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3 years ago

6

You serve a volleyball with a mass of 2.5 kg. The ball leaves your hand with a speed of 23 m/s. What is the kinetic energy of th

e ball?

1 answer:

Zarrin [17]3 years ago

8 0

Answer:

661.25, I believe!

Explanation:

KE = (1/2)(2.5)(23)^2

KE=(1.25)(529)

KE= 661.25

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This problem follows up on a discussion from lecture. A wind turbine with an efficiency of 45% for converting wind energy into e

Volgvan

Answer:

4.1 m

Explanation:

10 kW = 10000 W

20mi/h = 20*1.6 km/mi = 32 km/h = 32 * 1000 (m/km) *(1/3600) hr/s = 8.89 m/s

The power yielded by the wind turbine can be calculated using the following formula

$P = \frac{1}{2} \rho v^3 A C_p$

where $\rho = 1.2 kg/m^3$ is the air density, v = 8.89 m/s is the wind speed, A is the swept area and $C_p = 0.45$ is the efficiency

$10000 = 0.5 * 1.2 * 8.89^3 * A * 0.45$

$10000 = 190A$

$A = 10000 / 190 = 52.7 m^2$

The swept area is a circle with radius r being the blade length

$\pi r^2 = A = 52.7$

$r^2 = 52.7 / \pi = 16.79$

$r = \sqrt{16.79} = 4.1 m$

4 0

3 years ago

Speed=100m/sec<br> Frequency=10 Hz<br> Wavelength=?

elena-14-01-66 [18.8K]

Answer:

Wavelength = 10 m

Explanation:

Given:

Speed = 100 m $s^{-1}$

Frequency = 10 Hz = 10 $s^{-1}$

To find : Wavelength = ?

We know that the relationship between wavelength λ, frequency f and speed v is given by the equation

v = fλ

Therefore wavelength λ = v/f

= 100 m $s^{-1}$ / 10 m $s^{-1}$

= 10 m

Hence wavelength = 10 m

4 0

3 years ago

A billiard ball of mass m moving with speed v1=1 m/s collides head-on with a second ball of mass 2m which is at rest. What is th

Soloha48 [4]

Answer:

The velocity of mass 2m is $v_B = 0.67 m/s$

Explanation:

From the question w are told that

The mass of the billiard ball A is =m

The initial speed of the billiard ball A = $v_1$ =1 m/s

The mass of the billiard ball B is = 2 m

The initial speed of the billiard ball B = 0

Let the final speed of the billiard ball A = $v_A$

Let The finial speed of the billiard ball B = $v_B$

According to the law of conservation of Energy

$\frac{1}{2} m (v_1)^2 + \frac{1}{2} 2m (0) ^ 2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2$

Substituting values

$\frac{1}{2} m (1)^2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2$

Multiplying through by $\frac{1}{2}m$

$1 =v_A^2 + 2 v_B ^2 ---(1)$

According to the law of conservation of Momentum

$mv_1 + 2m(0) = mv_A + 2m v_B$

Substituting values

$m(1) = mv_A + 2mv_B$

Multiplying through by $m$

$1 = v_A + 2v_B ---(2)$

making $v_A$ subject of the equation 2

$v_A = 1 - 2v_B$

Substituting this into equation 1

$(1 -2v_B)^2 + 2v_B^2 = 1$

$1 - 4v_B + 4v_B^2 + 2v_B^2 =1$

$6v_B^2 -4v_B +1 =1$

$6v_B^2 -4v_B =0$

Multiplying through by $\frac{1}{v_B}$

$6v_B -4 = 0$

$v_B = \frac{4}{6}$

$v_B = 0.67 m/s$

4 0

3 years ago

A girl runs south 10 meters then turns around and walks back 3 meters

Svetlanka [38]

Answer:

12

Explanation:

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6 0

3 years ago

Two cylindrical resistors are made from the same material. The shorter one has length L, diameter D, and resistance R1. The long

nordsb [41]

Answer:

the resistance of the longer one is twice as big as the resistance of the shorter one.

Explanation:

Given that :

For the shorter cylindrical resistor

Length = L

Diameter = D

Resistance = R1

For the longer cylindrical resistor

Length = 8L

Diameter = 4D

Resistance = R2

So;

We all know that the resistance of a given material can be determined by using the formula :

$R = \dfrac{\rho L }{A}$

where;

A = πr²

$R = \dfrac{\rho L }{\pi r ^2}$

For the shorter cylindrical resistor ; we have:

$R = \dfrac{\rho L }{\pi r ^2}$

since 2 r = D

$R = \dfrac{\rho L }{\pi (\frac{2}{2 \ r}) ^2}$

$R = \dfrac{ 4 \rho L }{\pi \ D ^2}$

For the longer cylindrical resistor ; we have:

$R = \dfrac{\rho L }{\pi r ^2}$

since 2 r = D

$R = \dfrac{ \rho (8 ) L }{\pi (\frac{2}{2 \ r}) ^2}$

$R = \dfrac{32\rho L }{\pi \ (4 D) ^2}$

$R = \dfrac{2\rho L }{\pi \ (D) ^2}$

Sp;we can equate the shorter cylindrical resistor to the longer cylindrical resistor as shown below :

$\dfrac{R_s}{R_L} = \dfrac{ \dfrac{ 4 \rho L }{\pi \ D ^2}}{ \dfrac{2\rho L }{\pi \ (D) ^2}}$

$\dfrac{R_s}{R_L} ={ \dfrac{ 4 \rho L }{\pi \ D ^2}}* { \dfrac {\pi \ (D) ^2} {2\rho L}}$

$\dfrac{R_s}{R_L} =2$

${R_s}=2{R_L}$

Thus; the resistance of the longer one is twice as big as the resistance of the shorter one.

7 0

3 years ago