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3 years ago

True or false? When an object deforms, the change of shape is always permanent. false true

1 answer:

8 0

True. Because say if you go to a junk place where you can break anything and you decide to squish a printer with a car and when you go look at the printer it’s all deformed there’s really no way that you can fix it. Lol that was an example

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A car is idling at a stoplight. When the light turns green, the car takes 12 seconds to accelerate to 18.7 m/s. What is the aver

Sonbull [250]

Answer:

1.6 m/s^2

Explanation:

3 0

4 years ago

Two facing surfaces of two large parallel conducting plates separated by 8.5 cm have uniform surface charge densities such that

elena-s [515]

Answer:

positive plate

E = 5.764 KV / m

W = 490eV or 7.85 * 10^-17 J

E_p = 4.74 *10^(-12) eV

E_k = 490 eV

Explanation:

part a

The potential difference between two plates = 490 V

Distance between two plates = 8.5 cm

Answer: The positive plate is at higher potential because of convention.

part b

Electric Field between the plates

E = V / d

E = 490 / 0.085 = 5.764 KV / m

Answer: Electric Field between the plates E = 5.764 KV / m

part c

Work done by electric field

W = V*q

W = 490 * 1.602*10^-19

W = 7.85 * 10^-17 J

or W = 490 eV

Answer: Work done by electric field W = 490eV or 7.85 * 10^-17 J

part d

Potential Energy of an electron gained:

E_p = m_e * g * d / (1.602*10^-19)

E_p = 9.109*10^-31* 9.81 * 0.085 / (1.602*10^-19)

E_p = 4.74 *10^(-12) eV

Very very small E_p approximately 0

Answer: Potential Energy of an electron gained E_p = 4.74 *10^(-12) eV or 0.

part e

Kinetic Energy of an electron gained:

W - E_p = E_k

E_k = 490eV - 4.74*10^(-12)eV

E_k = 490 eV

Answer: Kinetic Energy of an electron gained E_k = 490 eV

7 0

4 years ago

A bowling ball with a mass of 8 kg is moving at a speed of 5m/s. What is its kinetic energy

dolphi86 [110]

Answer:The formula for kinetic energy is

(1/2) M V^2.

With M in kg and V in m/s, the answer will be in Joules.

K.E = 40joule

Explanation:

4 0

3 years ago

Suppose a 65 kg person stands at the edge of a 6.5 m diameter merry-go-round turntable that is mounted on frictionless bearings

Hatshy [7]

Answer:0.316 rad/s

Explanation:

Given

mass of Person $m=65 kg$

velocity of person $v=3.8 m/s$

diameter of turntable $d=6.5 m$

moment of Inertia of the table $I_0=1850 kg-m^2$

Moment of inertia of Person I

$I=mr^2=65\times (\frac{6.5}{2})^2$

$I=686.56 kg-m^2$

initial angular velocity $\omega _1=\frac{v}{r}$

$\omega _1=\frac{3.8}{3.25}=1.17 rad/s$

Conserving Angular momentum

$I\omega _1=(I+I_0)\omega _2$ , where $\omega _2=final\ angular\ velocity$

$686.56\times 1.17=(686.56+1850)\times \omega _2$

$\omega _2=\frac{686.56}{2536.56}\times 1.17$

$\omega _2=0.316 rad/s$

4 0

3 years ago

Two points charge of 4\mu C and 2\mu C are placed at theopposite corners of a rectangle. What is the potential difference Va- Vb

bulgar [2K]

Answer:

$Va-Vb=168KV$

Explanation:

From the question we are told that

Two points charge of 4\mu C and 2\mu C

Generally we find the Va and Vb individually to find there difference

Given a rectangle with two equal sides each,Assume lengths for bot sides

Length L=0.3

Breath B=0.4

Diagonal D= $\sqrt{0.3^2+0.4^2} =0.5$

at opposite sides

Mathematically Va can represented as

$Va =k(\frac{4*10^_-_6}{0.3} +\frac{-2*10^_-_6}{0.5} )$

$Va =9*10^9(\frac{4*10^_-_6}{0.3} +\frac{-2*10^_-_6}{0.5} )$

$Va =9*10^9(0.00001333333-0.000004} )$

$Va =84000V$

$Va =84KV$

Mathematically Vb is represented as

$Va =k(\frac{-4*10^_-_6}{0.3} +\frac{2*10^_-_6}{0.5} )$

$Va =9*10^9(\frac{-4*10^_-_6}{0.3} +\frac{+2*10^_-_6}{0.5} )$

$Va =9*10^9(-0.00001333333+0.000004} )$

$Va =-84000V$

$Va =-84KV$

Therefore

$Va-Vb=84-(-84)\\Va-Vb=84+84\\Va-Vb=168KV$

7 0

3 years ago