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Mkey [24]
3 years ago
5

An 12 kg object accelerating from 9 m/s to 14 m/s. What is the change in momentum of the object?

Physics
1 answer:
Trava [24]3 years ago
7 0

Answer:  8 kg*m/s and 2m/s

Explanation:   Momentum is mass times velocity. since the velocity changes by 2 m/s. you just take the difference from final velocity to beginning velocity.

4kg*12m/s = 48 kg*m/s

4kg*10m/s = 40 kg*m/s

48kg*m/s - 40 kg*m/s = 8kg*m/s

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A block of height 120mm is placed on top of another block with a height of 1.50 m. What is the height of the two blocks together
pav-90 [236]
We will first convert all units to meters and then solve the problem.
We are given that:
1000 mm = 1 m
120 mm = ?? meters
using cross multiplication:
120 mm = (120*1) / 1000 = 0.12 m

Now, when the two objects are placed over each other, their total height is the result of summation of both heights, therefore:
total height = 0.12 + 1.5 = 1.62 m

Based on the above calculations, the correct choice is:
<span>b) 1.62 m </span>
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When a net force of 17.0 newtons is applied to a dictionary placed on a frictionless table, it accelerates by 3.75 meters/second
4vir4ik [10]
Force = (mass) x (acceleration)    (Newton's second law of motion)

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Svetllana [295]

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A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle
xz_007 [3.2K]

Answer:

a) \Delta{t} = 5.39s

b) the motorcycle travels 155 m

Explanation:

Let t_2-t_1 = \Delta{t}, then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):

v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}

where:

v_{m2} is the speed of the motorcycle at time 2

v_{c} is the velocity of the car (constant)

v_{0} is the velocity of the car and the motorcycle at time 1

d is the distance between the car and the motorcycle at time 1

x is the distance traveled by the car between time 1 and time 2

Solving the system of equations:

\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]

v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s

For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:

x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933

3 0
3 years ago
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