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dusya [7]
3 years ago
11

A 9800-N automobile is pushed along a level road by four students who apply a total forward force of 600 N. Neglecting friction,

the acceleration of the automobile is: a) 0.0 m /s^2 b) 0.54 m/s^2 c) 0.6 m/s^2 d) 9.8 m/s^2
Physics
1 answer:
Alexandra [31]3 years ago
4 0

Answer:

c).  a = 0.60 m/s^2

Explanation:

As we know that weight of the automobile is given here

so weight = mass times gravity

W = mg

9800 = m(9.8)

m = 1000 kg

now from Newton's law

F = ma

600 = 1000 a

a = \frac{600}{1000}

a = 0.60 m/s^2

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When you jump, you exert a pushing force against the ground. Gravity pulls you back down. Why can a person jump higher on the mo
Dennis_Churaev [7]

Answer:

This is because the force of gravity is much less on the moon than on the earth, therefore the person wont be pulled down much and will jump higher

7 0
2 years ago
An alert driver can apply the brakes fully in about 0.5 seconds. How far would the car travel if it
dybincka [34]

Answer:

The car would travel after applying brakes is, d = 14.53 m

Explanation:

Given that,

The time taken to apply brakes fully is, t = 0.5 s

The velocity of the car, v = 29.06 m/s

The distance traveled by the car in 0.5 s, d = ?

The relation between the velocity, displacement, and time is given by the formula                

                                d = v x t    m

Substituting the values in the above equation,

                                  d = 29.06 m/s x 0.5 s

                                     = 14.53 m

Therefore, the car would travel after applying brakes is, d = 14.53 m

8 0
3 years ago
Suppose an object in freefall is dropped from a building it’s starting velocity is 0m/s. Ignoring the facts of air resistance wh
Alja [10]

v = v₀ + at

v = final speed, v₀ = initial speed, a = acceleration, t = elapsed time

Given values:

v₀ = 0m/s (starts from rest), a = 9.81m/s², t = 3s

Plug in and solve for v:

v = 0 + 9.81(3)

v = 29.4m/s

3 0
3 years ago
Rank these significant figures numbers from the least to the most
Mumz [18]

Answer:

0.006<357<700.003<6010<9256.0<9520.00

8 0
3 years ago
A circular cross section, d = 25 mm, experiences a torque load, T = 25 N·m, and a shear force, V = 85 kN. Calculate the shear st
Maru [420]

Answer:

The correct answer is 231 Mpa i.e option a.

Explanation:

using the equation of torsion we Have

\frac{T}{I_{p}}=\frac{\tau }{r}\\\\\therefore \tau =\frac{T}{I_{p}}\times r

where,

\tau= shear stress at a distance 'r' from the center

T = is the applied torque

I_{p} = polar moment of inertia of the section

r = radial distance from the center

Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.

We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals

\tau _{max}=\frac{4}{3}\times \frac{V}{A}

Applying values we get

\tau _{max}=\frac{4}{3}\times \frac{85\times 10^{3}}{0.25\times \pi \times (25\times 10^{-3})^{2}}\\\\\therefore \tau _{max}=230.88Mpa\approx 231Mpa

3 0
2 years ago
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