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2 years ago

A light ray strikes a plane mirror at an angle of 23° to the normal. What is the angle between the reflected ray and the mirror?

Physics

1 answer:

tiny-mole [99]2 years ago

4 0

Answer: $67^{\circ}$

Explanation:

Given

Ray of light incident on plane making an angle of $23^{\circ}$ with normal

Thus incident angle $i=23^{\circ}$

So according to the law of incidence is equal to angle of reflection

$\angle i=\angle r$

thus $\angle r=23^{\circ}$

and angle between mirror and reflected ray $=90^{\circ}-23^{\circ}$

$=67^{\circ}$

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which has a greater kinetic energy, a bowling ball that has a mass of 5kg travelling at 6m/s, or a ship that has a mass of 12000

AVprozaik [17]

Answer:bowling ball has greater kinetic energy

Explanation:

Kinetic energy of bowling ball:

mass=m=5kg

Velocity=v=6m/s

Kinetic energy =ke

Ke=0.5 x m x v x v

Ke=0.5 x 5 x 6 x 6

Ke=90J

Kinetic energy of ship:

mass=m=120000kg

velocity=v=0.02m/s

Ke=0.5 x m x v x v

Ke=0.5 x 120000 x 0.02 x 0.02

Ke=24J

6 0

2 years ago

A vertical spring of constant k = 400 N/m hangs at rest. When a 2 kg mass is attached to it, and it is released, the spring exte

Viefleur [7K]

Answer:

4.9 cm

Explanation:

From Hook's Law,

F = ke......................... Equation 1

Where F= force, e = extension, k = spring constant.

Note: the Force acting on the the spring is the weight of the mass.

W = mg.

F = mg.................... Equation 2

Where m = mass, g = acceleration due to gravity

Substitute equation 2 into equation 1

mg = ke

make e the subject of the equation

e = mg/k............... Equation 3.

Given: m = 2 kg, g = 9.8 m/s², k = 400 N/m

e = (2×9.8)/400

e = 19.6/400

e = 0.049 m

e = 4.9 cm

3 0

2 years ago

B. A hydraulic jack has a ram of 20 cm diameter and a plunger of 3 cm diameter. It is used for lifting a weight of 3 tons. Find

lozanna [386]

Answer:

option (b)

Explanation:

According to the Pascal's law

F / A = f / a

Where, F is the force on ram, A be the area of ram, f be the force on plunger and a be the area of plunger.

Diameter of ram, D = 20 cm, R = 20 / 2 = 10 cm

A = π R^2 = π x 100 cm^2

F = 3 tons = 3000 kgf

diameter of plunger, d = 3 cm, r = 1.5 cm

a = π x 2.25 cm^2

Use Pascal's law

3000 / π x 100 = f / π x 2.25

f = 67.5 Kgf

4 0

2 years ago

1. Two identical bowling balls of mass M and radius R roll side by side at speed v0 along a flat surface. Ball 1 encounters a ra

UNO [17]

Answer:

1/2

Explanation:

We need to make a couple of considerations but basically the problem is solved through the conservation of energy.

I attached a diagram for the two surfaces and begin to make the necessary considerations.

Rough Surface,

We know that force is equal to,

$F_r = mgsin\theta$

$F_r = \mu N$

$F_r = \mu mg cos\theta$

Matching the two equation we have,

$\mu N = \mu mg cos\theta$

$\mu = tan\theta$

Applying energy conservation,

$\frac{1}{2}mv^2_0+\frac{1}{2}I_w^2 = F_r*d+mgh_1$

$\frac{1}{2}mv^2_0+\frac{2}{5}mR^2\frac{V_0^2}{R^2} = F_r*d+mgh_1$

$\frac{1}{2}mv^2_0+\frac{mv_0^2}{5} = mgsin\theta \frac{h_1sin\theta}+mgh_1$

$\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_1+gh_1$

$h_1 = \frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})$

Frictionless surface

$\frac{1}{2}mv_0^2+\frac{1}{2}I\omega^2 = mgh_2$

$\frac{1}{2}m_v^2+\frac{1}{2}\frac{2}{5}mR^2\frac{v_0^2}{R^2} =mgh_2$

$\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_2$

$h_2 = \frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})$

Given the description we apply energy conservation taking into account the inertia of a sphere. Then the relation between $h_1$ and $h_2$ is given by

$\frac{h_1}{h_2} = \frac{\frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})}{\frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})}$

$\frac{h_1}{h_2} = \frac{1}{2}$

8 0

2 years ago

Norma kicks a soccer ball with an initial velocity of 10.0 meters per second at an angle of 30.0°. If the ball moves through the

enyata [817]

<u>Answer</u>

27.7

<u>Explanation</u>

The ball was hit at an angle of 30°, with the horizontal at a speed of 10 m/s. We have to find the horizontal component of speed.

cosx = adjacent/hypotenuse

cos 30 = adjacent / 10

adjacent = 10 cos30

= 8.66 m/s ⇒ This is the horizontal speed.

Now find the horizontal distance.

Distance = speed × time

= 8.66 × 3.2

= 27.71

Answer to the nearest tenth = 27.7

4 0

2 years ago

Read 2 more answers