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3 years ago

PLEASE HELP!!

Physics

1 answer:

IrinaK [193]3 years ago

7 0

Answer:

Explanation

i think its a not 100% sure though.

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Make a graph of the data. You may use a graphing program. Think about what data should be on the y-axis and the x-axis. Be sure

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yes it was a constant speed and the car traveled 10 meters in 20 seconds.

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3 years ago

Please help me with this question

ladessa [460]

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conduction

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3 years ago

Read 2 more answers

A 21.7kg child descends a slide 3.5 m high and reaches the bottom with a speed of 2.2m/s. How much thermal energy due to frictio

nadya68 [22]

Answer:

Subtract the kinetic energy at the bottom from the potential energy loss. The remainder becomes frictional heat.

Potential energy loss:

M g H = 21.7*9.81*3.5 = 745.1 J

Kinetic energy at bottom of slide:

= (1/2) M v^2 = 52.5 J

5 0

3 years ago

horizontal circular platform rotates counterclockwise about its axis at the rate of 0.919 rad/s. You, with a mass of 73.5 kg, wa

VARVARA [1.3K]

Answer:

The total angular momentum is 292.59 kg.m/s

Explanation:

Given that :

Rotation of the horizontal circular platform $\omega$ = 0.919 rad/s

mass of the platform (m) = 90.7 kg

radius (R) = 1.91 m

mass of the poodle $m_p$ = 20.5 kg

Your mass $m'$ = 73.5 kg

speed v = 1.05 m/s with respect to the platform

$V_p = \frac{1.05}{2} \ m /s \\ \\ = 0.525 \ m /s \\ \\r = \frac{R}{2}$

r = 0.955

Mass of the mutt $m_m$ = 18.5 kg

$r' = \frac{3}{4} \ R$

Your angular momentum is calculated as:

Your angular velocity relative to the platform is $\omega' = \frac{v}{R} = \frac{1.05}{1.91 } = 0.5497 \ rad/s$

$Actual \ \omega_y = \omega - \omega ' = (0.919 - 0.5497) \ rad/s = 0.3693 \ rad/s$

$I_y = m'R^2 = 73.5 *1.91^2= 268.14 \ kgm^2$

$L_Y = I_y*\omega_y= 268.14*0.3683= 98.76 \ kg.m/s$

For poodle :

$Relative \ \ \omega' = \frac{V_p}{R/2} = 0.550 \ rad/s$

Actual $\omega_p = \omega - \omega' = 0.919 -0.550 = 0.369 \ rad/s$

$I_p = m_p(\frac{R}{2} )^2 = 20.5(\frac{1.91}{2} )^2 = 18.70 \ kgm^2$

$L_p = I_p *\omega_p = 18.70*0.369 = 6.9003 \ kgm/s$

$I_M = m_m(\frac{3}{y4} R)^2 = 18.5 (\frac{3}{4} 1.91)^2 = 37.96 \ kgm^2$

$L_M = I_M \omega = 37.96 * 0.919= 34.89 \ kg.m/s$

Disk $I = \frac{mr^2}{2} = \frac{90.7*1.91^2}{2}= 165.44 \ kgm^2$

$L_D = I \omega = 165.44*0.919 =152.04 \ kg.m/s$

Total angular momentum of system is:

L = $L_D +L_Y+L_P+L_M$

= (152.04 + 98.76 + 6.9003 + 34.89) kg.m/s

= 292.59 kg.m/s

6 0

4 years ago

Why is nuclear energy able to be used for practical purposes? the reactions are controlled to regulate energy output. the reacti

Sonbull [250]

Nuclear energy is is used practically for many energy requirements because the reactions are controlled to regulate energy output

<h3>What is nuclear energy?</h3>

Nuclear energy is the energy reaelesed during nuclear relations.

Nuclear energy is a clean form of energy which is available in enormous amounts.

Nuclear energy is is used practically for many energy requirements because the reactions are controlled to regulate energy output.

Therefore, the control of nuclear reactions make the use if nuclear energy possible.

Learn more about nuclear reactions at: brainly.com/question/25387647

#SPJ4

3 0

2 years ago