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muminat
3 years ago
5

PLEASE HELPPPPPPPPPP

Physics
2 answers:
tekilochka [14]3 years ago
7 0

Answer:

sorry me computer is blocking the picture but i will try to unblock so i can answer

Explanation:

gayaneshka [121]3 years ago
7 0

Answer:

It looks like it’s the correct answer.

Explanation:

I hope this helps (:

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Can someone help me with these questions please. I will mark brainliest
VladimirAG [237]

Answer:

3: I can´t see the text/image, but it depend on the mass and the force applied to the ball, if both are too high, it will be harder to make a home run. (Second law)

4:It would be easier to make a home run because there is no interruption between the ball and the space the same travels. (Third law)

Explanation:

3 0
3 years ago
The continuous submarine mountain range which winds through all the oceans is called the mid-ocean
Artist 52 [7]
The continuous submarine mountain range which winds through all the oceans is called the mid-ocean <u>ridge.
</u>It is a form of a mountain which is found underwater, and it appeared there due to the movements of tectonic plates. It is responsible for the creation of new seafloor, meaning that the ground underwater changes constantly with the formation of these ridges. <u>
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3 years ago
Which statement relates most directly to the second law?
alukav5142 [94]
Statement three i do believe

7 0
3 years ago
Read 2 more answers
g power output of 87 W. At what distance will the decibel reading be 120 dB, which is noise level of a loud indoor rock concert
Kazeer [188]

Given that,

Output power = 87 W

Decibel reading = 120 dB

We need to calculate the intensity of sound

Using formula of intensity of sound

dB=10\log(\dfrac{I}{I_{0}})

Put the value into the formula

120=10\log(\dfrac{I}{1\times10^{-12}})

12=log(\dfrac{I}{1\times10^{-12}})

10^{12}=\dfrac{I}{1\times10^{-12}}

I=10^{12}\times1\times10^{-12}

I=1\ W/m^2

We need to calculate the distance

Using relation of power out[ut and intensity

I=\dfrac{P}{4\pi r^2}

Put the value into the formula

1=\dfrac{87}{4\pi r^2}

r=\sqrt{\dfrac{87}{4\pi}}

r=2.63\ m

Hence, The distance is 2.63 m

7 0
3 years ago
A car’s brakes decelerate it at a rate of -1.70 m/s2. What is the time, in seconds, required to slow the car from 15 m/s to 9.0
jeka94

Answer:

t = 3.52 s

Explanation:

Given that,

The deceleration of a car is, a = -1.7 m/s²

The initial velocity of the car, u = 15 m/s

Final velocity of the car, v = 9 m/s

We need to find the time that is required to slow the car from 15 m/s to 9 m/s. The definition of acceleration is :

a=\dfrac{v-u}{t}\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{9-15}{-1.7}\\\\t=3.52\ s

So, it will take 3.52 seconds to slow the car from 15 m/s to 9 m/s.

3 0
2 years ago
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