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Basile [38]
3 years ago
5

Serving at a speed of 164 km/h, a tennis player hits the ball at a height of 2.23 m and an angle θ below the horizontal. The ser

vice line is 11.6 m from the net, which is 0.99 m high. What is the angle θ in degrees such that the ball just crosses the net? Give a positive value for the angle.

Physics
1 answer:
sweet-ann [11.9K]3 years ago
7 0

Answer:

The angle θ is 6.1° below the horizontal.

Explanation:

Please, see the figure for a description of the situation.

The vector "r" gives the position of the ball and can be expressed as the sum of the vectors rx + ry (see figure).

We know the magnitude of these vectors:

magnitude rx = 11.6 m

magnitude ry = 2.23 m - 0.99 m = 1.24 m

Then:

rx = (11. 6 m, 0)

ry = (0, -1.24 m)

r = (11.6 m + 0 m, 0 m - 1.24 m) = (11.6 m, -1.24 m)

Using trigonometry of right triangles:

magnitude rx = r * cos θ = 11. 6 m

magnitude ry = r * sin θ = 2.23 m - 0.99 = 1.24 m

where r is the magnitude of the vector r

magnitude of vector r:

r = \sqrt{(11.6m)^{2} + (1.24m)^{2}} = 11.667m

Then:

cos θ = 11.6 m / 11.667 m

θ = 6.1°

Using ry, we should obtain the same value of θ:

sin θ = 1.24 m/ 11.667 m

θ = 6.1°

( the exact value is obtained if we do not round the module of r)

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Blizzard [7]

Answer:

The induced current and the power dissipated through the resistor are 0.5 mA and 7.5\times10^{-7}\ Watt.

Explanation:

Given that,

Distance = 1.0 m

Resistance = 3.0 Ω

Speed = 35 m/s

Angle = 53°

Magnetic field B=5.0\times10^{-5}\ T

(a). We need to calculate the induced emf

Using formula of emf

E = Blv\sin\theta

Where, B = magnetic field

l = length

v = velocity

Put the value into the formula

E=5.0\times10^{-5}\times1.0\times35\sin53^{\circ}

E=1.398\times10^{-3}\ V

We need to calculate the induced current

E =IR

I=\dfrac{E}{R}

Put the value into the formula

I=\dfrac{1.398\times10^{-3}}{3.0}

I=0.5\ mA

(b). We need to calculate the power dissipated through the resistor

Using formula of power

P=I^2 R

Put the value into the formula

P=(0.5\times10^{-3})^2\times3.0

P=7.5\times10^{-7}\ Watt

Hence, The induced current and the power dissipated through the resistor are 0.5 mA and 7.5\times10^{-7}\ Watt.

6 0
3 years ago
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The best way to cool soft and thick foods (such as beans, sauce or chili) when using the refrigerator is?
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Answer:

9.8 m/s2

Explanation:

In the first equation above, g is referred to as the acceleration of gravity. Its value is 9.8 m/s2 on Earth. That is to say, the acceleration of gravity on the surface of the earth at sea level is 9.8 m/s2.

Got it from the internet, hope it helps though ^^

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What happens if :<br> . The test charge is not tiny.
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The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.

<h3>How does test charge affect electric field?</h3>

As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.

Adjusting the amount of charge on the test charge will not change the electric field force.

<h3>What is a test charge used for?</h3>

The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.

To learn more about test charge, refer

brainly.com/question/16737526

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3 0
2 years ago
A car with a mass of 1600 kg is towing a trailer with a mass of 420 kg. The car
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Answer:

1.63366

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