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2 years ago

Susan drops her camera in the river from a bridge that 250 feet high. How long does it take the camera to fall 250 feet

Physics

2 answers:

fgiga [73]2 years ago

8 0

4 seconds by using the quadratic formula

Ksenya-84 [330]2 years ago

6 0

Answer:

It takes 4 seconds for the camera to fall 250 feet

Explanation:

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1. Determine the magnitude of two equal but opposite charges if they attract one another with a force of 0.7N when at distance o

adell [148]

Answer:

q = 2.65 10⁻⁶ C

Explanation:

For this exercise we use Coulomb's law

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

In this case they indicate that the load is of equal magnitude

q₁ = q₂ = q

the force is attractive because the signs of the charges are opposite

F = $k \ \frac{q^2}{r^2}$

q = $\sqrt{\frac{F \ r^2}{k} }$

we calculate

q = $\sqrt{\frac{0.7 \ 0.3^2 }{9 \ 10^9} }$

q = $\sqrt{7 \ 10^{-12} }$ Ra 7 10-12

q = 2.65 10⁻⁶ C

7 0

2 years ago

1. A listener stands 20.0 m from a speaker that pumps out music with a power output of 100.0 W.

marta [7]

(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².

(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².

(1.c) The relative intensity of the sound as heard by the listener is 103 dB.

(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.

(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.

<h3>Surface area being vibrated</h3>

The surface area being vibrated by the time the sound reaches the listener is calculated as follows;

A = 4πr²

A = 4π x (20)²

A = 5,026.55 m²

<h3>Intensity of the sound</h3>

The intensity of the sound is calculated as follows;

I = P/A

I = (100) / (5,026.55)

I = 0.02 W/m²

<h3>Relative intensity of the sound</h3>

$B = 10log(\frac{I}{I_0} )\\\\B = 10 \times log(\frac{0.02}{10^{-12}} )\\\\B = 103 \ dB$

<h3>Speed of sound at the given temperature</h3>

$v= 331.3\sqrt{1 + \frac{T}{273} } \\\\v = 331.3\sqrt{1 + \frac{15}{273} } \\\\v = 340.3 \ m/s$

<h3>Frequency of the sound</h3>

The frequency of the sound heard is determined by applying Doppler effect.

$f_o = f_s(\frac{v \pm v_0}{v \pm v_s} )$

where;

-v₀ is velocity of the observer moving away from the source
-vs is the velocity of the source moving towards the observer
fs is the source frequency
fo is the observed frequency
v is speed of sound

$f_0 = f_s(\frac{v-v_0}{v- v_s} )$

$f_0 = 512(\frac{340.3 - 10}{340.3 - 65} )\\\\f_0 = 614.3 \ Hz$

Learn more about intensity of sound here: brainly.com/question/17062836

3 0

1 year ago

A light-rail commuter train accelerates at a rate of 1.35 m/s. D A 33% Part (a) How long does it take to reach its top speed of

Dennis_Churaev [7]

Answer:

a) 17.49 seconds

b) 13.12 seconds

c) 2.99 m/s²

Explanation:

a) Acceleration = a = 1.35 m/s²

Final velocity = v = 85 km/h = $85\frac{1000}{3600}=23.61\ m/s$

Initial velocity = u = 0

Equation of motion

$v=u+at\\\Rightarrow 23.61=0+1.35t\\\Rightarrow t=\frac{23.61}{1.35}=17.49\ s$

Time taken to accelerate to top speed is 17.49 seconds.

b) Acceleration = a = -1.8 m/s²

Initial velocity = u = 23.61\ m/s

Final velocity = v = 0

$v=u+at\\\Rightarrow 0=23.61-1.8t\\\Rightarrow t=\frac{23.61}{1.8}=13.12\ s$

Time taken to stop the train from top speed is 13.12 seconds

c) Initial velocity = u = 23.61 m/s

Time taken = t = 7.9 s

Final velocity = v = 0

$v=u+at\\\Rightarrow 0=23.61+a7.9\\\Rightarrow a=\frac{-23.61}{7.9}=-2.99\ m/s^2$

Emergency acceleration is 2.99 m/s² (magnitude)

6 0

2 years ago

A 6 kg tennis ball moves at a velocity of 14 m/s. The ball is struck by a racket, causing it to rebound in the opposite directio

Sergeeva-Olga [200]

Answer and method on photo

6 0

2 years ago

QUESTION 7

ryzh [129]

Answer:

<em>The force required is 3,104 N</em>

Explanation:

<u>Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = ma

Where a is the acceleration of the object.

On the other hand, the equations of the Kinematics describe the motion of the object by the equation:

$v_f=v_o+at$

Where:

vf is the final speed

vo is the initial speed

a is the acceleration

t is the time

Solving for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

We are given the initial speed as vo=20.4 m/s, the final speed as vf=0 (at rest), and the time taken to stop the car as t=7.4 s. The acceleration is:

$\displaystyle a=\frac{0-20.4}{7.4}$

$a=-2.757\ m/s^2$

The acceleration is negative because the car is braking (losing speed). Now compute the force exerted on the car of mass m=1,126 kg:

$F = 1,126\ kg * 2.757\ m/s^2$

F= 3,104 N

The force required is 3,104 N

6 0

2 years ago