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3 years ago

You lift a 50 N object 2 meters off the ground what work did you do on the object

1 answer:

4 0

Work equals force × displacement (distance between initial point and end point is displacement)
if u follow this it becomes
work = 50 × 2 which is equal to 100

comment if u have more questions

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Which physical phenomenon causes the spherical shapes of raindrops

jeyben [28]

Answer:

Raindrops start to form in a roughly spherical structure due to the surface tension of water. This surface tension is the "skin" of a body of water that makes the molecules stick together. The cause is the weak hydrogen bonds that occur between water molecules.

6 0

3 years ago

A pitcher throws a 0.144-kg baseball toward the batter so that it crosses home plate horizontally and has a speed of 42 m/s just

Solnce55 [7]

(a) 12.8 kg m/s

The impulse delivered by the bat on the baseball is equal to the change in momentum of the baseball:

$I=\Delta p = m(v-u)$

where we have

m = 0.144 kg is the mass of the ball

v = -47 m/s is the final velocity of the ball

u = 42 m/s is the initial velocity of the ball

Substituting into the equation, we find

$I=(0.144 kg)(-47 m/s-(42 m/s))=-12.8 kg m/s$

And since we are interested in the magnitude only,

$I=12.8 kg m/s$

(b) 2.78 kN

The impulse exerted on the ball is also equal to the product between the average force and the contact time:

$I=F\Delta t$

where

F is the average force exerted on the ball

$\Delta t=0.0046 s$ is the contact time

Solving the formula for F, we find

$F=\frac{I}{\Delta t}=\frac{12.8 kg m/s}{0.0046 s}=2783 N = 2.78 kN$

(c) The force exerted on the ball is much larger (1988 times more) than the weigth of the ball

The weight of the ball is given by

$W=mg$

where

m = 0.144 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration due to gravity

Solving the equation for W, we find

$W=(0.144 kg)(9.8 m/s^2)=1.4 N$

So as we see, the force exerted on the ball (2783 N) is almost 2000 times larger than the weight of the ball (1.4 N):

$\frac{F}{W}=\frac{2783 N}{1.4 N}=1988$

8 0

3 years ago

Some examples of solar ejections

mylen [45]

Coronal Mass ejections or Solar ejections are activities on the surface of the sun termed to the reaction of the gas composition of the sun's surface attributing to explosion of these gases. One example most commonly known as Solar Flares, and another example is termed as erupting prominence.

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4 years ago

Two balls with equal masses, m, and equal speed, v, engage in a head on elastic collision. what is the final velocity of each ba

Allushta [10]

The collision is elastic. This means that both momentum and kinetic energy are conserved after the collision.

- Let's start with conservation of momentum. The initial momentum of the total system is the sum of the momenta of the two balls, but we should put a negative sign in front of the velocity of the second ball, because it travels in the opposite direction of ball 1. So ball 1 has mass m and speed v, while ball 2 has mass m and speed -v:
$p_i = p_1-p_2 = mv-mv =0$
So, the final momentum must be zero as well:
$p_f = 0$
Calling v1 and v2 the velocities of the two balls after the collision, the final momentum can be written as
$p_f = mv_1 + mv_2 = 0$
From which
$v_1 = -v_2$

- So now let's apply conservation of kinetic energy. The kinetic energy of each ball is $\frac{1}{2} mv^2$ . Therefore, the total kinetic energy before the collision is
$K_i = \frac{1}{2} mv^2 + \frac{1}{2} mv^2 = mv^2$
the kinetic energy after the collision must be conserved, and therefore must be equal to this value:
$K_f = K_i = mv^2$ (1)
But the final kinetic energy, Kf, is also
$K_f = \frac{1}{2} mv_1^2 + \frac{1}{2}mv_2^2$
Substituting $v_1 = -v_2$ as we found in the conservation of momentum, this becomes
$K_f = mv_2 ^2$
we also said that Kf must be equal to the initial kinetic energy (1), therefore we can write
$mv_2^2 = mv^2$

Therefore, the two final speeds of the balls are
$v_2 = v$
$v_1 = -v_2 = -v$

This means that after the collision, the two balls have same velocity v, but they go in the opposite direction with respect to their original direction.

8 0

3 years ago

A 1500 kg car skids to a halt on a wet road where μk = 0.47. You may want to review (Pages 141 - 145) . Part A How fast was the

shusha [124]

The car travels at a speed of 25m/s.

<u>Explanation:</u>

Given-

Mass, m = 1500kg

Coefficient of friction, μk = 0.47

Distance, x = 68m

Speed, s = ?

We know,

$Force, F = ma$

and

F = μ X m X g

Therefore,

μ * m * g = m * a

μ * g = a

Let, g = 9.8m/s²

So,

$a = 0.47 * 9.8 m/s^2$

$a = 4.606m/s^2$

We know,

$v^2 - u^2 = 2as$

where, v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

If the car comes to rest, the final velocity, v becomes 0.

So,

$u^2 = 2 * 4.606 * 68\\\\u^2 = 626.416m/s\\\\u = 25m/s$

The car travels at a speed of 25m/s.

6 0

3 years ago