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3 years ago

8

7. Katie and her best friend liam play tennis every saturday morning. When katie serves the ball to liam, it travels 9.5 meters

south in 2.1 seconds. What is the velocity of the tennis ball

1 answer:

Ksenya-84 [330]3 years ago

6 0

Answer:

The velocity of the tennis ball is 4.52 m/s.

Explanation:

Given that,

The distance covered by ball, d = 9.5 meters (due south)

Time, t = 2.1 sec

Let v is the velocity of the tennis ball. We know that the velocity of an object is given by the total distance covered divided by total time taken. It is given by :

$v=\dfrac{d}{t}\\\\v=\dfrac{9.5\ m}{2.1\ s}\\\\v=4.52\ m/s$

So, the velocity of the tennis ball is 4.52 m/s. Hence, this is the required solution.

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Read 2 more answers

A capacitance C and an inductance L are operated at the same angular frequency.

Levart [38]

A) $\omega = \frac{1}{\sqrt{LC}}$

The magnitude of the capacitive reactance is given by

$X_C = \frac{1}{\omega C}$

where

$\omega$ is the angular frequency

C is the capacitance

While the magnitude of the inductive capacitance is given by

$X_L = \omega L$

where L is the inductance.

Since we want the two reactances to be equal, we have

$X_C = X_L$

So we find

$\frac{1}{\omega C}= \omega L\\\omega^2 = \frac{1}{LC}\\\omega = \frac{1}{\sqrt{LC}}$

B) 7449 rad/s

In this case, we have

$L=5.30 mH = 5.3\cdot 10^{-3}H$ is the inductance

$C= 3.40 \mu F= 3.40 \cdot 10^{-6}F$ is the capacitance

Therefore, substituting in the formula for the angular frequency, we find

$\omega=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{(5.30\cdot 10^{-3}H)(3.40\cdot 10^{-6} F)}}=7449 rad/s$

C) $39.5 \Omega$

Now we can us the formulas of the reactances written in part A). We have:

- Capacitive reactance:

$X_C = \frac{1}{\omega C}=\frac{1}{(7449 rad/s)(3.40\cdot 10^{-6}F)}=39.5 \Omega$

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7 0

3 years ago

A golf ball strikes a hard, smooth floor at an angle of 28.2 ° and, as the drawing shows, rebounds at the same angle. The mass o

WITCHER [35]

Answer:

I = 1.06886 N s

Explanation:

The expression for momentum is

I = F t = Δp

therefore the momentum is a vector quantity, for which we define a reference system parallel to the floor

Let's find the components of the initial velocity

sin 28.2 = v_y / v

cos 28.2= vₓ / v

v_y = v sin 282

vₓ = v cos 28.2

v_y = 42.8 sin 28.2 = 20.225 m / s

vₓ = 42.8 cos 28.2 = 37.72 m / s

since the ball is heading to the ground, the vertical velocity is negative and the horizontal velocity is positive, it can also be calculated by making

θ = -28.2

v_y = -20.55 m / s

v_x = 37.72 m / s

X axis

Iₓ = Δpₓ = $p_{fx} - p_{ox}$

since the ball moves in the x-axis without changing the velocity, the change in moment must be zero

Δpₓ = m $v_{fx}$ - m v₀ₓ = 0

v_{fx} = v₀ₓ

therefore

Iₓ = 0

Y axis

I_y = Δp_y = p_{fy} -p_{oy}

when the ball reaches the floor its vertical speed is downwards and when it leaves the floor its speed has the same modulus but the direction is upwards

v_{fy} = - v_{oy}

Δp_y = 2 m v_{oy}

Δp_y = 2 0.0260 (20.55)

$\Delta p_{y}$ = 1.0686 N s

the total impulse is

I = Iₓ i ^ + I_y j ^

I = 1.06886 j^ N s

7 0

2 years ago

A charge of -3.30 nC is placed at the origin of an xy-coordinate system, and a charge of 2.05 nC is placed on the y axis at y =

Elis [28]

Answer:

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis is the net horizontal force

$F_v=80062.47\times 10^9$ attractive toward +y axis is the net vertical force

Explanation:

Given:

charge at origin, $Q_0=-3.35\times 10^{-6}\ C$

magnitude of second charge, $Q_2=2.05\times 10^{-6}\ C$

magnitude of third charge, $Q_3=5\times 10^{-6}\ C$

position of second charge, $(x_2,y_2)\equiv(0,4.35)\ cm$

position of third charge, $(x_3,y_3)\equiv(3.1,3.8)\ cm$

<u>Now the distance between the charge at at origin and the second charge:</u>

$d_2=\sqrt{(x_2-0)^2+(y_2-0)^2}$

$d_2=\sqrt{(0-0)^2+(4.35-0)^2}$

$d_2=0.0435\ m$

<u>Now the distance between the charge at at origin and the third charge:</u>

$d_3=\sqrt{(x_3-0)^2+(y_3-0)^2}$

$d_3=\sqrt{(3.1-0)^2+(3.8-0)^2}$

$d_3=0.04904\ m$

<u>Now the force due to second charge:</u>

$F_2=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_2}{d_2^2}$

$F_2=9\times 10^9\times \frac{3.3\times 2.05}{0.0435^2}$

$F_2=32175.98\times 10^9\ N$ attractive towards +y

<u>Now the force due to third charge:</u>

$F_3=\frac{1}{4\pi.\epsilon_0} \times \frac{Q_0.Q_3}{d_3^2}$

$F_3=9\times 10^9\times \frac{3.3\times 5}{0.04904^2}$

$F_3=61748.38\times 10^9\ N$ attractive

<u>Now the its horizontal component:</u>

$F_{3h}=\frac{3.1}{4.9} \times 61748.38\times 10^9$

$F_{3h}=39065.298\times 10^9\ N$ attractive toward +x axis

<u>Now the its vertical component:</u>

$F_{3v}=\frac{3.8}{4.9} \times 61748.38\times 10^9$

$F_{3v}=47886.49\times 10^9\ N$ upwards attractive

Now the net vertical force:

$F_v=F_{3v}+F_2$

$F_v=47886.49\times 10^9+32175.98\times 10^9$

$F_v=80062.47\times 10^9$

3 0

3 years ago