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Doss [256]
3 years ago
8

A ball is thrown straight up from a bridge at a speed of 11.0 m/s. If it takes 5.5 seconds to hit the water below, what is the v

elocity just before it hits the water?
Physics
1 answer:
seropon [69]3 years ago
3 0

Answer:

 v = 42.92 m/s

Explanation:

Given,

initial speed of the ball, v = 11 m/s

time taken to hit the ground = 5.5 m/s

velocity of the ball just before it hit the ground, v = ?

time taken by the ball to reach the maximum height

using equation of motion

 v = u + at

final velocity = 0 m/s  

 0 = 11 - 9.8 t

  t = 1.12 s.

time taken by the ball to reach the water from the maximum height

 t' - 5.5 -1.12 = 4.38 s

using equation of motion for the calculation of speed just before it hit the water.

 v  = u + a t

 v = 0 + 9.8 x 4.38

 v = 42.92 m/s

Velocity of the ball just before it reaches the water is equal to v = 42.92 m/s

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50g of ice at 0°C is mixed with 50g of water at 80°C, what will be the final temperature of a mixture in
xxTIMURxx [149]

Answer:

0° C

Explanation:

Given that

Mass of ice, m = 50g

Mass of water, m(w) = 50g

Temperature of ice, T(i) = 0° C

Temperature of water, T(w) = 80° C

Also, it is known that

Specific heat of water, c = 1 cal/g/°C

Latent heat of ice, L(w) = 89 cal/g

Let us assume T to be the final temperature of mixture.

This makes the energy balance equation:

Heat gained by ice to change itself into water + heat gained by melted ice(water) to raise its temperature at T° C = heat lost by water to reach at T° C

m(i).L(i) + m(i).c(w)[T - 0] = m(w).c(w)[80 - T], on substituting, we have

50 * 80 + 50 * 1(T - 0) = 50 * 1(80 - T)

4000 + 50T = 4000 - 50T

0 = 100 T

T = 0° C

Thus, the final temperature is 0° C

3 0
3 years ago
a runner increases her speed from 3.1 m/s to 3.5 m/s during the last 15 seconds of her run, what was her acceleration during her
Rama09 [41]
Acceleration = velocity/time
A= 3.5m/s/15s
A= 0.23m/s^2
5 0
3 years ago
Read 2 more answers
What is one watt ? Write the relation of watt with kilowatt , megawatt , and horsepower .​
alina1380 [7]

Answer:

See the explanation below

Explanation:

The watt (the power) is equal to the relationship between the work and the time in which that work is performed.

P = W/t

where:

W = work [J] (units of Joules)

t = time [s].

Now 1000 [W] are equal to 1 [kW]

And 1000000 [W] are equal to 1 [MW]

The horsepower is the unit of power in the imperial system of units.

And 745.7 [W] are equal to 1 [Hp]

3 0
3 years ago
You charge an initially uncharged 89.9-mf capacitor through a 30.5-ω resistor by means of a 9.00-v battery having negligible int
blsea [12.9K]
<span>1) The differential equation that models the RC circuit is :

(d/dt)V_capacitor </span>+ (V_capacitor/RC)​ = (V_source/<span>RC)​​</span>

<span>Where the time constant of the circuit is defined by the product of R*C

Time constant = T = R*C = (</span>30.5 ohms) * (89.9-mf) = 2.742 s


2)
C<span>harge of the capacitor 1.57 time constants

1.57*(2.742) = 4.3048 s

The solution of the differential equation is

</span>V_capac (t) = (V_capac(0) - V_capac(∞<span>))e ^(-t /T)  +  </span>V_capac(∞)

Since the capacitor is initially uncharged V_capac(0) = 0

And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery  V_capac(∞) = 9V 

This means,

V_capac (t) = (-9V)e ^(-t /T)  +  9V

The charge in a capacitor is defined as Q = C*V

Where C is the capacitance and V is the Voltage across

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /T)  +  9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s)  +  9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s)  +  9V = -1.87V +9V

V_capac (4.3048 s) = 7.1275 V

Q (4.3048 s)  = 89.9mF*(7.1275V) = 0.6407 C

3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V

Q (∞)  = 89.9mF*(9V) = 0.8091 C
7 0
3 years ago
If our planet was twice as far away how would the Earth’s orbit change?
Tomtit [17]

Answer:

One of the indirect proofs that orbits change is actually in the growth of our own teeth when we were children. our teeth are some of the most basic, and primitive

parts of our bodies. They grow on a 9 day cycle, which was an ancient full moon to full moon cycle when the Earth and the Moon were a lot smaller, and closer together, and the co-orbital period was only 9 days, not the 29.5 days that it is currently.

So Given any two " Planets " that co-orbit a common gravitational center, the larger planet will grow larger far faster than the smaller planet, and the larger planet will accelerate the smaller planet to a higher orbit with a longer period, and decelerate itself to a lower orbit with a longer period, and the absolute value of the center to center distance will increase, and the orbital period will increase. The two orbs and their common gravitational center will remain co-linear through out the gradual growing and changing process.

This is an important process for the enlargement of the solar system as time passes, and an important process for larger galaxies as they attract and merge with smaller galaxies.

All of the planets grow larger at an accelerating rate, and thus systems spiral outward concentrating mass into larger and fewer galaxies, solar systems, and planet - moon systems.

5 0
2 years ago
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