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NISA [10]
3 years ago
14

An object travels 7.5 m/s toward the west . Under the influence of a constant net force of 5.2 kN, it comes to rest in 3.2 s. Wh

at is its mass?
Physics
1 answer:
Softa [21]3 years ago
5 0

Answer:

m = 2218.67 kg

Explanation:

It is given that,

Initial velocity, u = 7.5 m/s

Final speed of an object, v = 0 (at rest)

Force, F = 5.2 kN

Time, t = 3.2 s

We need to find the mass of the object. Force acting on an object is given by :

F = ma

m is mass, a is acceleration

F=\dfrac{m(v-u)}{t}\\\\m=\dfrac{Ft}{v-u}\\\\m=\dfrac{5.2\times 10^3\times 3.2}{0-7.5}\\\\m=2218.67\ kg

So, the mass of the object is 2218.67 kg

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Select all the correct answers.
amid [387]

Answer:

A

Explanation:

because in the moving object there's a certain energy applied

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The orientation of which of the following does not influence the phases of the moon? a. Earth c. Sun b. the moon d. Stars Please
beks73 [17]
I'm pretty sure it's D. The stars don't influence the moon's phases.
6 0
3 years ago
A 1,700 kg car is being used to give a 1,400 kg car a push start by exerting a force of
Gwar [14]

Answer:

- 670 kg.m/s

Explanation:

Newton's third law states that to every action, there is equal and opposite reaction force. Since the force will be same but different in direction and acted in the same time then the impulses ( force multiply by time) of the two car be same in magnitude but different in direction - 670 kg.m/s

5 0
3 years ago
The Hubble Space Telescope has an aperture of 2.4 m and focuses visible light (400-700 nm). The Arecibo radio telescope in Puert
stealth61 [152]

Answer:

y_{hubble} = 77\ \ m

y_{aceribo} = 1.1*10^6 \ \ m

Explanation:

what is the smallest crater that each of these telescopes could resolve on our moon?

For moon ;

s = 3.8 × 10 ⁸ m

y = 1.22 λs/D

where;

λ = 400 nm = 400× 10 ⁻⁹

D = 2.4 m

The smallest crater for the hubble space is calculated as follows:

y_{hubble} = 1.22*400*10^{-9}*3.8*10^8/2.4

y_{hubble} = 77\ \ m

For Aceribo ;

y = 1.22 λs/D

where :

λ = 75 cm = 0.75 m

D = 305 m

y_{acerbo} = 1.22*0.75 *3.8*10^8/305

y_{aceribo} = 1.1*10^6 \ \ m

5 0
3 years ago
Suppose a car of mass m is moving at a constant speed v of
SIZIF [17.4K]

Answer:

The angle of banked curve that makes the reliance on friction unnecessary is

\arcsin(v^2/(gR))

Explanation:

In order the car to stay on the curve without friction, the net force in the direction of radius should be equal or smaller than the centripetal force. Otherwise the car could slide off the curve.

The only force in the direction of radius is the sine component of the weight of the car

w_r = mg\sin(\theta)

The cosine component is equivalent to the normal force, which we will not be using since friction is unnecessary.

Newton’s Second Law states that

F_{net} = ma = mg\sin(\theta)\\\sin(\theta) = a/g

Also, the car is making a circular motion:

a = \frac{v^2}{R}

Combining the equations:

\sin(\theta) = \frac{a}{g} = \frac{v^2/R}{g} = \frac{v^2}{gR}

Finally the angle is

\arcsin(v^2/(gR))

4 0
3 years ago
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