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Lostsunrise [7]
2 years ago
7

Akito pushes a wheelbarrow with 800 W of power. How much work is required to get the wheelbarrow across the yard in 12 s? Round

your answer to the nearest whole number.
It takes blank J of work to get the wheelbarrow across the yard
Physics
2 answers:
masha68 [24]2 years ago
5 0

Answer:

9600 J

Explanation:

The power is given by the ratio between the work done and the time taken:

P=\frac{W}{t}

where W is the work done and t the time taken.

In this problem, we know the power:

P = 800 W

and the time taken:

t = 12 s

So, we can re-arrange the equation to get the work required:

W=Pt=(800 W)(12 s)=9600 J

12345 [234]2 years ago
4 0
800 x 12 = 9,600
9,600 is your answer
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Does Earth´s magnetic field move?
AlekseyPX

Answer:

Yes it does.

Explanation:

"The North Magnetic Pole moves over time due to magnetic changes in Earth's core. " - Wikipedia.

It does move around as the magnetic north does.

6 0
3 years ago
A 52 kg and a 95 kg skydiver jump from an airplane at an altitude of 4750 m, both falling in the pike position. Assume all value
Scilla [17]

Answer: 52 kg skydiver: 9.09 m/s and 522.55 s

              95 kg skydiver: 12.3 m/s and 386.2 s

Explanation: <u>Drag</u> <u>Force</u> is an opposite force when an object is moving in a fluid.

For skydivers, when falling through the air, the forces acting on it are gravitational and drag forces. At a certain point, drag force equals gravitational force, which is constant on any part of the planet, producing a net force that is zero. Since there is no net force, there is no acceleration and, consequently, velocity is constant. When that happens, the person reached the <u>Terminal</u> <u>Velocity</u>.

Drag Force and Velocity are proportional to the squared speed. So, terminal velocity is given by:

F_{G}=F_{D}

mg=\frac{1}{2}C \rho Av_{T}^{2}

v_{T}=\sqrt{\frac{2mg}{\rho CA} }

where

m is mass in kg

g is acceleration due to gravitational force in m/s²

ρ is density of the fluid in kg/m³

C is drag coefficient

A is area of the object in the fluid in m²

Calculating:

The 52kg skydiver has terminal velocity of:

v_{T}=\sqrt{\frac{2(52)(9.8)}{(1.21)(0.7)(0.14)} }

v_{T}= 9.09

The 95kg skydiver's terminal velocity is

v_{T}=\sqrt{\frac{2(95)(9.8)}{(1.21)(0.7)(0.14)} }

v_{T}= 12.3

The 52 kg and 95kg skydivers' terminal velocity are 9.09m/s and 12.3m/s, respectively.

The time each one will reach the floor will be:

52 kg at 9.09 m/s:

t=\frac{4750}{9.09}

t = 522.5

95 kg at 12.3 m/s:

t=\frac{4750}{12.3}

t = 386.2

The 52 kg and 95kg skydivers' time to reach the floor are 522.5 s and 386.2 s, respectively.

3 0
2 years ago
A lighthouse is located on a small island, 3 km away from the nearest point on a straight shoreline, and its light makes four re
lbvjy [14]

Answer:

The beam of light is moving at the peed of:

\frac{dy}{dt} = \frac{80\pi}{3} km/min

Given:

Distance from the isalnd, d = 3 km

No. of revolutions per minute, n = 4

Solution:

Angular velocity, \omega = \frac{d\theta'}{dt} = 2\pi n = 2\pi \times 4 = 8\pi    (1)

Now, in the right angle in the given fig.:

tan\theta' = \frac{y}{3}

Now, differentiating both the sides w.r.t t:

\frac{dtan\theta'}{dt} = \frac{dy}{3dt}

Applying chain rule:

\frac{dtan\theta'}{d\theta'}.\frac{d\theta'}{dt} = \frac{dy}{3dt}

sec^{2}\theta'\frac{d\theta'}{dt} = \frac{dy}{3dt} = (1 + tan^{2}\theta')\frac{d\theta'}{dt}

Now, using tan\theta = \frac{1}{m} and y = 1 in the above eqn, we get:

(1 + (\frac{1}{3})^{2})\frac{d\theta'}{dt} = \frac{dy}{3dt}

Also, using eqn (1),

8\pi\frac{10}{9})\theta' = \frac{dy}{3dt}

\frac{dy}{dt} = \frac{80\pi}{3}

7 0
2 years ago
Thanks + BRAINLIST <br><br> Please need correct answer asappp
nirvana33 [79]

Answer:

  1. Standing waves can be thought of as a sin wave and a cos wave overlapping each other. They go in different direction hence C is correct
  2. Wave interference can be thought of as the opposite of destructive ---> constructive anda hencd meet and interact on the same medium such that answer B is correct
6 0
2 years ago
Each inner energy level of an atom has a maximum number of it can hold
CaHeK987 [17]
That would be a maximum of 4 atoms 
3 0
3 years ago
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