All categories

3 years ago

If you are in a car that is being pulled down a 56 m path with a force of 12.5

Physics

1 answer:

Elis [28]3 years ago

7 0

Answer:

answer is 700 joules

Explanation:

work is equal to force into displacement

W=12.5*56

700 joules

You might be interested in

How to get a + b on a graph

Nataly [62]

The first positively essential requirement is that
you absolutely have to know what 'a' and 'b' are.

I have no clue, so this is as far as I can go.

3 0

4 years ago

Why does the cyclist have less kinetic energy at position A than at position B?

Semmy [17]

Suppose that the cyclist begins his journey from the rest from the top of a wedge with a slope of a degree above the horizontal.
At point A (where it starts its journey), the energy is:
Ea = m * g * h
In other words, energy is only potential.
At point B (located at the bottom of the wedge), the energy is:
Eb = (1/2) * (m) * (v ^ 2)
In other words, the energy is only kinetic.
For energy conservation we have:
Ea = Eb
That is, we have that all potential energy is transformed into kinetic energy.
Which means that the cyclist has less kinetic energy at point A because that's where he has more potential energy.
answer:
the cyclist has less kinetic energy at point A because that's where he has more potential energy.

6 0

4 years ago

A rifle fires a 2.01 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the t

Zina [86]

Answer:

The value of spring constant is 266.01 $\frac{N}{m}$

Explanation:

Given:

Mass of pellet $m = 2.01 \times 10^{-2}$ kg

Height difference of pellet rise $h_{f} - h_{o} = 6.03$ m

Spring compression $x = 9.45 \times 10^{-2}$ m

From energy conservation law,

Spring potential energy is stored into potential energy,

$mg(h_{f} -h_{o}) = \frac{1}{2} kx^{2}$

Where $k =$ spring constant, $g = 9.8 \frac{m}{s^{2} }$

$k = \frac{2mg(h_{f} -h_{o} )}{x^{2} }$

$k = \frac{2 \times 9.8 \times 6.03\times 2.01 \times 10^{-2} }{(9.45\times 10^{-2} )^{2} }$

$k = 266.01$ $\frac{N}{m}$

Therefore, the value of spring constant is 266.01 $\frac{N}{m}$

6 0

3 years ago

A 2.00-m long uniform beam has a mass of 4.00 kg. The beam rests on a fulcrum that is 1.20 m from its left end. In order for the

Shalnov [3]

Answer:

x ’= 1,735 m, measured from the far left

Explanation:

For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.

Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive

They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,

the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar

x_{cm} = 1.2 -1

x_ {cm} = 0.2 m

Σ τ = 0

w₁ 1.2 + mg 0.2 - W₂ x = 0

x = $\frac{m_1 g\ 1.2 \ + m g \ 0.2}{M_2 g}$