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2 years ago

How much time is needed to produce 720 joules of work if 90 watts of power is used

Physics

1 answer:

Sladkaya [172]2 years ago

4 0

Answer:

8 seconds

Explanation:

power (P) is defined as the rate at which work is done.

power is measured in Watts (W) , when the work done is measured in Joules (J) and time in seconds

by the definition of power,

$power=\frac{work.done}{time}\\ time=\frac{work.done}{power}\\ =\frac{720J}{90W} \\=8 seconds$

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in a closed system three objects have the following momentum: 11 kg* m/s, -65 kg*m/s and -100 kg m/s. the objects collide and mo

guajiro [1.7K]

Explanation:

The momentum of the three objects are as follow :

11 kg-m/s, -65 kg-m/s and -100 kg-m/s

Before collision, the momentum of the system is :

$P_i=11+(-65)+(-100)\\\\P_i=-154\ kg-m/s$

After collison, they move together. It means it is a case of inelastic collision. In this type of collision, the momentum of the system remains conserved.

It would mean that, after collision, momentum of the system is equal to the initial momentum.

Hence, final momentum = -154 kg-m/s.

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3 years ago

an ice skater starts with a velocity of 2.25 m/s in a 50.0 direction. after 8.33 m/s in a 120 direction. what is the y-component

Nata [24]

The y-component of the acceleration is $0.28 m/s^2$

Explanation:

The y-component of the ice skater acceleration can be calculated with the equation

$a_y = \frac{v_y-u_y}{t}$

where

$v_y$ is the y-component of the final velocity

$u_y$ is the y-component of the initial velocity

t is the time elapsed

Here we have:

Initial velocity is $u=2.25 m/s$ at $\theta_1=50.0^{\circ}$ , so its y-component is $u_y = u sin \theta_1 = (2.25)(sin 50.0^{\circ})=1.72 m/s$

Final velocity is $v=4.65 m/s$ at $\theta_2=120.0^{\circ}$ , so its y-component is $v_y = v sin \theta_2 = (4.65)(sin 120.0^{\circ})=4.03 m/s$

The time elapsed is

t = 8.33 s

Therefore, the y-component of the acceleration is

$a_y = \frac{4.03-1.72}{8.33}=0.28 m/s^2$

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