Q3. A bridge is built without expansion gaps.

The Sun is about 150 million km from earth. how long does it take light from the sun to reach earth? (Speed of light is 3x10^8m/

madam [21]

Time =      (distance) / (speed)

Time = (150 x 10⁹ m) / (3 x 10⁸ m/s) =

   50 x 10¹ sec =

       <em>500 sec</em> = 8 min 20 sec

3 0

3 years ago

A rollercoaster car has 2500 J of potential energy and 160 J of

Mrrafil [7]

Answer:

xaubUajnaai ajn AJ au aun a

Explanation:

ahayba uabah an aj

6 0

3 years ago

Calculate the work done if a boy lifts a bag of cement 500N to the floor of a lorry 2.5m above the ground

eduard

Answer:

W = 1250 J = 1.25 KJ

Explanation:

The work done by the boy is due to the change in the position of the cement vertically. Hence, the work done in this case will be equal to the potential energy of the cement:

$W = P.E\\W = mgh$

where,

W = Work done = ?

mg = W = weight of cement = 500 N

h = height covered = 2.5 m

Therefore,

$W = (500\ N)(2.5\ m)$

4 0

3 years ago

The boy soon loses his balance and falls backwards off the board at a velocity of 1.0 m/s. Assuming momentum is conserved in thi

Andrei [34K]

Answer:

(a) W(s) = M*g

(b) 7.84kg

(c) - 10.07m/s

Explanation:

Parameters given:

Mass of boy, m = 79kg

Initial velocity of boy before jumping on the skateboard, u(b) = 1.55m/s

Final velocity of boy after jumping on the skateboard, v(b) = 1.41m/s

Mass of skateboard, M = ?

Initial velocity of skateboard = u(s)

Final velocity of skateboard = v(s)

(a) The weight of the skateboard is given as the mass of the skateboard multiplied by acceleration due to gravity. Mathematically,

W = M * g

Where g = acceleration due to gravity

(b) Using the law of conservation of momentum, we have that:

m*u(b) + M*u(s) = m*v(b) + m*v(s)

We know that after jumping on the skateboard, the boy and the skateboard have the same final velocity, hence,

v(b) = v(s) = v

=> m*u(b) + M*u(s) = (m + M) * v

Making M the subject of the formula, we have that:

M = m{u(b) - v} / v

M = 79 * (1.55 - 1.41) / 1.41

M = 7.84kg

(c) According to Newton's third law of motion, the momentum of the boy after falling must be equal to but opposite the momentum of the skateboard. This is the concept that birthed the law of conservation of momentum.

Momentum is given as:

P = m*v

Where P = momentum

m = mass of object

v = velocity of object

Therefore, applying Newton's third law of motion:

m * v(b) = -M * v(s)

Where

v(b) is now the velocity of the boy after failing off the skateboard = 1.0m/s

v(s) is now the velocity of the skateboard after the boy falls off.

Making v(s) subject of the formula, we have that:

v(s) = m * v(b) / M

v(s) = -(79 * 1.0) / 7.8

v(s) = -10.07m/s

The negative sign means that that velocity of the boy is opposite the velocity of the skateboard.

4 0

3 years ago

A charge of 3.20 μC is held fixed at the origin. A second charge of 3.20 μC is released from rest at the position (1.25 m, 0.570

siniylev [52]

Answer:

a) $v = 7.137 m/s$ and b) $r = 1.832\ m$

Explanation:

We have to analyze this problem from the point of view of energy conservation. In this case there are two kind of energy, electric potential energy and kinetic energy. First, at $t = t_1$ there isn't relative movement between the two charges, so kinetic energy is zero and the total energy ( $E_T = E_p + E_k$ ) is just potential.

$E_T = E_p = \frac{k q_1 q_2}{r}$

where $k$ is the Coulomb constant, $q_1$ and $q_2$ are the two interacting charges, and $r$ is the distance between them.

Considering the fixed charge at (x1,y1) = (0,0) and the second one at (x2, y2) = (1.25, 0.57), the initial distance is

$r = \sqrt{(x_2-x_1)^2 + (y_2- y_1)^2} = \sqrt{(1.25)^2 + (0.57)^2} = 1.374\ m$ , then if

$k =8.987\times 10^9 N m^2/C^2$ and $q_1 = q_2 =3.2\times 10^{-6} C$ ,

$E_T = E_p = \frac{k q_1 q_2}{r} = \frac{8.987\times 10^9 * 3.2\times 10^{-6}*3.2\times 10^{-6}}{1.374} = 6.698 \times 10^{-2} Nm$ .

Now, at $t = t_2$ , $r \rightarrow \infty$ and $E_p \rightarrow 0$ . This means all the energy is kinetic

$E_T = E_k = \frac{1}{2}mv_f^2$ , so

$v_f = \sqrt{2E_T/m} =\sqrt{2 *6.698 \times 10^{-2} /0.00263} = 7.137 m/s$ (mass in Kg).

That would be the velocity when the second charge moves infinitely far from the origin.

For the second part we have that $v = v_f/2$ , so kinetic energy is

$E_k = \frac{1}{2}mv^2 = 1.674 \times 10^{-2} Nm$

and potential energy is

$E_p = E_T - E_k = 6.698 \times 10^{-2} - 1.674 \times 10^{-2} = 5.024 \times 10^{-2} Nm$

so the distance is

$r = \frac{k q_1 q_2}{E_p} = 1.832\ m$

6 0

3 years ago