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3 years ago

Please send me solution of the question pls

Physics

1 answer:

ELEN [110]3 years ago

4 0

Answer:

20m

Explanation:

P.E=mgh

2000=10×10×h

2000=100h

Divide both side by 100

2000/100=20

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A) On which hemisphere does Bhutan lie on the earth?

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3 years ago

An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10

Alexeev081 [22]

Answer:

a) $c=1822.3214\ J.kg^{-1}.K^{-1}$

b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c) The material is peat, possibly.

d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.

Explanation:

Given:

mass of aluminium, $m_a=0.1\ kg$

mass of water, $m_w=0.25\ kg$

initial temperature of the system, $T_i=10^{\circ}C$

mass of copper block, $m_c=0.1\ kg$

temperature of copper block, $T_c=50^{\circ}C$

mass of the other block, $m=0.07\ kg$

temperature of the other block, $T=100^{\circ}C$

final equilibrium temperature, $T_f=20^{\circ}C$

We have,

specific heat of aluminium, $c_a=910\ J.kg^{-1}.K^{-1}$

specific heat of copper, $c_c=390\ J.kg^{-1}.K^{-1}$

specific heat of water, $c_w=4186\ J.kg^{-1}.K^{-1}$

Using the heat energy conservation equation.

The heat absorbed by the system of the calorie-meter to reach the final temperature.

$Q_{in}=m_a.c_a.(T_f-T_i)+m_w.c_w.(T_f-T_i)$

$Q_{in}=0.1\times 910\times (20-10)+0.25\times 4186\times (20-10)$

$Q_{in}=11375\ J$

The heat released by the blocks when dipped into water:

$Q_{out}=m_c.c_c.(T_c-T_f)+m.c.(T-T_f)$

where

$c=$ specific heat of the unknown material

For the conservation of energy : $Q_{in}=Q_{out}$

so,

$11375=0.1\times 390\times (50-20)+0.07\times c\times (100-20)$

$c=1822.3214\ J.kg^{-1}.K^{-1}$

This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

The material is peat, possibly.

The material cannot be ice because ice doesn't exists at a temperature of 100°C.

7 0

3 years ago

A 3.00-kg model airplane has velocity components of 5.00 m/s due east and 8.00 m/s due north. What is the plane’s kinetic energy

GalinKa [24]

Answer:

Kinetic energy, E = 133.38 Joules

Explanation:

It is given that,

Mass of the model airplane, m = 3 kg

Velocity component, v₁ = 5 m/s (due east)

Velocity component, v₂ = 8 m/s (due north)

Let v is the resultant of velocity. It is given by :

$v=\sqrt{v_1^2+v_2^2}$

$v=\sqrt{5^2+8^2}=9.43\ m/s$

Let E is the kinetic energy of the plane. It is given by :

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2$

E = 133.38 Joules

So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.

5 0

3 years ago

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Please send me solution of the question pls​

Please send me solution of the question pls