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IceJOKER [234]
3 years ago
7

Suppose that the most massive car ever built is moving down a 10.00 slope when the driver applies the brakes. The net force acti

ng on the car as it stops is -2.00X104 N. If the coefficient of kinetic friction between the car’s tires and pavement is 0.797, find the car’s mass. What is the normal force that the pavement exerts on the car?
Physics
1 answer:
slava [35]3 years ago
6 0

Hey there!

Use the question and search for a quizlet or science solving program

Here is a link:

www.scienceslovingprogram.com

Or

www.sciencehelp.com

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Would the frequency of the angular simple harmonic motion (SHM) of the balance wheel increase or decrease if the dimensions of t
storchak [24]

Answer:

Yes the frequency of the angular simple harmonic motion (SHM) of the balance wheel increases three times if the dimensions of the balance wheel reduced to one-third of original dimensions.

Explanation:

Considering the complete question attached in figure below.

Time period for balance wheel is:

T=2\pi\sqrt{\frac{I}{K}}

I=mR^{2}

m = mass of balance wheel

R = radius of balance wheel.

Angular frequency is related to Time period as:

\omega=\frac{2\pi}{T}\\\omega=\sqrt{\frac{K}{I}} \\\omega=\sqrt{\frac{K}{mR^{2}}

As dimensions of new balance wheel are one-third of their original values

R_{new}=\frac{R}{3}

\omega_{new}=\sqrt{\frac{K}{mR_{new}^{2}}}\\\\\omega_{new}=\sqrt{\frac{K}{m(\frac{R}{3})^{2}}}\\\\\omega_{new}={3}\sqrt{\frac{K}{mR^{2}}}\\\\\omega_{new}={3}\omega

5 0
3 years ago
One of your summer lunar space camp activities is to launch a 1090 kg rocket from the surface of the Moon. You are a serious spa
Ludmilka [50]

Answer:

ΔP.E = 6.48 x 10⁸ J

Explanation:

First we need to calculate the acceleration due to gravity on the surface of moon:

g = GM/R²

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of moon = 7.36 x 10²² kg

R = Radius of Moon = 1740 km = 1.74 x 10⁶ m

Therefore,

g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²

g = 2.82 m/s²

now the change in gravitational potential energy of rocket is calculated by:

ΔP.E = mgΔh

where,

ΔP.E = Change in Gravitational Potential Energy = ?

m = mass of rocket = 1090 kg

Δh = altitude = 211 km = 2.11 x 10⁵ m

Therefore,

ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)

<u>ΔP.E = 6.48 x 10⁸ J</u>

7 0
3 years ago
How do land and sea breeze work?
Studentka2010 [4]
 by the wind and air flow in the wind

7 0
3 years ago
Please help I literally don’t understand
Veseljchak [2.6K]

Answer:

A= 2

B=3

C=4

D=5

E=7

F=8

H=12

7 0
3 years ago
Number of waves that pass a given point in one second
Studentka2010 [4]
<em>number of waves that pass a given point in one second is called <u>frequency..</u></em>
5 0
3 years ago
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