All categories

2 years ago

When a block slides a certain distance down an incline, the work done by

1 answer:

7 0

That is -300 J, because the work = the force times the distance. When the distance is a positive value when it is going down, the distance is a negative value when it is going up. The only thing that change is the minus, the absolute value of the work remains the same

You might be interested in

4. The electric force between electric charges is much larger than the

Illusion [34]

The electric force is small in comparison because charges are VERY small but the

gravitational force has to do with the mass of the Earth and Moon.

4 0

3 years ago

Atoms that emit particles and energy from their nuclei are

saw5 [17]

Atoms that emit particles and energy from their nuclei are (Radioactive).

4 0

3 years ago

Imagine that I have a ping-pong ball and a bowling ball resting on the floor of our classroom. I go up to the bowling ball and g

MatroZZZ [7]

Answer:

the speed and aceleration of the ping pong ball is greater than that of the bowling ball.

Explanation:

We can analyze this exercise from several points of view, if we use Newton's second law

Bowling ball

F = M a₁

pingpongg ball

F = m a₂

as the forces the same

M a₁ = m a₂

a₂ = $\frac{M}{m}$ a₁

Since the mass of the bowling ball is much greater than the ping pong ball,

a₂ »a₁

so the acceleration of the ping pong ball is much greater than the acceleration of the bowling ball.

If we use the relationship of momentum and momentum, assuming that the time for the two cases is the same and that both start from rest

Bowling ball

I = F t = Δp

I = M (v₁ - v₀)

Ping pong ball

I = F t = Δp

I = m (v₂ -v₀)

the impulse itself

M v₁ = m v₂

v₂ = $\frac{M }{ m}$ v₁

so we conclude that the speed of the ping pong ball is much greater than the speed of the bowling ball.

In conclusion the speed and aceleration of the ping pong ball is greater than that of the bowling ball.

4 0

2 years ago

A 70.0-kg person throws a 0.0420-kg snowball forward with a ground speed of 35.0 m/s. A second person, with a mass of 57.0 kg, c

ollegr [7]

Answer:

so throwers new velocity = 2.18032m/s

so catchers new velocity = 0.02577m/s

Explanation:

Directly by conservation of momentum we can write

$m_1u_1+m_2u_2= m_1v_1+m_2v_2$

let x be the thrower's new velocity

(70+0.042)×2.2 + 57×0 = 70× x +0.042×35 +57×0

x = 2.18032m/s

so the velocity of 70 kg man = 2.18032m/s

so throwers new velocity = 2.18032m/s

now again by conservation of momentum

0.042×35 = (57+0.042) ×y

y = 0.02577m/s

so catchers new velocity = 0.02577m/s

3 0

3 years ago

A 8,000 kg amazon truck carrying a 3000 kg package is moving at 20 m/s along a level road as shown to the left. The driver appli

Reil [10]

Answer:

(a) -6.67 $m/s^2$

(b) 0.68

Explanation:

Given that the mass of the truck, $m_1=8000$ kg.

Mass of the package, $m_2=3000$ kg.

As the package does not slide, so the acceleration of both, truck as well as the package, is the same.

Let $a\;\; m/s^2$ is the acceleration of the combined mass, m.

$m=m_1+m_2= 8000+3000=11000$ kg.

The initial velocity of the combined mass, u= 20 m/s.

Time required to stop, t=3 seconds.

Final velocity, v=0.

Displacement traveled, s=50 m.

(a) As $a=\frac {v-u}{t}$

$\Rightarrow a=\frac{0-20}{3}=-\frac{20}{3} =6.67 m/s^2$

Hence the acceleration of the truck is $-6.67 m/s^2.$

(b) Now, $a=-6.67 m/s^2$ is the acceleration of the package, this acceleration is due the frictional force, f.

Due to inertia, on application of break, the package have tendency to slide to left (in the direction of velocity). But the package does not slides, this is only due to the frictional force, f, which acts in the right direction ( opposite to the direction of velocity).

So, the magnitude of frictional force required on $m_2$ to avoid slide is