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2 years ago

A compact car moving at 60 m/s has 320,000 J of kinetic energy. What is its mass?

Physics

1 answer:

joja [24]2 years ago

8 0

Answer:

320,000J

Explanation:

mass=320,000J

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6. A girl pushes her little brother on his sled with a force of 300 N for 750 m.

Oduvanchick [21]

w=225000 J

E=225000 J

P=9000 W

P=22500 W

Ew=Fd=E

P=E/t

5 0

2 years ago

Compute the torque about the origin of the gravitational force F--mgj acting on a particle of mass m located at 7-xî+ yj and sho

Andrews [41]

Answer:

Explanation:

Force, F = - mg j

r = - 7x i + y j

Torque is defined as the product f force and the perpendicular distance.

It is also defined as the cross product of force vector and the displacement vector.

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$

$\overrightarrow{\tau }=(- 7 x i + yj)\times (-mgj)$

[tex]\overrightarrow{\tau }= 7 m g x k

Here, we observe that the torque is independent of y coordinate.

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2 years ago

Why does it take significantly stronger magnetic and electric field strengths to move the beam of alpha particles compared with

wlad13 [49]

It takes significantly stronger magnetic and electric field strengths to move a beam of alpha particles compared with the beam of electrons(betaparticles) because the charge of an alpha particle is twice stronger than a beta particle. Therefore, more energy is needed to move the alpha particle.

4 0

3 years ago

Questions for 1.21 physics lab report

Lapatulllka [165]

Ok cool dude bro I just need to answer a question

6 0

3 years ago

A person throws a ball straight up. He releases the ball at a height of 1.75 m above the ground and with a velocity of 12.0 m/s.

Lyrx [107]

Answer:

a) 1.22 s

b) 9.089 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-12}{-9.81}\\\Rightarrow t=1.22\ s$

Time taken by the ball to reach the highest point is 1.22 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=12\times 1.22+\frac{1}{2}\times -9.81\times 1.22^2\\\Rightarrow s=7.339\ m$

The maximum height the ball will reach above the ground is 1.75+7.339 = 9.089 m

8 0

2 years ago