Does anybody know the formula needed to calculate distance given the values for mass, power, and time?

A man throws a ball straight up to his friend on a balcony who catches it at its highest point. The ball was thrown with an init

Ede4ka [16]

Answer:

The maximum height reached by the ball is 16.35 m.

Explanation:

Given;

initial velocity of the ball, u = 17.9 m/s

the final velocity of the ball at the maximum height, v = 0

The maximum height reached by the ball is given by;

v² = u² + 2gh

During upward motion, gravity is negative

v² = u² + 2(-g)h

v² = u² - 2gh

0 = u² - 2gh

2gh = u²

h = u² / 2g

h = (17.9)² / (2 x 9.8)

h = 16.35 m

Ttherefore, the maximum height reached by the ball is 16.35 m.

3 0

2 years ago

What will happen to the charge and potential difference if the plate area were increased while the plate separation remains unch

Alenkinab [10]

Answer:

Potential difference and charge will also increase.

Explanation:

Asking that :

What will happen to the charge and potential difference if the plate area were increased while the plate separation remains unchanged?

The charge is directly proportional to area of the plate. That is, increase in area of the plate of a capacitor will lead to the increase in the charges between the plates.

And since charge is also proportional to the magnitude of potential difference between the plates from the definition of capacitance of a capacitor which says that:

Q = CV

Therefore, increase in the area of the plate will also lead to increase in potential difference between the plates.

Therefore, if the plate area were increased while the plate separation remains unchanged, the charge and potential difference between them will also increase.

5 0

3 years ago

A thermos contains m1 = 0.89 kg of tea at T1 = 31° C. Ice (m2 = 0.075 kg, T2 = 0° C) is added to it. The heat capacity of both w

lorasvet [3.4K]

Answer:

a) T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

b) T = 295.37 K

Explanation:

Given;

Initial temperature of tea T1 = 31 C

Initial temperature of ice T2 = 0 C

Mass of tea m1 = 0.89 kg

Mass of ice m2 = 0.075kg

The heat capacity of both water and tea c = 4186 J/(kg⋅K)

the latent heat of fusion for water is Lf = 33.5 × 10^4 J/kg

And T = the final temperature of the mixture

Heat loss by tea = heat gained by ice

m1c∆T1 = m2c∆T2 + m2Lf

m1c(T1-T) = m2c(T-T2) + m2Lf

m1cT1 - m1cT = m2cT - m2cT2 + m2Lf

m1cT + m2cT = m1cT1 + m2cT2 - m2Lf

T(m1c + m2c) = m1cT1 + m2cT2 - m2Lf

T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

Substituting the values;

T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

T = (0.89×4186×31 + 0.075×4186×0 - 0.075×33.5 × 10^4)/(0.89×4186 + 0.075×4186)

T = 22.37 °C

T = 273 + 22.37 K

T = 295.37 K

4 0

3 years ago

Read 2 more answers

HELP PLEASE

-BARSIC- [3]

Angel ! You have a formula, and you have an example that's
completely worked out. The ONLY POSSIBLE reason that you
could still need help is that you're letting math scare you.

I'll do 'A' for you, 'B' most of the way, and get 'C' set up.
If THAT's not enough for you to run with and finish them all,
then you and I should both be embarrassed.

Write the formula on the wall:

   °F = (9/5) °C + 32°

A). Convert 35° C °F = (9/5)(35°) + 32°

(9/5)(35) = 63 °F = 63° + 32°

   °F = 95°
____________________________________

B). Convert 80°F to °C
       The formula: °F = (9/5) °C + 32°

°F = 80    80 = (9/5)°C + 32

Subtract 32 from each side: 48 = (9/5)°C

Multiply each side by 5 : 240 = (9) C

Now you take over:
_________________________________________

C). Convert 15°C to °F.
       The formula: °F = (9/5) °C + 32°

°C = 15 °F = (9/5) 15° + 32

(9/5) (15) = 27
   Go !    °F =

7 0

3 years ago

A boy kicks a football from ground level. The ball takes 3 seconds to reach its maximum height. What is the angle of the initial

ruslelena [56]

<h2>The angle of the initial velocity with respect to the horizontal is 85.14°</h2>

Explanation:

Given that the ball takes 3 seconds to reach its maximum height.

Consider the vertical motion of ball till maximum height.

We have equation of motion v = u + at

Acceleration, a = -9.81 m/s²

Final velocity, v = 0 m/s

Time, t = 3 s

Substituting

v = u + at

0 = u + -9.81 x 3

u = 29.43 m/s

Initial vertical velocity is 29.43 m/s.

Now consider horizontal motion of ball.

Time of flight of ball = 2 x Time to reach maximum height = 2 x 3 = 6 s

Displacement = 15 m

We have equation of motion s = ut + 0.5 at²

Displacement, s = 15 m

Acceleration, a = 0 m/s²

Time, t = 6 s

Substituting

s = ut + 0.5 at²

15 = u x 6 + 0.5 x 0 x 6²

u = 2.5 m/s

Initial horizontal velocity is 2.5 m/s

Let r be the initial velocity and θ be the angle with horizontal

Initial vertical velocity = rsinθ = 29.43 m/s

Initial horizontal velocity = rcosθ = 2.5 m/s

Dividing

$\frac{rsin\theta }{rcos\theta }=\frac{29.43}{2.5}\\\\tan\theta=11.77\\\\\theta=85.14^0$

The angle of the initial velocity with respect to the horizontal is 85.14°

4 0

2 years ago