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3 years ago

HELPP WILL MARK BRAINLIEST!!!

Physics

2 answers:

seraphim [82]3 years ago

4 0

Answer: the answer is c

guapka [62]3 years ago

3 0

I guess it’s B cause that maybe is the output

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By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 d

lyudmila [28]

Answer:

d) $7.94\times 10^{9}$

Explanation:

β₁ = sound level of sound at rock concert = 120 dB

β₂ = sound level of sound due to whisper = 21 dB

I₁ = Intensity of sound at rock concert

I₂ = Intensity of sound due to whisper

sound level of sound at rock concert is given as

$\beta _{1} = 10 log\left ( \frac{I_{1}}{10^{-12}} \right )$

$120 = 10 log\left ( \frac{I_{1}}{10^{-12}} \right )$

$12 = log\left ( \frac{I_{1}}{10^{-12}} \right )$ Eq-1

sound level due to whisper is given as

$\beta _{2} = 10 log\left ( \frac{I_{2}}{10^{-12}} \right )$

$21 = 10 log\left ( \frac{I_{2}}{10^{-12}} \right )$

$2.1 = log\left ( \frac{I_{2}}{10^{-12}} \right )$ Eq-2

subtracting Eq-2 from Eq-1

$12 - 2.1 = log\left ( \frac{I_{1}}{10^{-12}} \right ) - log\left ( \frac{I_{2}}{10^{-12}} \right )$

$9.9 = log\left ( \frac{I_{1}}{I_{2}} \right )$

$\left ( \frac{I_{1}}{I_{2}} \right ) = 7.94\times 10^{9}$

6 0

3 years ago

Following a collision between a large spacecraft and an asteroid, a copper disk of radius 28.0 m and thickness 1.20 m, at a temp

coldgirl [10]

Answer:

A. 9.31 x10^10J

B. -8.47x10 ^ 12J

C. 8.38x 10^12J

Explanation:

See attached file pls

3 0

3 years ago

A car initially traveling at 27.7 m/s undergoes a constant negative acceleration of magnitude 2.00 m/s2 after its brakes are app

lys-0071 [83]

The car's velocity at time t is given by

$v=27.7\dfrac{\rm m}{\rm s}+\left(-2.00\dfrac{\rm m}{\mathrm s^2}\right)t$

It comes to a stop when v = 0, which happens when

$0=27.7\dfrac{\rm m}{\rm s}+\left(-2.00\dfrac{\rm m}{\mathrm s^2}\right)t\implies t=13.85\,\mathrm s$

or after about 13.9 s.

In this time, the car travels a distance x given by

$x=\left(27.7\dfrac{\rm m}{\mathrm s}\right)(13.85\,\mathrm s)+\dfrac12\left(-2.00\dfrac{\rm m}{\mathrm s^2}\right)(13.85\,\mathrm s)^2=191.823\,\mathrm m$

or about 192 m.

In one complete revolution, each tire covers a distance equal to its circumference,

$2\pi(0.340\,\mathrm m)\approx2.13628\,\mathrm m$

or about 2.14 m.

This means each tire will complete approximately 192/2.14 ≈ 90 revolutions.

8 0

3 years ago

How long will it take a basketball starting from rest to roll without slipping 2.6 m down an incline that makes an angle of 30.8

Romashka-Z-Leto [24]

Answer:

t = 1.02 s

Explanation:

friction = 0 N (no slipping)

downward force along the inclined plane = F = mgsin(30.8)

F = ma

⇒ a = 9.81 sin(30.8)

a = 5.02 m/s^2

now we have

displacement = s = 2.6 m

acceleration = a = 5.02m/s^2

initial velocity = Vi = 0m/s

time = t = ?

APPLYING THE 2ND EQUATION OF MOTION:

S = (Vi)(t)+0.5at^2

2.6 = 0(t) + 0.5 * 5.02 * t^2

t = 1.02 s

7 0

3 years ago

The moon is closest to earth at ____.

Paraphin [41]

The point of the orbit closest to Earth is called perigee, while the point farthest from Earth is known as apogee

8 0

3 years ago

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