 30\ s.

Thus, required time for the elevator in order to travel 120 m upwards is 30 s.

3 0
3 years ago
Along a horizontal snow-covered track, a sled, of mass m = 105 kg, slides by the action of a horizontal force of 230 N. The coef
Andrew [12]

Answer:

Explanation:

The only thing I can figure you need here is the accleration of the sled. The equation we need to find this is Newton's Second Law that says that sum of the forces acting on an object is equal to the object's mass times its acceleration. For us, that looks like this because of the friction working against the sled:

F - f = ma but of course it's much more involved than that simple equation! We have the F value as 230 N, and we have the mass as 105, but we do not have the frictional force, f, and we need it to solve for a in the above equation. We know that

f = μF_n where μ is the coefficient of friction, and F_n is the normal force, aka weight of the object. We will use the coefficient of friction and find the weight in order to fill in for f:

F_n=mg so

F_n=(105)(9.8) so the weight of the sled is

F_n= 1.0 × 10³ with the correct number of sig dig there. Now to find f:

f = (.025)(1.0 × 10³) so

f = 25 to the correct number of sig fig. Now on to our "real" equation:

F - f = ma and

230 - 25 = 105a. We have to do the subtraction first, round, and then divide since the rules for addition and subtraction are different from the rules for dividing and multiplying.

230 - 25 will round to the tens place giving us 210. Then

210 = 105a. 210 has 2 sig figs in it while 105 has 3, so we will divide and round to 2 sig fig:

a = 2.0 m/sec²

3 0
2 years ago
Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rock
pashok25 [27]

Answer:

a) s_a=98100\ m is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) v_a=1962\ m.s^{-1} is speed of the rocket going when it stops accelerating.

c) H=294300\ m

d) t_T=544.95\ s

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, a=2g=2\times 9.81=19.62\ m.s^{-2}

time for which the rocket accelerates, t_a=100\ s

<u>For the course of upward acceleration:</u>

using eq. of motion,

s_a=ut+\frac{1}{2}at_a^2

where:

u= initial velocity of the rocket at the launch =0

s_a= height the rocket travels just before its fuel finishes off

so,

s_a=0+\frac{1}{2}\times 19.62\times 100^2

a) s_a=98100\ m is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

<u>Now the velocity of the rocket just after the fuel is finished:</u>

v_a=u+at_a

v_a=0+19.62\times 100

b) v_a=1962\ m.s^{-1} is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

v^2=v_a^2-2gh

where:

g= acceleration due to gravity

v= final velocity of the rocket at the top height

0^2=1962^2-2\times 9.81\times h

h=196200\ m

c) So the total height at which the rocket gets:

H=h+s

H=196200+98100

H=294300\ m

d)

Time taken by the rocket to reach the top height after the fuel is over:

v=v_a+g.t

0=1962-9.81t

t=200\ s

Now the time taken to fall from the total height:

H=v.t'+\frac{1}{2}\times gt'^2

294300=0+0.5\times 9.81\times t'^2

t'=244.95\ s

Hence the total time taken by the rocket to strike back on the earth:

t_T=t_a+t+t'

t_T=100+200+244.95

t_T=544.95\ s

e)

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

8 0
2 years ago
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