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Anit [1.1K]
3 years ago
11

Which of the following is equal to an impulse of 15 units?

Physics
2 answers:
Anna11 [10]3 years ago
4 0

It totally depends on WHAT the units are.

IF the force is 7.5 Newtons AND the time is 2 seconds, THEN the impulse is 15 newton-seconds or 15 kg-m/sec. (choice-B)

If either of them isn't, then it isn't.

strojnjashka [21]3 years ago
3 0

Answer:

B) Force = 7.5, Time = 2 is equal to an impulse of 15 units

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What factor affects weight( what force causes weight to change)
Alex Ar [27]

Answer:

Gravitational force affects weight, weight changes with change in gravity

Hope it helped u,

mark as the brainliest

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3 0
3 years ago
1. What distance is required for a train to stop if its initial velocity is 23 m/s and its
Irina-Kira [14]

Answer:

x=?

dt=?

vi=23m/s

vf=0m/s (it stops)

d=0.25m/s^2

time =

vf=vi+d: 0=23m/s+(0.25m/s^2)t

t=92s

displacement=

vf^2=vi^2+2a(dx)

23^2=0^2+2(0.25m/s^2)x =-1058m

Explanation:

you can find time from vf = vi + a(Dt): 0 = 23 m/s + (0.25 m/s/s)t so t = 92 s and you can find the displacement from vf2 = vi2 + 2a(Dx) and find the answer in one step: 232 = 02 + 2(0.25 m/s/s)x so x = -1058 m

6 0
3 years ago
A baseball is released at rest from the top of the Washington Monument. It hits the ground after falling for 6 s. What was the h
alukav5142 [94]

Answer:

Total height (s) = 176.4 m

Explanation:

Given:

Initial velocity (u) = 0 m/s

Time taken (t) = 6 sec

Acceleration due to gravity = 9.8 m/s²

Find:

Total height (s)

Computation:

s = ut + [1/2]gt²

s = (0)(6) + [1/2][9.8][6²]

s = 176.4 m

Total height (s) = 176.4 m

6 0
3 years ago
A phonograph record accelerates from rest to 28.0 rpm in 5.73 s.
Arturiano [62]

Answer:

a) \alpha=0.5117\ rad.s^{-2}

b) \theta=8.4\ rad

Explanation:

Given:

  • initial rotational speed of phonograph, \omega_i=0\ rad.s^{-1}
  • final rotational speed of phonograph, N_f=28\ rpm \Rightarrow \omega_f=2\pi\times\frac{28}{60} =2.932\ rad.s^{-1}
  • time taken for the acceleration, t=5.73\ s

a)

Now angular acceleration:

\alpha=\frac{\omega_f-\omega_i}{t}

\alpha=\frac{2.932-0}{5.73}

\alpha=0.5117\ rad.s^{-2}

b)

Using eq. of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=0+\frac{1}{2}\times 0.5117\times 5.73^2

\theta=8.4\ rad

5 0
3 years ago
1-Find the gravitational force
butalik [34]

g = \frac{GMm}{r^2}

where G is the gravitational constant and is about 6.67 * 10^{-11}

g = \frac{6.67 * 10^{-11} * 25 * 0.55}{35^2}

g = \frac{6.67 * 10^{-11} * 25 * 0.55}{1225} = 0.000000000000749 N :D

5 0
3 years ago
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