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Gemiola [76]
3 years ago
8

A 15.0 cm object is 12.0 cm from a convex mirror that has a focal length of -6.0 cm. What is the height of the image produced by

the mirror? A. –5.0 cm B. 7.5 cm C. -7.5 cm D. 5.0 cm
Physics
1 answer:
Vlad [161]3 years ago
6 0

Answer: D. 5cm

Explanation:

Given the following :

Focal length (f) = - 6.0 cm

Height of object = 15.0cm

Distance of object from mirror (u) = 12.0cm

Height of image produced by the mirror =?

Firstly, we calculate the distance of the image from the mirror.

Using the mirror formula

1/f = 1/u + 1/v

1/v = 1/f - 1/u

1/v = 1/-6 - 1/12

1/v = - 1/6 - 1/12

1/v = (- 2 - 1) / 12

1/v = - 3 / 12

v = 12 / - 3

v = - 4

Using the relation :

(Image height / object height) = (- image distance / object distance)

Image height / 15 = - (-4) / 12

Image height / 15 = 4 / 12

Image height = (15 × 4) / 12

Image height = 60 / 12

Image height = 5cm

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concave <span>ray diagrams were constructed in order to determine the general location, size, orientation, and type of image formed by concave mirrors. Perhaps you noticed that there is a definite relationship between the image characteristics and the location where an object placed in front of a concave mirror. but, convex</span><span>ray diagrams were constructed in order to determine the location, size, orientation, and type of image formed by concave mirrors. The ray diagram constructed earlier for a convex mirror revealed that the image of the object was virtual, upright, reduced in size and located behind the mirror. </span>
3 0
3 years ago
A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calcula
Luda [366]

Answer:

a) 0.1832 A

b) 11.91 Volts

c) 2.18 Watt , 0.0168 Watt

Explanation:

(a)

R = external resistor connected to the terminals of the battery = 65 Ω

E = Emf of the battery = 12.0 Volts

r = internal resistance of the battery = 0.5 Ω

i = current flowing in the circuit

Using ohm's law

E = i (R + r)

12 = i (65 + 0.5)

i = 0.1832 A

(b)

Terminal voltage is given as

V_{ab} = i R

V_{ab} = (0.1832) (65)

V_{ab} = 11.91 Volts

(c)

Power dissipated in the resister R is given as

P_{R} = i²R

P_{R} = (0.1832)²(65)

P_{R} = 2.18 Watt

Power dissipated in the internal resistance is given as

P_{r} = i²r

P_{r} = (0.1832)²(0.5)

P_{r} = 0.0168 Watt

5 0
3 years ago
A pendulum has 895 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o
LuckyWell [14K]

Newton's law of conservation states that energy of an isolated system  remains a constant. It can neither be created nor destroyed but can be transformed  from one form to the other.

Implying the above law of conservation of energy in the case of pendulum we can conclude that at the bottom of the swing the entire potential energy gets converted to kinetic energy. Also the potential energy is zero at this point.

Mathematically also potential energy is represented as

Potential energy= mgh

Where m is the mass of the pendulum.

g is the acceleration due to gravity

h is the height from the bottom z the ground.

At the bottom of the swing,the height is zero, hence the potential energy is also zero.

The kinetic energy is represented mathematically as

Kinetic energy= 1/2 mv^2

Where m is the mass of the pendulum

v is the velocity of the pendulum

At the bottom the pendulum has the maximum velocity. Hence the kinetic energy is maximum at the bottom.

Also as it has been mentioned energy can neither be created nor destroyed hence the entire potential energy is converted to kinetic energy at the bottom and would be equivalent to 895 J.

7 0
3 years ago
A ball of mass 0.160 kg is dropped from a height of 2.25 m. When it hits the ground it compresses 0.087 m.
Studentka2010 [4]

A) 6.64 m/s downward

B) 0.026 s

C) -40.9 N

Explanation:

A)

We can solve this problem by using the law of conservation of energy.

In fact, since the total mechanical energy of the ball must be conserved, this means that the initial gravitational potential energy of the ball before the fall is entirely converted into kinetic energy just before it reaches the floor.

So we can write:

PE=KE\\mgh = \frac{1}{2}mv^2

where

m = 0.160 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration due to gravity

h = 2.25 m is the initial height of the ball

v is the final velocity of the ball before hitting the ground

Solving for v, we find:

v=\sqrt{2gh}=\sqrt{2(9.8)(2.25)}=6.64 m/s

And the direction of the velocity is downward.

B)

The motion of the ballduring the collision is a uniformly accelerated motion (= with constant acceleration), so the time of impact can be found by using a suvat equation:

s=(\frac{u+v}{2})t

where:

v is the final velocity

u is the initial velocity

s is the displacement of the ball during the impact

t is the time

Here we have:

u = 6.64 m/s is the velocity of the ball before the impact

v = 0 m/s is the final velocity after the impact (assuming it comes to a stop)

s = 0.087 m is the displacement, as the ball compresses by 0.087 m

Therefore, the time of the impact is:

t=\frac{2s}{u+v}=\frac{2(0.087)}{0+6.64}=0.026 s

C)

The force exerted by the floor on the ball can be found using the equation:

F=\frac{\Delta p}{t}

where

\Delta p is the change in momentum of the ball

t is the time of the impact

The change in momentum can be written as

\Delta p = m(v-u)

So the equation can be rewritten as

F=\frac{m(v-u)}{t}

Here we have:

m = 0.160 kg is the mass of the ball

v = 0 is the final velocity

u = 6.64 m/s is the initial velocity

t = 0.026 s is the time of impact

Substituting, we find the force:

F=\frac{(0.160)(0-6.64)}{0.026}=-40.9 N

And the sign indicates that the direction of the force is opposite to the direction of motion of the ball.

4 0
3 years ago
A solid sphere is released from the top of a ramp that is at a height h1 = 2.30 m. It rolls down the ramp without slipping. The
Oksi-84 [34.3K]

Answer:

The horizontal distance d does the ball travel before landing is 1.72 m.

Explanation:

Given that,

Height of ramp h_{1}=2.30\ m

Height of bottom of ramp h_{2}=1.69\ m

Diameter = 0.17 m

Suppose we need to calculate the horizontal distance d does the ball travel before landing?

We need to calculate the time

Using equation of motion

h_{2}=ut+\dfrac{1}{2}gt^2

t=\sqrt{\dfrac{2h_{2}}{g}}

t=\sqrt{\dfrac{2\times1.69}{9.8}}

t=0.587\ sec

We need to calculate the velocity of the ball

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times(\dfrac{2}{5}mr^2)\times(\dfrac{v}{r})^2

K.E=\dfrac{7}{10}mv^2

Using conservation of energy

K.E=mg(h_{1}-h_{2})

\dfrac{7}{10}mv^2=mg(h_{1}-h_{2})

v^2=\dfrac{10}{7}\times g(h_{1}-h_{2})

Put the value into the formula

v=\sqrt{\dfrac{10\times9.8\times(2.30-1.69)}{7}}

v=2.922\ m/s

We need to calculate the horizontal distance d does the ball travel before landing

Using formula of distance

d =vt

Where. d = distance

t = time

v = velocity

Put the value into the formula

d=2.922\times 0.587

d=1.72\ m

Hence, The horizontal distance d does the ball travel before landing is 1.72 m.

8 0
3 years ago
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