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LenaWriter [7]
3 years ago
12

What is force x time is called

Physics
1 answer:
Law Incorporation [45]3 years ago
4 0

Answer:

In words, it could be said that the force times the time equals the mass times the change in velocity. In physics, the quantity Force • time is known as impulse. And since the quantity m•v is the momentum, the quantity m•Δv must be the change in momentum. The equation really says that the Impulse = Change in momentum.

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Quick please and will give Brainliest!!!
masha68 [24]

25 nC

That is the answer

3 0
3 years ago
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A positively charged body exerts an electric field that causes a positively charged particle to move away from it. If work is do
ollegr [7]

Answer:

b. electric potential energy.

Explanation:

The energy required to move a charge against the electric field is known as the electric potential energy. As in above case positively charged body is exerting an electric field on the positive charge. As the same charges repel so the charge tend to move away. In order to push it towards the body we need a work done. As it is hard to push the positive charged particle towards the positive electric field. So in the cases like these particle occupies the electric potential energy.

8 0
3 years ago
A 0.0414 kg ingot of metal is heated to 243◦C
tino4ka555 [31]

Answer:

448 J/kg/°C

Explanation:

m₁ C₁ (T₁ − T) + m₂ C₂ (T₂ − T) = 0

(0.0414 kg) C (243°C − 20.4°C) + (0.411 kg) (4186 J/kg/°C) (18°C − 20.4°C) = 0

(9.22 kg°C) C − 4129 J = 0

C = 448 J/kg/°C

3 0
3 years ago
2
BigorU [14]
Could you maybe show us the diagram
3 0
3 years ago
In the Hunger Games movie, Katniss Everdeen fires a 0.0200-kg arrow from ground level to pierce an apple up on a stage. The spri
egoroff_w [7]

Answer:

a) v=99.8584\ m.s^{-1}

b) v'=99.366\ m.s^{-1}

Explanation:

Given:

mass of the arrow, m=0.02\ kg

stiffness constant of the bow, k=330\ N,m^{-1}

distance of pulling back the arrow on the bow from its mean position, \Delta x=0.55\ m

height of the apple targeted, h=5\ m

<u>Force on the arrow due to the stiffness of the bow:</u>

F=k.\Delta x

F=330\times 0.55

F=181.5\ N

Now the acceleration of the arrow upwards:

a=\frac{F}{m}

a=\frac{181.5}{0.02}

a=9075\ m.s^{-2}

a) For the course of motion when the arrow leaves the bow after the stretch is relaxed we consider that the arrow left the bow after its string goes to the mean position. During this phase the arrow also faces gravity in the downward direction.

Using the equation of motion:

v^2=u^2+2(a-g).\Delta x

where:

v= velocity with which the arrow leaves the bow

u= initial velocity of the arrow after it left

v^2=0^2+2\times (9075-9.81)\times 0.55

v=99.8584\ m.s^{-1}

b) Now when the arrow travels up then it is under a constant gravitational force acting opposite to the motion.

<u>Using eq. of motion:</u>

v'^2=v^2-2\times g.h

where:

v'= final velocity when the arrow hits the target

v= initial velocity after the arrow has been launched

v'^2=99.8584^2-2\times 9.81\times 5

v'=99.366\ m.s^{-1}

3 0
3 years ago
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