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2 years ago

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A student on a tower 49 m height drops a stone. One second later he throws a second stone after the first. They both hit the gro

und at the same time, with what speed did he throw the second stone. [Ans: 10.1m/s]

1 answer:

ohaa [14]2 years ago

6 0

By applying the second equation of motion, the speed at which he threw the second stone is equal to 12.10 m/s.

<h3>How to determine the speed?</h3>

First of all, we would calculate the time taken by the first stone to reach a height of 49 meters by applying the second equation of motion as follows:

S = ut + ½gt²

49 = 0(t) + ½ × 9.8 × t²

49 = 4.9t²

t² = 49/4.9

t = √10

t = 3.16 seconds.

Now, we can determine the speed at which he threw the second stone:

<u>Note:</u> Time = 3.16 - 1 = 2.16 seconds.

S = ut + ½gt²

49 = u(2.16) + ½ × 9.8 × 2.16²

49 = 2.16u + 22.86

2.16u = 49 - 22.86

u = 26.14/2.16

u = 12.10 m/s.

Read more on initial speed here: brainly.com/question/19365526

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You are at the carnival with you your little brother and you decide to ride the bumper cars for fun. You each get in a different

Aliun [14]

Answer:

<em>a) The equation is </em> $(m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}$

<em>b) Your velocity after collision is 2.64 m/s</em>

<em>c) The force you felt is 7392 N</em>

<em>d) you and your brother undergo an equal amount of acceleration</em>

<em></em>

Explanation:

Your mass $m_{y}$ = 60 kg

your brother's mass $m_{b}$ = 30 kg

mass of the car $m_{c}$ = 80 kg

your initial speed $u_{y}$ = 0 m/s (since you've not started moving yet)

your brother's initial velocity $u_{b}$ = 3 m/s

your final speed $v_{y}$ after collision = ?

your brother's final speed $v_{b}$ after collision = ?

a) equations you need to use to figure out how fast you and your brother are moving after the collision is

$(m_{y}+m_{c} )u_{y} + (m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}$

but $u_{y}$ = 0 m/s

the equation reduces to

$(m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}$

b) if your little brother reverses with velocity of 0.36 m/s it means

$v_{b}$ = -0.36 m/s (the reverse means it travels in the opposite direction)

then, imputing values into the equation, we'll have

$(m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}$

(30 + 80)3 = (60 + 80) $v_{y}$ + (30 + 80)(-0.36)

330 = 140 $v_{y}$ - 39.6

369.6 = 140 $v_{y}$

$v_{y}$ = 369.6/140 = <em>2.64 m/s</em>

This means you will also reverse with a velocity of 2.64 m/s

c) your initial momentum = 0 since you started from rest

your final momentum = (total mass) x (final velocity)

==> (60 + 80) x 2.64 = 369.6 kg-m/s

If the collision lasted for 0.05 s,

then force exerted on you = (change in momentum) ÷ (time collision lasted)

force on you = ( 369.6 - 0) ÷ 0.05 =<em> 7392 N</em>

<em></em>

d) you changed velocity from 0 m/s to 2.64 m/s in 0.05 s

your acceleration is (2.64 - 0)/0.05 = 52.8 m/s^2

your brother changed velocity from 3 m/s to 0.36 m/s in 0.05 s

his deceleration is (3 - 0.36)/0.05 = 52.8 m/s

<em>you and your brother undergo an equal amount of acceleration. This is because you gained the momentum your brother lost</em>

<em></em>

7 0

3 years ago

A proton in a high-energy accelerator is given a kinetic energy of 50.0 GeV. Determine (a) the momentum and (b) the speed of the

koban [17]

(a) The momentum of the proton is determined as 5.17 x 10⁻¹⁸ kgm/s.

(b) The speed of the proton is determined as 3.1 x 10⁹ m/s.

<h3>Momentum of the proton</h3>

The momentum of the proton is calculated as follows;

K.E = ¹/₂mv²

where;

m is mass of proton = 1.67 x 10⁻²⁷ kg
v is speed of the proton = ?

<h3>Speed of the proton</h3>

v² = 2K.E/m

v² = (2 x 50 x 10⁹ x 1.602 x 10⁻¹⁹ J)/(1.67 x 10⁻²⁷)

v² = 9.6 x 10¹⁸

v = 3.1 x 10⁹ m/s

<h3>Momentum of the proton</h3>

P = mv = (1.67 x10⁻²⁷ x 3.1 x 10⁹) = 5.17 x 10⁻¹⁸ kgm/s

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4 0

2 years ago

A hiker walks 2.00 km north and then 3.00 km east, all in 2.50 hours. Calculate the magnitude and direction of the hiker’s (a) d

tino4ka555 [31]

Answer:

Incomplete third question

I think it should be

C. What was her average speed

Explanation:

Check attachment for solution

5 0

3 years ago

A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 22.0 V across the plates. A pie

-BARSIC- [3]

a) $3.27\cdot 10^{-3} J$

b) $11.60\cdot 10^{-3} J$

c) $8.33\cdot 10^{-3} J$

Explanation:

a)

The energy stored in a capacitor is given by

$U=\frac{1}{2}CV^2$

where

C is the capacitance of the capacitor

V is the potential difference across the plates of the capacitor

For the capacitor in this problem, before insering the dielectric, we have:

$C=13.5 \mu F = 13.5\cdot 10^{-6}F$ is its capacitance

V = 22.0 V is the potential difference across it

Therefore, the initial energy stored in the capacitor is:

$U=\frac{1}{2}(13.5\cdot 10^{-6})(22.0)^2=3.27\cdot 10^{-3} J$

b)

After the dielectric is inserted into the plates, the capacitance of the capacitor changes according to:

$C'=kC$

where

k = 3.55 is the dielectric constant of the material

C is the initial capacitance of the capacitor

Therefore, the energy stored now in the capacitor is:

$U'=\frac{1}{2}C'V^2=\frac{1}{2}kCV^2$

where:

$C=13.5\cdot 10^{-6}F$ is the initial capacitance

V = 22.0 V is the potential difference across the plate

Substituting, we find:

$U'=\frac{1}{2}(3.55)(13.5\cdot 10^{-6})(22.0)^2=11.60\cdot 10^{-3} J$

C)

The initial energy stored in the capacitor, before the dielectric is inserted, is

$U=3.27\cdot 10^{-3} J$

The final energy stored in the capacitor, after the dielectric is inserted, is

$U'=11.60\cdot 10^{-3} J$

Therefore, the change in energy of the capacitor during the insertion is:

$\Delta U=11.60\cdot 10^{-3}-3.27\cdot 10^{-3}=8.33\cdot 10^{-3} J$

So, the energy of the capacitor has increased by $8.33\cdot 10^{-3} J$

8 0

3 years ago

When stated as a percentage, the useful energy transferred by a device divided by the total energy supplied to the device is kno

Verdich [7]

Efficiency

Explanation:

Efficiency is a ratio/percentage that is useful in comparing the energy transferred by a device to the total energy supplied to it.

Percentage efficiency = $\frac{energy input}{total energy supplied}$ x 100

As with most system, none is 100% efficient.
During energy is transferred some are lost and only a little portion is used in doing actual work by the machine.
This validates the third law of thermodynamics which proposes that no system is 100% efficient.
A 100% efficiency implies total energy input is used doing all the work.
This is impossible. The bulk of the energy goes into heating the system.

learn more:

Third law of thermodynamics brainly.com/question/3564634

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3 years ago