All categories

3 years ago

You serve a volleyball with a mass of 0.77 kg. The ball leaves your hand with a velocity of

Physics

1 answer:

dimaraw [331]3 years ago

7 0

<h2>Given :</h2>

mass (m) = 0.77 kg

velocity (v) = 2.33 m/s

<h2>Solution :</h2>

$\boxed{ \mathrm{Kinetic \: Energy = \frac{1}{2} m {v}^{2} }}$

$\dfrac{1}{2} \times 0.77 \times (2.33) {}^{2}$

$\dfrac{418.0253}{2}$

$209.01265 \: \: joules$

$\mathrm{209.01 \: \: joules} \: \: \: (approx)$

_____________________________

$\mathrm{ \#TeeNForeveR}$

You might be interested in

How much work is done against gravity when lowering a 16 kg box 0.50 m? (g = 9.8 m/s2)

leonid [27]

Answer:

The work done against gravity is 78.4 J

Explanation:

The work is calculated by multiplying the force by the distance that the

object moves

W = F × d, where W is the work , F is the force and d is the distance

The SI unit of work is the joule (J)

We need to find the work done against gravity when lowering a

16 kg box 0.50 m

→ F = mg

→ m = 16 kg, and g = 9.8 m/s²

Substitute these value in the rule

→ F = 16 × 9.8 = 156.8 N

→ W = F × d

→ F = 156.8 N and d = 0.50

Substitute these values in the rule

→ W = 78.4 J

<em>The work done against gravity is 78.4 J</em>

6 0

3 years ago

Location C is 0.02 m from a small sphere which has a charge of 3 nanocoulombs uniformly distributed on its surface. Location D i

kkurt [141]

The change in potential along a path from C to D due to a small charged sphere is 900 V.

Given:

Charge, Q = 3 nC = 3 × 10⁻⁹ C

Distance between the sphere and point C, r₁ = 0.02 m

Distance between the sphere and point D, r₂ = 0.06 m

Calculation:

We know that the electric potential is given as:

V = k Q/r - (1)

where, V is the electric potential

k is Coulomb's force constant

Qis the charge on the sphere

r is the separation distance

The electric potential at point C due to charged sphere can be given as:

V₁ = k Q/r₁

= (9×10⁹ Nm²/C²) [(3 × 10⁻⁹ C)/(0.02 m)]

= 1350 V

The electric potential at point D due to charged sphere can be given as:

V₂ = k Q/r₂

= (9×10⁹ Nm²/C²) [(3 × 10⁻⁹ C)/(0.06 m)]

= 450 V

Now, the change in potential along the path from C to D can be calculated as:

ΔV = V₂ - V₁

= 450 V - 1350 V

= -900 V

The negative sign indicates that the work is done against the electric field in moving the charge from C to D.

Therefore, the change in potential along a path from C to D is 900 V against the direction of the electric field.

Learn more about the electric potential here:

<u>brainly.com/question/12645463</u>

#SPJ4

8 0

1 year ago

Read 2 more answers

a car accelerates from 4 meters/second to 16 meter/second in 4 seconds. The cars acceleration is how many meter/seconds.

Allushta [10]

Answer:

=3 metre per second ^2

Explanation:

Formula for acceleration is

V-U÷T

In the given information

V=16

U=4

T=4

Acceleration =16-4/4

=3 metre per second ^2

4 0

3 years ago

2. A car is moving with a constant speed of 10m/s. What is the total distance the car travels

Tems11 [23]

Answer:

1200 meters

Explanation:

there are 60 seconds in a minute times 2 is 120 ten times 120 is 1200

6 0

3 years ago

To make a given sound seem twice as loud, how should a musician change the intensity of the sound?

Serhud [2]

Answer:

C. Quadruple the intensity

Explanation:

The intensity of the sound is proportional to square of amplitude of the sound.

I ∝ A²

$\frac{I_1}{A_1^2} = \frac{I_2}{A_2^2}\\\\I_2 = \frac{I_1A_2^2}{A_1^2}$

When the given sound is twice loud as the initial value, then the new amplitude is twice the former.

A₂ = 2A₁

$I_2 = \frac{I_1A_2^2}{A_1^2} \\\\I_2 = \frac{I_1(2A_1)^2}{A_1^2} \\\\I_2 = \frac{4I_1A_1^2}{A_1^2}\\\\ I_2 = 4I_1$

Thus, to make a given sound seem twice as loud, the musician should Quadruple the intensity

3 0

3 years ago