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ycow [4]
3 years ago
5

Practice questions, will mark brainliest!

Physics
1 answer:
andrew-mc [135]3 years ago
4 0

Answer:

266.67Watts

Explanation:

Time = 2.5hr to seconds

3600s = 1hr

2.5hrs = 3600×2.5= 9000s

Force = 32N

Distance = 75km  to m

1000m = 1km

75km = 1000×75 = 75000m

Power = workdone / time

Work = force × distance

Therefore work = 32N × 75000m

Work = 2400000Nm

Power = work ➗ time

Power = 2400000Nm ➗ 9000s

Power = 266.67Watts

Watts is the S. i unit of power

I hope this was helpful, please mark as brainliest

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Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
DiKsa [7]

Answer:

a.\rm -1.49\ m/s^2.

b. \rm 50.49\ m.

Explanation:

<u>Given:</u>

  • Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).

At time t = 3 seconds,

\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.

For the time interval of 2 seconds,

\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.

It is the displacement of the particle in 2 seconds.

7 0
3 years ago
One normal afternoon in undisclosed City X, the chocolate factory workers overload the Easter egg machine. A group of angsty tee
musickatia [10]

Answer:

The correct option is;

C. 1,715 m

Explanation:

We are given the information from the group of teen at the City edge

Time of arrival of explosion sound = 5 s after sighting

Time of sighting explosion = 5 s before hearing the boom

Speed of sound in air ≈ 343 m/s

Speed of light = 299,792 km/s

Therefore, distance covered by sound in 5 seconds is given by the following equation;

Speed = \frac{Distance}{Time}

\therefore 343 \ m/s= \frac{Distance}{5 \, s}

Hence Distance = 343 m/s × 5 s = 1715 m

To check, we compare the time it would take for the light to cover 1715 m

That is Time = \frac{Distance}{Speed} =  \frac{1715}{299,792,000} = 0.00000572 \, s which is instantaneous hence the distance can be approximated by the time duration for the speed of sound.

Therefore, the distance of the students from the factory is approximately 1,715 m

8 0
3 years ago
Energy from solar radiation may be ________ or taken in by a surface or an object.
tester [92]

Answer:

Absorbed

Explanation:

I hope this helps you

7 0
3 years ago
Hi gir_ls join nkd-mbja-nuj​
GREYUIT [131]

Answer:

never lol

studying is your work

but why all are doing I don't know=_=

3 0
3 years ago
The average force of a baseball is 18.9 N . It’s mass is 0.145kg fine the acceleration in m/s^2
Artyom0805 [142]

Answer:

the acceleration is 130.3m/s²

Explanation:

Given data

Force F= 18.9N

Mass of ball m= 0.145kg

Acceleration a=?

Applying the Newton's second law of motion

"The rate of change of momentum of a body is proportional to the external force".

F=ma

a= F/m

a= 18.9/0.142

a= 130.3m/s²

3 0
3 years ago
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