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Neporo4naja [7]

3 years ago

14

Explain how heat is transferred between a hot and cold object by conduction.

2 answers:

vredina [299]3 years ago

8 0

Heat is the internal kinetic energy of the atoms or molecules.

When they bump into each other they transfer that energy (like dodgem cars at a fairground).
Conduction is when heat is transferred by touch

ludmilkaskok [199]3 years ago

3 0

When something hot touches something cold the heat touches the surface of the cold thing making it burn and spread the heat around making it warm.

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As the moon revolves around the earth,it also rotated on its axis why is that the same side of the moon is always visible from e

Alina [70]

Answer:

Explanation:

Our lunar companion rotates while it orbits Earth. It's just that the amount of time it takes the moon to complete a revolution on its axis is the same it takes to circle our planet — about 27 days. As a result, the same lunar hemisphere always faces Earth.

7 0

3 years ago

Read 2 more answers

Question 20 of 20

Sphinxa [80]

Answer:

a

Explanation:

hI I

6 0

2 years ago

A yo-yo of mass M has an axle of radius b and a spool of radius R. Its moment of inertia can be taken to be MR2/2 and the thickn

kow [346]

Answer:

The tension in the cord is $T=\frac{MR^{2}g }{2b^{2}+R^{2} }$

Explanation:

Given:

M = mass

b = radius

R = spool of radius

The equation is:

$bT=(\frac{MR^{2} }{2} )(\frac{a}{b} )\\T=\frac{MR^{2}a }{2b^{2} }$ (eq. 1)

The sum of forces in y:

∑Fy = Mg - T = Ma

$Mg=(M+\frac{MR^{2} }{2b^{2} } )a\\a=\frac{2b^{2}g }{2b^{2}+R^{2} }$

Replacing in eq. 1

$T=\frac{MR^{2} }{2b^{2} } (\frac{2b^{2}g }{2b^{2} +R^{2} } )\\T=\frac{MR^{2}g }{2b^{2}+R^{2} }$

3 0

3 years ago

Read 2 more answers

A car is traveling at a velocity of 22 m/s when the driver puts on the brakes

Brums [2.3K]

The car’s velocity at the end of this distance is <em>18.17 m/s.</em>

Given the following data:

Initial velocity, U = 22 m/s

Deceleration, d = 1.4 $m/s^2$

Distance, S = 110 meters

To find the car’s velocity at the end of this distance, we would use the third equation of motion;

Mathematically, the third equation of motion is calculated by using the formula;

$V^2 = U^2 + 2dS$

Substituting the values into the formula, we have;

$V^2 = 22 + 2(1.4)(110)\\\\V^2 = 22 + 308\\\\V^2 = 330\\\\V^2 = \sqrt{330}$

<em>Final velocity, V = 18.17 m/s</em>

Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>

<em></em>

Read more: brainly.com/question/8898885

8 0

2 years ago

The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca

d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

$U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}$

the elastic potencial energy is given by:

$U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J$

The work is defined as:

$W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J$

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

$U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J$

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

$8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s$

3 0

3 years ago