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3 years ago

If an object doubles in mass, what happens to its inertia?

1 answer:

5 0

Inertia is proportional to mass. It is a measure of the resistance to changes in velocity. Inertia is a property of mass and cannot change. Momentum changes as an object changes its velocity.

HENCE The inertia will also be doubled

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3000 N is exerted for 4.0 seconds on a 9500 kg object.<br><br> What is the change in momentum?

WARRIOR [948]

Force is the change in momentum over a specific time. The change of momentum is therefore the force multiplied by the time that the force acts, so 3000x4.0=12000 N s=12000 kg m/s

8 0

2 years ago

If an object travels at a constant speed in a circular path, the acceleration of the object is:

Vilka [71]

Answer:

1- The acceleration of the object is larger in magnitude the smaller the radius of the circle.

Explanation:

The acceleration of an object in a circular path is

$a = \frac{v^2}{r}$

As can be seen from the equation, if the radius of the circle is decreases, the magnitude of the acceleration increases.

As for the direction of the acceleration, it is always towards the center, and it is always perpendicular to the direction of the velocity.

6 0

3 years ago

A driver in a 2290-kg car car traveling at 42.7 m/s slams on the brakes and skids to a stop. If the coefficient of friction betw

8_murik_8 [283]

Answer:

126.56 m

Explanation:

Applying,

-F = ma............. Equation 1

Where F = frictional force, m = mass of the car, a = acceleration.

Note: Frictional force is negative because it act in opposite direction to motion

But,

F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

-mgμ = ma

a = -gμ.............. Equation 3

From the question,

Given: μ = 0.735

Constant: 9.8 m/s²

Substitute these values in equation 3

a = -9.8×0.735

a = -7.203 m/s²

Finally,

Applying

v² = u²+2as.............. Equation 4

Where v = final velocity, u = initial velocity, s = distance

From the question,

Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²

Substitute these values into equation 4

0² = 42.7²+2(-7.203)s

-1823.29 = -14.406s

s = -1823.29/-14.406

s = 126.56 m

4 0

3 years ago

A 2500-lb vehicle has a drag coefficient of 0.35 and a frontal area of 20 ft2. What is the minimum tractive effort required for

emmasim [6.3K]

Answer:

i put this in the calculator and my answer is 600. hope this helps

Explanation:

4 0

3 years ago

Calculate the minimum amount of work needed to move a 500-kg rocket payload from Earth's surface to the ISS in orbit 400,000m ab

WINSTONCH [101]

Answer:

2.000.000.000

Explanation:

Apply the formula:

Work = Force . Displacement

W = 500.10 . 400.000 (the 10 come from gravity)

W = 5000 . 400.000

W = 2.000.000.000 Joules

I think it is that, can be wrong.

7 0

3 years ago