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3 years ago

A 5 kg ball is sitting on top of a hill. It has a Potential Energy of 6000J. What is the height of the ball?

Physics

1 answer:

natali 33 [55]3 years ago

6 0

Explanation:

mass=5 kg

potential energy=6000j

height=?

Now

potential energy =m.g.h

or 6000=5*9.8*h

or 6000=49h

or 6000÷49=h

or h= 122.45m

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A raindrop or partially melted snowflake that freezes into a pellet of ice in a deep subfreezing layer of air near the surface i

Morgarella [4.7K]

A raindrop or partially melted snowflake that freezes into a pellet of ice in a deep subfreezing layer of air near the surface is called sleet.

Answer: Sleet

Hope that helps! ★ If you have further questions about this question or need more help, feel free to comment below or leave me a PM. -UnicornFudge aka Nadia

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3 years ago

What two things does light provide?

labwork [276]

Answer:

Light provides brightness to see and also light carries energy

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2 years ago

A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7.

dalvyx [7]

Answer:

(a) $W_c=127.008 J$

(b) $W_g=148.176 J$

(c) K.E. = 21.168 J

(d) $v=3.4293m.s^{-1}$

Explanation:

Given:

mass of a block, M = 3.6 kg

initial velocity of the block, $u=0 m.s^{-1}$

constant downward acceleration, $a_d= \frac{g}{7}$

$\Rightarrow$ That a constant upward acceleration of $\frac{6g}{7}$ is applied in the presence of gravity.

∴ $a=- \frac{6g}{7}$

height through which the block falls, d = 4.2 m

(a)

Force by the cord on the block,

$F_c= M\times a$

$F_c=3.6\times (-6)\times\frac{9.8}{7}$

$F_c=-30.24 N$

∴Work by the cord on the block,

$W_c= F_c\times d$

$W_c= -30.24\times 4.2$

We take -ve sign because the direction of force and the displacement are opposite to each other.

$W_c=-127.008 J$

(b)

Force on the block due to gravity:

$F_g= M.g$

∵the gravity is naturally a constant and we cannot change it

$F_g=3.6\times 9.8$

$F_g=35.28 N$

∴Work by the gravity on the block,

$W_g=F_g\times d$

$W_g=35.28\times 4.2$

$W_g=148.176 J$

(c)

Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.

mathematically:

$K.E.= W_g+W_c$

$K.E.=148.176-127.008$

K.E. = 21.168 J

(d)

From the equation of motion:

$v^2=u^2+2a_d\times d$

putting the respective values:

$v=\sqrt{0^2+2\times \frac{9.8}{7}\times 4.2 }$

$v=3.4293m.s^{-1}$ is the speed when the block has fallen 4.2 meters.

6 0

3 years ago

A conducting bar moves along a circuit with a constant velocity. A constant magnetic field is perpendicular to the bar and circu

lana [24]

Answer:

0.500 T

Explanation:

Since the change in time and the number of coils are both 1, I set the problem up to be 1.3=(1.5(x)-13(x)). I then plugged in numbers for x until I got the answer to be 1.3 V.

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3 years ago

Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 13

svet-max [94.6K]

Answer:

5945.27 W per meter of tube length.

Explanation:

Let's assume that:

Steady operations exist;
The heat transfer coefficient (h) is uniform over the entire fin surfaces;
Thermal conductivity (k) is constant;
Heat transfer by radiation is negligible.

First, let's calculate the heat transfer (Q) that occurs when there's no fin in the tubes. The heat will be transferred by convection, so let's use Newton's law of cooling:

Q = A*h*(Tb - T∞)

A is the area of the section of the tube,

A = π*D*L, where D is the diameter (5 cm = 0.05 m), and L is the length. The question wants the heat by length, thus, L= 1m.

A = π*0.05*1 = 0.1571 m²

Q = 0.1571*40*(130 - 25)

Q = 659.73 W

Now, when the fin is added, the heat will be transferred by the fin by convection, and between the fin and the tube by convection, thus:

Qfin = nf*Afin*h*(Tb - T∞)

Afin = 2π*(r2² - r1²) + 2π*r2*t

r2 is the outer radius of the fin (3 cm = 0.03 m), r1 is the radius difference of the fin and the tube ( 0.03 - 0.025 = 0.005 m), and t is the thickness ( 0.001 m).

Afin = 0.006 m²

Qfin = 0.97*0.006*40*(130 - 25)

Qfin = 24.44 W

The heat transferred at the space between the fin and the tube will be:

Qspace = Aspace*h*(Tb - T∞)

Aspace = π*D*S, where D is the tube diameter and S is the space between then,

Aspace = π*0.05*0.003 = 0.0005

Qspace = 0.0005*40*(130 - 25) = 1.98 W

The total heat is the sum of them multiplied by the total number of fins,

Qtotal = 250*(24.44 + 1.98) = 6605 W

So, the increase in heat is 6605 - 659.73 = 5945.27 W per meter of tube length.

5 0

3 years ago

A 5 kg ball is sitting on top of a hill. It has a Potential Energy of 6000J. What is the height of the ball?​

A 5 kg ball is sitting on top of a hill. It has a Potential Energy of 6000J. What is the height of the ball?