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Marizza181 [45]
2 years ago
13

A shopper in a supermarket pushes a cart with a force of 35Ndirected at an angle of 25 degree below the horizontal. The forceis

just sufficient to overcome various frictional forces, so thecart moves at constant speed.
(A) Find the work done by the shopper as she moves down a 50.0mlength aisle.
(B) What is the net work done on the cart? why?
(C) The shopper goes down the next aisle, pushing horizontally andmaintaining the same speed as before. If the work don't byfrictional force doesn't chance, would the shopper's applied forcebe larger, smaller, or the same? What about the work done on thecart by the shopper?
Physics
1 answer:
Delicious77 [7]2 years ago
8 0
C the shipped hose down the next aisku pushing horizontally andmanitaining the sama speed as before if the work dont byfrictional Dorce dosent chance would the shopper appplied Dorce be larger smaller or the same what about the work dont on thecart BH the shopper
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A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in
Leno4ka [110]

Answer:

The  electric field  is 35\cos(90\pi t)\ mV/m

Explanation:

Given that,

Radius = 2.00 cm

Number of turns per unit length n= 1.65\times10^{3}

Current I = 6.00\sin 90\pi t

We need to calculate the induced emf

\epsilon =\mu_{0}nA\dfrac{dI}{dt}

Where, n = number of turns per unit length

A = area of cross section

\dfrac{dI}{dt}=rate of current

Formula of electric field is defined as,

E=\dfrac{\epsilon}{2\pi r}

Where, r = radius

Put the value of emf in equation (I)

E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}....(II)

We need to calculate the rate of current

I=6.00\sin 90\pi t....(III)

On differentiating equation (III)

\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)

Now, put the value of rate of current in equation (II)

E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}

E=35\cos(90\pi t)\ mV/m

Hence, The  electric field  is 35\cos(90\pi t)\ mV/m

7 0
2 years ago
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An echo is heard from a cliff 3.71 s after a rifle is fired. How many feet away is the cliff
LenKa [72]

The cliff is 2042 ft away.

We know that the speed of sound in air is directly proportional to the absolute temperature.

First convert the Fahrenheit temperature to Celsius;

 °C = 5/9(44.5 - 32)

°C = 6.9 °C

Applying the formula;

V1/V2 = √T1/T2

Where; V1 = velocity of sound in air at  0°C

V2 = Velocity of sound in air at 6.9 °C

1087/V2 = √273/279.9

V2= 1101 ft/s

Given that; V = 2s/t

Where s is the distance of the cliff

t is the time taken

1101 ft/s = 2s/3.71 s

s = 1101 ft/s × 3.71 s/2

s = 2042 ft

Learn more:brainly.com/question/15381147

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2 years ago
The contact force that acts on objects in a liquid or gas and allows objects to float is called
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A 58 kg skier is going down a 35 degree slope. The areaof each
maxonik [38]

To solve this problem we will use a free body diagram that allows us to determine the Normal Force.

In general, the normal force would be equivalent to

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Since the skier is standing on two skis, his weight will be divide by two

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Our values are given as,

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A = 0.3m^2

Replacing we have that

P = \frac{\frac{(58)(9.8)cos(35)}{2}}{0.3}

P = 776.01Pa

Therefore the pressure exerted by each ski on the snow is 776.01Pa

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Which vector best represents the force that could act concurrently with force A to produce force B
viva [34]
Chose answer 2

cheers

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