Ask question

Ask question

All categories

2 years ago

8

WHO LIKES ROGUE ONE AND CLONE WARS AND IS ADDICTED TO STAR WARS AND WOULD WISH THAT YOU COULD WORK WITH DAVE FILONI. Same bro Sa

me

1 answer:

scoray [572]2 years ago

3 0

Answer:

heck ya its so fire

Explanation:

You might be interested in

Write a hypothesis for Part II of the lab, which is about the relationship described by F = ma. In the lab, you will use a toy c

Temka [501]

Answer:

F=ma is the relationship where, F is force, m is mass and a is acceleration.

Newton's second law states that the unbalanced force applied to the object accelerates the object which is directly proportional to the force and inversely to the mass.

If we apply force to a toy car then It will accelerate.

This is how Newton's second law of motion is verified.

6 0

2 years ago

Read 2 more answers

Imagine that a bowling ball needs to be lifted 1.5 meters, and its gravitational potential energy is 90 joules. How much does th

Vikki [24]

Answer: 6.12 kg

Explanation:

Since Mass of ball = ? (let the unknown value be Z)

Acceleration due to gravity, g= 9.8m/s^2

Height, h = 1.5 metres

Gravitational potential energy GPE = 90J

Gravitational potential energy depends on the weight of the ball, the action of gravity and height.

Thus, GPE = Mass m x Acceleration due to gravity g x Height h

90J = Z x 9.8m/s^2 x 1.5m

90 = Z x 14.7

Z = 90/14.7

Z = 6.12 kg

Thus, the bowling ball weigh 6.12 kilograms

8 0

3 years ago

A 3500 kg truck is parked on a 7.0∘ slope. How big is the friction force on the truck?

Arada [10]

Answer:

4180 N

Explanation:

In order to keep the truck balanced and stationary, the force of friction on the truck must be equal to the component of the weight of the truck acting parallel to the slope.

The component of the weight of the truck acting parallel to the slope is:

$mg sin \theta$

where

m = 3500 kg is the mass of the truck

g = 9.8 m/s^2 is the acceleration of gravity

$\theta=7.0^{\circ}$ is the angle of the slope

Therefore, the force of friction must be equal to

$F=(3500)(9.8)(sin 7.0^{\circ})=4180 N$

6 0

3 years ago

Read 2 more answers

How many photons strike the cd surface during the playing of a cd 69 minutes in length?

Reptile [31]

Answer:

Photons strike the CD $n=1.624 x10^{18}$

Explanation:

First of all, it is necessary to calculate the energy produced by a semiconductor laser for 69 minutes:

$69minutes*\frac{60s}{1 minute}=4140s$

$E=0.1x10^{-3}w*4140s=0.414J$

Next, it is possible to calculate the number of photons striking the CD surface during this time:

$E=\frac{n*h*c}{l}$

$l=780x10^{-9}m$ long in CD

$c=3x10^8 m/s$ light velocity

$h=6.626x10^{-34} J*s$

Solve to n'

$n=\frac{E*l}{c*h}=\frac{0.414 J*780x10^{-9}m}{3x10^8m/s*6.626x10^{-34}}$

$n=1.624 x10^{18}$

8 0

2 years ago

What happens when light from the sun passes through any type of matter ?

dsp73

Most of the light is absorbed.

5 0

3 years ago

Read 2 more answers