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3 years ago

6

if the time is taken to bring a ball to rest from a certain velocity 'v'. is reduced to half, what will be the change in the val

ues of a) initial momentum b) final momentum c) rate of change of momentum?

1 answer:

777dan777 [17]3 years ago

6 0

The answer is C) rate of change of momentum. The answer is not initial or final momentum as the start and end points are not changing. On the other hand, the time it takes for the ball to change velocity is. This change relates to the change of momentum. Hope this helped :))

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velikii [3]

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4 0

3 years ago

The Earth is 1.5 × 1011 m from the Sun and takes a year to make one complete orbit. It rotates on its own axis once per day. It

katrin2010 [14]

Answer:

Part a)

$v_{cm} = 2.98 \times 10^4 m/s$

Part b)

$K_{trans} = 2.68 \times 10^{33} J$

Part c)

$\omega = 7.27 \times 10^{-5} rad/s$

Part d)

$KE_{rot} = 2.6 \times 10^{29} J$

Part e)

$KE_{tot} = 2.68 \times 10^{33} J$

Explanation:

Time period of Earth about Sun is 1 Year

so it is

$T = 1 year = 3.15 \times 10^7 s$

now we know that angular speed of the Earth about Sun is given as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{3.15 \times 10^7}$

now speed of center of Earth is given as

$v_{cm} = r\omega$

$r = 1.5 \times 10^{11} m$

$v_{cm} = (1.5 \times 10^{11})(\frac{2\pi}{3.15 \times 10^7})$

$v_{cm} = 2.98 \times 10^4 m/s$

Part b)

now transnational kinetic energy of center of Earth is given as

$K_{trans} = \frac{1}{2}mv^2$

$K_{trans} = \frac{1}{2}(6 \times 10^{24})(2.98 \times 10^4)^2$

$K_{trans} = 2.68 \times 10^{33} J$

Part c)

Angular speed of Earth about its own axis is given as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{24 \times 3600}$

$\omega = 7.27 \times 10^{-5} rad/s$

Part d)

Now moment of inertia of Earth about its own axis

$I = \frac{2}{5}mR^2$

$I = \frac{2}{5}(6 \times 10^{24})(6.4 \times 10^6)^2$

$I = 9.83 \times 10^{37} kg m^2$

now rotational energy is given as

$KE_{rot} = \frac{1}{2}I\omega^2$

$KE_{rot} = \frac{1}{2}(9.83 \times 10^{37})(7.27 \times 10^{-5})^2$

$KE_{rot} = 2.6 \times 10^{29} J$

Part e)

Now total kinetic energy is given as

$KE_{tot} = KE_{trans} + KE_{rot}$

$KE_{tot} = 2.68 \times 10^{33} + 2.6 \times 10^{29}$

$KE_{tot} = 2.68 \times 10^{33} J$

6 0

3 years ago

Two lightbulbs work on a 120-V circuit One 50 W and the other is 100 W. Which bulb has a higher resistance? Explain pls!!!!

Sladkaya [172]

Answer:

Bulb 1 has more resistance.

Explanation:

Given that,

Two lightbulbs work on a 120-V circuit.

The power of circuit 1, P₁ = 50 W

The power of circuit 2, P₂ = 100 W

We need to find the bulb that has a higher resistance.

The power of the bulb is given by :

$P = \dfrac{V^2}{R}$

For bulb 1,

$R_1=\dfrac{V^2}{P_1}\\\\R_1=\dfrac{(120)^2}{50}\\\\R_1=288\ \Omega$

For bulb 2,

$R_2=\dfrac{V^2}{P_2}\\\\R_2=\dfrac{(120)^2}{100}\\\\R_2=144\ \Omega$

So, bulb 1 has higher resistance.

8 0

3 years ago

A motorcycle traveling at 36 m/s slams on the brakes to avoid an accident. The motorcycle skids 23m before stoping. What is the

Svetach [21]

Since the motor cycle was moving at 36 m/s before the brakes were applied, the initial velocity
$u = 36 {ms}^{ - 1}$
The motorcycle traveled 23m before stopping, that means
$s = 23m$
Since the motorcycle stopped moving the final velocity is zero.

$v = 0{ms}^{ - 1}$
We can use this 'suvat' equation of linear motion.

${v}^{2} = {u}^{2} + 2as$
Plugging in the above values gives,

${0}^{2} = {36}^{2} + 2a(23)$

$0 = 1296+ 46a$
$46a = - 1296$
$a = - 28.2 {ms}^{ - 2}$

We can use the equation
$v = u + at$
to find the time the motorcycle took to stop.

$0 = 36 + - 28.2t$
$- 36 = - 28.2t$
$t = 1.3s$

8 0

3 years ago

A system gains 1500 J of heat, while the internal energy of the system increases by 4500 J and the volume decreases by . Assume

Assoli18 [71]

Answer:

Hence the pressure is $3\times 10^5 Pa$

Explanation:

Given data

Q=1500 J system gains heat

ΔV=- 0.010 m^3 there is a decrease in volume

ΔU= 4500 J internal energy decrease

We know work done is

W= Q- ΔU

=1500-4500= -3000 J

The change in the volume at constant pressure is

ΔV= W/P

there fore P = W/ΔV= -3000/-0.01= 3×10^5

Hence the pressure is $3\times 10^5 Pa$

3 0

4 years ago