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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.

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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Does the sun sun traces shortest path across local sky on june solstice

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3 years ago

To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3

iren [92.7K]

Answer:

$372.3 J/^{\circ}C$

Explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

$P=VI=(3.6)(2.6)=9.36 W$

Then, this power is delivered for a time of

t = 350 s

Therefore, the energy supplied is

$E=Pt=(9.36)(350)=3276 J$

Finally, the change in temperature of an object is related to the energy supplied by

$E=C\Delta T$

where in this problem:

E = 3276 J is the energy supplied

C is the heat capacity of the object

$\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C$ is the change in temperature

Solving for C, we find:

$C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C$

5 0

3 years ago

A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav

Vinvika [58]

<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

$y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}$ (1)

$V=V_{o}-g.t$ (2)

Where:

$y$ is the rock's final position

$y_{o}=0$ is the rock's initial position

$V_{o}=30\frac{m}{s}$ is the rock's initial velocity

$V$ is the final velocity

$t$ is the time the parabolic movement lasts

$g=1.62\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of the moon

As we know $y_{o}=0$ , equation (2) is rewritten as:

$y=V_{o}.t+\frac{1}{2}g.t^{2}$ (3)

On the other hand, the maximum height is accomplished when $V=0$ :

$V=V_{o}-g.t=0$ (4)

$V_{o}-g.t=0$

$V_{o}=g.t$ (5)

Finding $t$ :

$t=\frac{V_{o}}{g}$ (6)

Substituting (6) in (3):

$y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}$ (7)

$y_{max}=\frac{{V_{o}}^{2}}{2g}$ (8) Now we can calculate the maximum height of the rock

$y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}$ (9)

Finally:

$y_{max}=277.777m$

4 0

3 years ago

belka [17]

Answer:

2156 J

Explanation:

From the question,

Work done = Combined mass of the bucket and water×height×gravity.

W = (M+m)hg............................. Equation 1

Where M = mass of water, m = mass of the bucket, h = height, g = acceleration due to gravity.

Given: M = 20 kg, m = 2 kg, h = 10 m

Constant: g = 9.8 m/s²

Substitute these value into equation 1

W = (20+2)×10×9.8

W = 22×98

W = 2156 J

4 0

2 years ago