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3 years ago

A ball of mass 12 kg is released from a height of 2.5 m

Physics

1 answer:

VMariaS [17]3 years ago

4 0

Answer:

300 J

Explanation:

From conservation of energy;

Gravitational potential energy at top = Kinetic Energy at bottom

mgh = ½mv²

We are given;

Mass; m = 12 kg.

Height; h = 2.5 m

g = 10 m/s²

Thus;

mgh = 12 × 2.5 × 10 = 300 J

Since mgh = Kinetic Energy at bottom.

Thus; kinetic energy of the ball once it reach the ground = 300 J

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The critical angle for a substance is measured at 53.7 degrees. light enters from air at 45.0 degrees. at what angle it will con

faust18 [17]

When light travels from a medium with greater refractive index $n_1$ to a medium with smaller refractive index $n_2$ , there exists an angle (called critical angle) above which the light is totally reflected, and the value of this angle is given by
$\theta_c = \arcsin ( \frac{n_2}{n_1} )$
In this problem, we know that the critical angle is $\theta_c = 53.7^{\circ}$ , so we can find the ratio between the refractive indices of the two mediums:
$\frac{n_2}{n_1} = \sin \theta_c = \sin 53.7^{\circ} =0.81$
and since the second medium is air (n=1.00), the refractive index of the first medium is
$n_1= \frac{n_2}{0.81}= \frac{1.00}{0.81}=1.23$

In the second part of the problem, we have light entering from air ( $n_i = 1.00$ ) at angle of incidence of $\theta_i = 45.0 ^{\circ}$ , into the second medium with $n_r = 1.23$ . By using Snell's law, we can find the angle of refraction of the light inside the medium:
$n_i \sin \theta_i = n_r \sin \theta_r$
$\sin \theta_r = \frac{n_i}{n_r} \sin \theta_i = \frac{1.00}{1.23} \sin 45^{\circ}=0.574$
$\theta_r = \arcsin(0.574)=35.1^{\circ}$

3 0

3 years ago

A truck travels to and from a stone quarry that is located 2.5 km to the east. What is the total distance traveled by the truck?

scoray [572]

Answer:

5 km

Explanation:

The question says the truck travelled to and fro from the stone quarry.

When going to the stone quarry, the distance was 2.5 km to the east.

On returning, the distance back is also 2.5km

Therefore total distance travelled is 2.5km + 2.5km

= 5km

The displacement 8s dependent on what is the initial and also the final position. The initial and the final position are one and the same thing. Because the truck starts and goes back to the same position, the displacement is zero.

5 0

3 years ago

Two frisky grasshoppers collide in midair at the top of their respective trajectories and grab onto each other, holding tight th

Svetlanka [38]

Answer:

The decrease in Kinetic energy is 0.0107 Joules

Explanation:

Given

Mass of grasshoppers

Let m1 = Mass of grasshopper 1

Let m2 = Mass of grasshopper 2

Let u1 = initial speed of grasshopper 1

Let u2 = initial speed of grasshopper 2

m1 = 250g = 0.25kg

m2 = 130g = 0.13kg

u1 = 15cm/s = 0.15m/s

u2 = 65cm/s = 0.65m/s

First, we calculate the final velocity of the grasshoppers after collision using conservation of momentum.

Using

m1u1 + m2u2 = (m1 + m2) * v

Where v = final velocity

By substituton

0.25 * 0.15 + 0.13 * 0.65 = (0.25 + 0.13) * v

0.0375 + 0.0845 = 380v

0.122 = 0.38v

Make v the subject of formula

v = 0.122/0.38

v = 0.321 m/s

Calculating the Kinetic energies before and after impact.

Before collision;

KE = ½m1u1²+ ½m2u2²

KE = ½(m1u1² + m2u2²)

By substituton;

KE = ½(0.25 * 0.15² + 0.13 * 0.65²)

KE = 0.030275J

After collision:

KE = ½(m1+m2)v²

KE = ½(0.25 + 0.13) * 0.321²

KE = 0.01957779 J

Change in kinetic energy = ∆KE

∆KE = 0.030275J - 0.01957779J

∆KE = 0.01069721J

∆KE = 0.0107 J --- Approximately

Hence the decrease in Kinetic energy is 0.0107 Joules

7 0

3 years ago

How are sound waves different from light (and other electromagnetic, waves?

Blababa [14]

Sound waves travel faster

8 0

3 years ago

When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.6

Bezzdna [24]

Answer:

Distance between the charges, r = 0.8 meters

Explanation:

Given that,

Charge 1, $q_1=+8.4\ \mu C=+8.4\times 10^{-6}\ C$

Charge 2, $q_2=+5.6\ \mu C=+5.6\times 10^{-6}\ C$

Repulsive force between charges, F = 0.66 N

Let r is the distance between charges. The formula for the electrostatic force is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$r=\sqrt{\dfrac{kq_1q_2}{F}}$

$r=\sqrt{\dfrac{9\times 10^9\times 8.4\times 10^{-6}\times 5.6\times 10^{-6}}{0.66}}$

r = 0.8009 meters

r = 0.8 meters

So, the distance between the charges i 0.8 meters. Hence, this is the required solution.

4 0

3 years ago