What isthe correct 1/4=20%

Which processes transfer energy from the core to the photosphere?

nikklg [1K]

-radiation and convection

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7 0

3 years ago

A person drives north 8 blocks, then turns west and drives 4 blocks. The driver then turns south and drives 8 blocks. How could

stealth61 [152]

A person would be driving 4 blocks west from the starting point to make the shorter distance while maintaining same displacement.

Explanation:

The measurement of an object’s position change from its point is called displacement. It is usually calculated from starting to the end points and represented by ‘delta s’. In the given scenarios, the person drove in the way that he finals the driving by 4 blocks away from the west.

Means, the persons drive to 8 blocks north and then to 8 blocks south get cancelled. Hence, to make the shorter distance with maintaining same displacement he would be driving 4 blocks west from the starting point.

7 0

3 years ago

• How can you find the temperature in Fahrenheit when its given in the Celsius scale? I

Ray Of Light [21]

Answer:

there are conversion scales used for doing that

to convert Celsius to Fahrenheit we use the formula,

T(°F) = T(°C) × 9/5 + 32

and to convert Fahrenheit to Celsius we use,

T(°F) = T(°C) × 5/9 + 32

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Explanation:

4 0

3 years ago

A toy cannon uses a spring to project a 5.38-g soft rubber ball. The spring is originally compressed by 5.08 cm and has a force

yawa3891 [41]

(a) 1.43 m/s

We can solve this problem by using the law of conservation of energy.

The initial total energy stored in the spring-mass system is

$E=U=\frac{1}{2}kx^2$

where

k = 7.91 N/m is the spring constant

$x=5.08 cm = 0.0508 m$

Substituting,

$E=\frac{1}{2}(7.91)(0.0508)^2=0.0102 J$

The final kinetic energy of the ball is equal to the energy released by the spring + the work done by friction:

$E+W_f=K$

where

$K_f=\frac{1}{2}mv^2$ is the kinetic energy of the ball, with

$m=5.38 g = 5.38\cdot 10^{-3} kg$ being the mass of the ball

v being the final speed

$W_f = -F_f d$ is the work done by friction (which is negative since the force of friction is opposite to the motion), where

$F_f = 0.0323 N$ is force of friction

d = 14.5 cm = 0.145 m is the displacement

Substituting,

$W_f = -(0.0323)(0.145)=-4.68\cdot 10^{-3} J$

So, the kinetic energy of the ball as it leaves the cannon is

$K_f = E+W_f = 0.0102 - 4.68\cdot 10^{-3}=0.00552 J$

And so the final speed is

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.00552)}{0.00538}}=1.43 m/s$

(b) +5.08 cm

The speed of the ball is maximum at the instant when all the elastic potential energy stored in the spring has been released: in fact, after that moment, the spring does no longer release any more energy, so the kinetic energy of the ball from that moment will start to decrease, due to the effect of the work done by friction.

The elastic potential energy of the spring is

$U=\frac{1}{2}kx^2$

And this has all been released when it becomes zero, so when x = 0 (equilibrium position of the spring). However, the spring was initially compressed by 5.08 cm, so the ball has maximum speed when

x = +5.08 cm

with respect to the initial point.

(c) 1.78 m/s

The maximum speed is the speed of the ball at the moment when the kinetic energy is maximum, i.e. when all the elastic potential energy has been released.

As we calculated in part (a), the total energy released by the spring is

E = 0.0102 J

The work done by friction here is just the work done to cover the distance of

d = 5.08 cm = 0.0508 m

Therefore

$W_f = -(0.0323)(0.0508)=-1.64\cdot 10^{-3} J$