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Musya8 [376]
3 years ago
13

A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/

s. what will be the momentum of this ball just as it reaches the ground
Physics
1 answer:
Anastasy [175]3 years ago
3 0

First we will find the speed of the ball just before it will hit the floor

so in order to find the speed of the cart we will first use energy conservation

KE_i + PE_i = KE_f + PE_f

\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 + 0

\frac{1}{2}(1.25)(52)^2 + 1.25(9.8)(225) = \frac{1}{2}(1.25)v_f^2

So by solving above equation we will have

v_f = 84.3 m/s

now in order to find the momentum we can use

P = mv

P = 1.25 \times 84.3

P = 105.4 kg m/s

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11.0 second hope i helped
6 0
3 years ago
21) A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest
zlopas [31]

Answer:

-22,150 N

Explanation:

When the youngster jumps off the platform, during the fall her initial potential energy is converted into kinetic energy, according to the law of conservation of energy. Therefore, we can write:

mgh=\frac{1}{2}mu^2

where the term on the left is the potential energy while the term on the right is the kinetic energy, and where

m = 50.0 kg is the mass of the youngster

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

h = 1.00 m is the heigth of the platform

u is the speed of the youngster as she reaches the floor

Solving for u,

u=\sqrt{2gh}=\sqrt{2(9.8)(1.00)}=4.43 m/s

Then, when the youngster hits the floor, the force exerted on her during the deceleration is given by:

F=\frac{\Delta p}{\Delta t}=\frac{m(v-u)}{\Delta t}

where \Delta p is her change in momentum, and where

m is the mass

v = 0 is the final velocity (she comes to a stop)

u = 4.43 m/s is the initial velocity

\Delta t=10.0 ms =0.010 s is the duration of the collision

Substituting,

F=\frac{(50.0)(0-4.43)}{0.010}=-22150 N

And the negative sign means the direction of the force is opposite to the motion (so, upward).

6 0
4 years ago
A series circuit has two 10-ohm bulb is added in a series. Technician A says that the three bulbs will be dimmer than when only
ANTONII [103]

Answer:

Technician  A  is right. The situation will happens even with only two bulbs in series

Explanation:

We must take into account that

1.- All electric device need its nominal voltage to operate

2.-Any and all electric device means an electric load for the source in terms of equation that means any device will implies a drop voltage of V = I*R ( I the flows current and R  the resistance of the device)

3.-Nominal voltage for bulbs are specify for houses voltages you find between fase and neutral wires for instance in Venezuela 120 (v).

4.-In a imaginary circuit of only one bulb, the nominal voltage will be applied and the bulb will operates correctly, but when you add another bulb (in series) the nominal voltage will split  between the two bulbs ( we  could find a situation such as the first bulb work properly but the second one does not). The voltage split according to Ohms law (in such way that the sum of voltage between the terminal of the first bulb plus the voltage at terminals of the second one are equal to nominal voltage.

For that reason all the bulbs are connected in parallel in wich case all of them will operate with the common voltage

4 0
3 years ago
A prodect is defined as
Keith_Richards [23]
<span>Product can be defined as a thing or person that is the result of an action or process</span>
5 0
4 years ago
2. A certain object revolves at a rate of 30 rpm. Please determine the frequency and period of this
schepotkina [342]

Answer:

The time period of the motion is, T = 0.03 s

The frequency of the rotation is, f = 30 Hz

Explanation:

Given data,

The rotational speed of an object, ω = 30 rpm

                                                          ω = 188.5 rad/s

The time period of motion is,

                            T = 2π / ω

Substituting the given values in the above equation

                               = 2π / 188.5

                            T = 0.03 s

The time period of the motion is, T = 0.03 s

The frequency of rotation,

                                f = 1 /T

                                   = 1 / 0.03

                                   = 30 Hz

Hence, the frequency of the rotation is, f = 30 Hz

7 0
4 years ago
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