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2 years ago

8

The gravitational force between two objects is most affected by their

1 answer:

Oxana [17]2 years ago

5 0

Answer: mass and distance

Explanation:

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a 2kg marble is moving at 3 m/s when it strikes another 4kg marble moving in the opposite direction at -3 m/s. What will be the

Elan Coil [88]

Momentum is conserved after the collision

Momentum of 2 Kg before collision = 2 * 3 = 6
Momentum of 4 kg before collision = 4 * -3 = -12

so 6 + -12 = 2 * -4 + 4 *x where x is velocity of 4kg marble.

4x - 8 = -6
4x = 2
x = 0.5

Velocity of 4 kg marble is 0.5 m/s after collision

The 2 kg marble will move in the opposite direction to which it was moving before the collision.

4 0

3 years ago

A boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s .

zimovet [89]

Answer:

Force exerted, F = 1.5 N

Explanation:

It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.

i.e. u = 0

v = 30 m/s

Time taken, t = 0.06 s

Mass of the paper, m = 0.003 kg

We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :

$F=m\times a$

$F=m\times (\dfrac{v-u}{t})$

$F=0.003\times (\dfrac{30}{0.06})$

F = 1.5 N

So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.

6 0

2 years ago

A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. After the wheel has undergone 14 rotations, what is the radi

Mamont248 [21]

Answer:

$a=368.97\ m/s^2$

Explanation:

Given that,

Initial angular velocity, $\omega=0$

Acceleration of the wheel, $\alpha =7\ rad/s^2$

Rotation, $\theta=14\ rotation=14\times 2\pi =87.96\ rad$

Let t is the time. Using second equation of kinematics can be calculated using time.

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s$

Let $\omega_f$ is the final angular velocity and a is the radial component of acceleration.

$\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s$

Radial component of acceleration,

$a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2$

So, the required acceleration on the edge of the wheel is $368.97\ m/s^2$ .

6 0

2 years ago

A pair of toy freight cars, one twice the mass of the other, fly apart when a compressed spring that joins them is released. Acc

saul85 [17]

Answer:

greater acceleration is experienced by the car with lower mass

Explanation:

Since both the toys are connected by same spring so the force due to spring on both the toys will be same and it is given as

$F = kx$

now we know by Newton's II law

$F = ma$

so here we have

$a = \frac{F}{m}$

here we have same force on both the blocks

so acceleration will be more if mass is less

so greater acceleration is experienced by the car with lower mass

8 0

2 years ago

Goldberg's sleigh currently runs at 203mph, but he needs it to reach 400mph with all the packages he has to deliver.

svp [43]

Answer:

c

Explanation:

6 0

2 years ago

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