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3 years ago

How long does it take light from a camera to travel to the face of someone standing 7 meters away?

Physics

1 answer:

mestny [16]3 years ago

8 0

The time taken for the light to travel from the camera to someone standing 7 m away is 2.33×10¯⁸ s

Speed is simply defined as the distance travelled per unit time. Mathematically, it is expressed as:

<h3>Speed = distance / time </h3>

With the above formula, we can obtain the time taken for the light to travel from the camera to someone standing 7 m away. This can be obtained as follow:

Distance = 7 m

Speed of light = 3×10⁸ m/s

Time = Distance / speed

Time = 7 / 3×10⁸

Therefore, the time taken for the light to travel from the camera to someone standing 7 m away is 2.33×10¯⁸ s

Learn more: brainly.com/question/14988345

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A pitcher throws a 0.144-kg baseball toward the batter so that it crosses home plate horizontally and has a speed of 42 m/s just

Solnce55 [7]

(a) 12.8 kg m/s

The impulse delivered by the bat on the baseball is equal to the change in momentum of the baseball:

$I=\Delta p = m(v-u)$

where we have

m = 0.144 kg is the mass of the ball

v = -47 m/s is the final velocity of the ball

u = 42 m/s is the initial velocity of the ball

Substituting into the equation, we find

$I=(0.144 kg)(-47 m/s-(42 m/s))=-12.8 kg m/s$

And since we are interested in the magnitude only,

$I=12.8 kg m/s$

(b) 2.78 kN

The impulse exerted on the ball is also equal to the product between the average force and the contact time:

$I=F\Delta t$

where

F is the average force exerted on the ball

$\Delta t=0.0046 s$ is the contact time

Solving the formula for F, we find

$F=\frac{I}{\Delta t}=\frac{12.8 kg m/s}{0.0046 s}=2783 N = 2.78 kN$

(c) The force exerted on the ball is much larger (1988 times more) than the weigth of the ball

The weight of the ball is given by

$W=mg$

where

m = 0.144 kg is the mass of the ball

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

Solving the equation for W, we find

$W=(0.144 kg)(9.8 m/s^2)=1.4 N$

So as we see, the force exerted on the ball (2783 N) is almost 2000 times larger than the weight of the ball (1.4 N):

$\frac{F}{W}=\frac{2783 N}{1.4 N}=1988$

8 0

3 years ago

What unit of electric current,the ampere,is equivalent to

Paladinen [302]

1 ampere is 1 coulomb of charge moving through a cross section in one second.

3 0

3 years ago

1.00kg of ice at -24 degrees Celsius is placed in contact with a 1.00kg block of a metal at 5.00 degrees Celsius. They come to e

Alona [7]

Answer:

C₂ = 2.22 KJ/kg °C

Explanation:

Since, both objects are in thermal contact. Therefore, the law of conservation of energy tells us that:

$Heat\ Lost\ by\ Metal\ Block = Heat\ Gained\ by\ Ice\\m_{1}C_{1} \Delta T_{1} = m_{2}C_{2} \Delta T_{2}$

where,

m₁ = mass of ice = 1 kg

C₁ = specific heat of ice = 2.04 KJ/kg.°C

ΔT₁ = Change in Temperature of Ice = -8.88°C - (-24°C) = 15.12°C

m₂ = mass of metal block = 1 kg

C₂ = specific heat of metal = ?

ΔT₂ = Change in Temperature of Metal Block = 5°C - (-8.88°C) = 13.88°C

Therefore, using these values in the equation, we get:

$(1\ kg)(2.04\ KJ/kg.^0C)(15.12\ ^0C) = (1\ kg)C_{2}(13.88\ ^0C) \\C_{2} = \frac{30.84\ KJ}{13.88\ kg/^0C}$

4 0

3 years ago

Read 2 more answers

A piano tuner stretches a steel piano wire with a tension of 1070 N . The wire is 0.400 m long and has a mass of 4.00 g . A. Wha

pychu [463]

A. 409 Hz

The fundamental frequency of a string is given by:

$f_1=\frac{1}{2L}\sqrt{\frac{T}{m/L}}$

where

L is the length of the wire

T is the tension in the wire

m is the mass of the wire

For the piano wire in this problem,

L = 0.400 m

T = 1070 N

m = 4.00 g = 0.004 kg

So the fundamental frequency is

$f_1=\frac{1}{2(0.400)}\sqrt{\frac{1070}{(0.004)/(0.400)}}=409 Hz$

B. 24

For this part, we need to analyze the different harmonics of the piano wire. The nth-harmonic of a string is given by

$f_n = nf_1$

where $f_1$ is the fundamental frequency.

Here in this case

$f_1 = 409 Hz$

A person is capable to hear frequencies up to

$f = 1.00 \cdot 10^4 Hz$

So the highest harmonics that can be heard by a human can be found as follows:

$f=nf_1\\n= \frac{f}{f_1}=\frac{1.00\cdot 10^4}{409}=24.5 \sim 24$

8 0

4 years ago

*PLEASE HELP*

Westkost [7]

Answer:

640.5

Explanation: i got it right on acellus

5 0

3 years ago

Read 2 more answers