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s344n2d4d5 [400]
3 years ago
12

A car has a mass of 900 kg and a truck has a mass of 1800 kg. In which of the following situations would they have the same mome

ntum?
The car travels at twice the velocity of the truck.

The car and truck travel at the same velocity.

The truck has a negative acceleration.

The car has a positive acceleration.
Physics
2 answers:
Kaylis [27]3 years ago
7 0

a) since force = mass * acceleration  

f= 900 * 0 (because constant speed = 0 acceleration)  

similarly b) f = 0

Yakvenalex [24]3 years ago
7 0

Answer:

The car travels at twice the velocity of the truck.

Explanation:

As we know that momentum is defined as the product of mass and velocity

so here we know

mass of the car = 900 kg

mass of the truck = 1800 kg

now let say that the speed of car = v_1

speed of truck = v_2

so if the momentum of car and truck is same

900 v_1 = 1800 v_2

v_1 = 2 v_2

so we can say that car will move with double the speed of truck

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Answer:

the distance that the object is raised above its initial position is 5.625 m.​

Explanation:

Given;

applied effort, E = 15 N

load lifted by the ideal pulley system, L = 16 N

distance moved by the effort, d₁ = 6 m

let the distance moved by the object = d₂

For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.

M.A = V.R

M.A = \frac{Load}{Effort} = \frac{L}{E} \\\\V.R = \frac{disatnce \ moved \  by \ the \ effort}{disatnce \ moved \  by \ the \ load} = \frac{d_1}{d_2} \\\\For \ ideal \ machine; \ M.A = V.R\\\\\frac{L}{E} = \frac{d_1}{d_2} \\\\d_2 = \frac{E \times d_1}{L} \\\\d_2 = \frac{15 \times 6}{16} \\\\d_2 = 5.625 \ m

Therefore, the distance that the object is raised above its initial position is 5.625 m.​

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What is motivation in your own world
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Answer:

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(8c8p49) A 115g Frisbee is thrown from a point 1.00 m above the ground with a speed of 12.00 m/s. When it has reached a height o
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Answer:

The work done on the Frisbee is 1.36 J.

Explanation:

Given that,

Mass of Frisbee, m = 115 g = 0.115 kg

Initial speed of Frisbee, u = 12 m/s at a point 1 m above the ground

Final speed of Frisbee , v = 10.9674 m/s when it has reached a height of 2.00 m. Let W is the work done on the Frisbee by its weight. According to work energy theorem, the work done is equal to the change in its kinetic energy. So,

W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times0.115\times\left((10.9674)^{2}-(12)^{2})\right)\\\\W=-1.36\ J

So, the work done on the Frisbee is 1.36 J. Hence, this is the required solution.

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