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Sever21 [200]
2 years ago
10

Find the average power Pavg created by the force F in terms of the average speed vavg of the sled.

Physics
1 answer:
Katena32 [7]2 years ago
3 0

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The power created  is  P_{avg} =  F  *  v_{avg}

Explanation:

From the question we are told that

    The that the average power is  mathematically represented as

            P_{avg} =  \frac{W }{\Delta  t  }

Where W is  is the Workdone which is  mathematically represented as

         W =  F  *  s

      Where F is  the applies force and  s  is the displacement  due to the force  

        So  

                P_{avg} =  \frac{F *s  }{\Delta  t  }

Now this  displacement can be represented mathematically as  

            s =  v_{avg} *  \Delta  t

Where v_{avg } is the average  velocity and \Delta  t is the time  taken  

So  

            P_{avg} =  \frac{F *v_{avg} *  \Delta  t    }{\Delta  t  }

=>         P_{avg} =  F  *  v_{avg}

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Leno4ka [110]

Answer: a=9.8*10^-10s

b=9.8*10^-13s

c=1.7*10^-8s

d=5.57*10^-4s

Explanation:

a) given 980 ps

Expected answer is 980 * 10^-12

Therefore, 980ps = 9.8*10^-10s

b) given 980 fs

Expected answer is 980 * 10^-15

Therefore, 980fs = 9.8*10^-13s

c) given 17 ns

Expected answer is 17 * 10^-9

Therefore, 17ns = 1.7*10^-8s

d) given 577 μs

Expected answer is 577 * 10^-6

Therefore, 577μs = 5.57*10^-4s

a=9.8*10^-10s

b=9.8*10^-13s

c=1.7*10^-8s

d=5.57*10^-4s

8 0
2 years ago
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A bee wants to fly to a flower located due North of the hive on a windy day. The wind blows from East to West at speed 6.68 m/s.
Aleonysh [2.5K]

Answer:  53.31\° East of North

Explanation:

We have the following data:

Speed of the wind from East to West: 6.68 m/s

Speed of the bee relative to the air:  8.33 m/s

If we graph these speeds (which in fact are velocities because are vectors) in a vector diagram, we will have a right triangle in which the airspeed of the bee (its speed relative to te air) is the hypotense and the two sides of the triangle will be the <u>Speed of the wind from East to West</u> (in the horintal part) and the <u>speed due North relative to the ground</u> (in the vertical part).

Now, we need to find the direction the bee should fly directly to the flower (due North):

sin \theta=\frac{Windspeed-from-East-to-West}{Speed-bee-relative-to-air}

sin \theta=\frac{6.68 m/s}{8.33 m/s}

Clearing \theta:

\theta=sin^{-1} (\frac{6.68 m/s}{8.33 m/s})

\theta=53.31\°

6 0
2 years ago
4. A cylindrical tube has a length of 14.4cm and a radius of 1.5cm and is filled with a colorless gas. If the density of the gas
professor190 [17]

Answer:

Mass, m of gas is 0.2504 grams.

Explanation:

First, we need to solve for the volume of the cylindrical tube.

Volume of cylinder is given by the formula;

V = 2\Pi r^{2}h

Where, V represents volume.

π represents pie

r represents radius.

h represents height or length.

Given the following data;

Radius, r = 1.5cm

Length, h = 14.4cm

Density, d = 0.00123g/cm³

Substituting into the equation;

V = 2 * 3.142 * (1.5)^{2}14.4

V = 2 * 3.142 * 2.25 * 14.4

V = 203. 6016

Therefore, the volume of the cylindrical tube is 203. 6016cm³

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the formula;

Density = \frac{mass}{volume}

Mass = density  *  volume

Substituting into the equation, we have;

Mass = 0.00123 * 203. 6016

Mass = 0.2504g.

5 0
2 years ago
The weight of Earth's atmosphere exerts an average pressure of 1.01 ✕ 105 Pa on the ground at sea level. Use the definition of p
zloy xaker [14]

Answer:

The weight of Earth's atmosphere exert is 516.6\times10^{17}\ N

Explanation:

Given that,

Average pressure P=1.01\times10^{5}\ Pa

Radius of earth R_{E}=6.38\times10^{6}\ m

Pressure :

Pressure is equal to the force upon area.

We need to calculate the weight of earth's atmosphere

Using formula of pressure

P=\dfrac{F}{A}  

F=PA

F=P\times 4\pi\times R_{E}^2

Where, P = pressure

A = area

Put the value into the formula

F=1.01\times10^{5}\times4\times\pi\times(6.38\times10^{6})^2

F=516.6\times10^{17}\ N

Hence, The weight of Earth's atmosphere exert is 516.6\times10^{17}\ N

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2 years ago
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A simple pendulum 0.64m long has a period of 1.2seconds. Calculate the period of a similar pendulum 0.36m long in the same locat
weqwewe [10]

The period of the second pendulum is 0.9 s

Explanation:

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T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration of gravity at the location of the pendulum

For the first pendulum, we have

L = 0.64 m

T = 1.2 s

Therefore we can find the value of g at that location:

g=(\frac{2\pi}{T})^2 L=(\frac{2\pi}{1.2})^2 (0.64)=17.5 m/s^2

Now we can find the period of the second pendulum at the same location, which is given by

T=2\pi \sqrt{\frac{L}{g}}

where we have

L = 0.36 m (length of the  second pendulum)

g=17.5 m/s^2

Substituting,

T=2\pi \sqrt{\frac{0.36}{17.5}}=0.9 s

#LearnwithBrainly

8 0
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