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A 1.0-kg ball has a velocity of 12 m/s downward just before it strikes the ground and bounces up with a velocity of 12 m/s upwar

Nezavi [6.7K]

Answer:

The change in momentum of the ball is 24 kg-m/s

Explanation:

It is given that,

Mass of the ball, m = 1 kg

Initial velocity of the ball, u = -12 m/s (in downwards)

Final velocity of the ball, v = +12 m/s (in upward)

We need to find the change in momentum of the ball.

Initial momentum of the ball, $p_i=mu=1\ kg\times (-12\ m/s)=-12\ kg-m/s$

Final momentum of the ball, $p_f=mv=1\ kg\times (12\ m/s)=12\ kg-m/s$

Change in momentum of the ball, $\Delta p=p_f-p_i$

$\Delta p=12-(-12)=24\ kg-m/s$

So, the change in momentum of the ball is 24 kg-m/s. Hence, this is the required solution.

3 0

3 years ago

Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen

Stella [2.4K]

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

$\lambda$ = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

$tan\theta=\frac{y}{D}\\\Rightarrow \theta=tan^{-1}{\frac{y}{D}}\\\Rightarrow \theta=tan^{-1}{\frac{4.84\times 10^{-3}}{3}}\\\Rightarrow \theta=0.09243\ ^{\circ}$

$sin\theta=\frac{\lambda}{d}\\\Rightarrow d=\frac{\lambda}{sin\theta}\\\Rightarrow d=\frac{598\times 10^{-9}}{sin0.09243}\\\Rightarrow d=0.00037066\ m$

For first dark fringe

$dsin\theta=\frac{\lambda'}{2}\\\Rightarrow \lambda'=2dsin\theta\\\Rightarrow \lambda'=2\times 0.00037066\times sin0.09243\\\Rightarrow \lambda'=1.196\times 10^{-6}\\\Rightarrow \lambda'=1.196\ \mu m$

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm

3 0

2 years ago

A solid cylinder with a mass of 2.72 kg and a radius of 0.083 m starts from rest at a height of 4.20 m and rolls down a 88.7 ◦ s

Bingel [31]

Explanation:

According to the law of conservation of energy ,

Potential energy = kinetic energy

$mgh = \frac{1}{2} \times mv^{2} + \frac{1}{2} \times I \times \omega^{2}$

I = $\frac{mr^{2}}{2}$

$\omega = \frac{v}{r}$

$mgh = [\frac{1}{2} \times mv^{2}] + [\frac{1}{2} \times (\frac{mr^{2}}{2}) \frac{v^{2}}{r^{2}}]$

mgh = $[\frac{1}{2} \times mv^{2}] + [\frac{1}{4} \times mv^{2}]$

$g \times h = \frac{3}{4} \times v^{2}$

$9.8 \times 4.2 = \frac{3}{4} \times v^{2}$

v = 7.4 m/s

thus, we can conclude that the translational speed of the cylinder when it leaves the incline is 7.4 m/s.

5 0

2 years ago

A jet taxiing down the runway receives word that it must return to the gate. The jet is traveling 37.6 m/s when the pilot receiv

IrinaVladis [17]

Answer:

7 m/s^2

Explanation:

Given that the jet is traveling 37.6 m/s when the pilot receives the message.

And it takes the pilot 5.37 s to bring the plane to a halt.

Acceleration of the plane can be calculated by using first equation of motion

V = U - at

Since the plane is going to stop, the final velocity V = zero.

And the acceleration will be negative

Substitute all the parameters into the formula

0 = 37.6 - 5.37a

5.37a = 37.6

Make a the subject of formula

a = 37.6 / 5.37

a = 7.0 m/s^2

Therefore, the acceleration of the plane to bring the plane to a halt is 7 m/s^2

5 0

3 years ago

A diamond with a mass of 45 g hangs motionless from a chain. what is the upward force of the chain on the diamond?

Oksi-84 [34.3K]

The upward force of the chain on the diamond would be the tension in the chain, and this tension would have to support the weight of the 45g that hangs from the chain.

mass = 45 g = 45/1000 kg = 0.045kg

Weight = mg = 0.045 * 10 ≈ 0.45N, g ≈ 10 m/s²

<span>So the upward force is ≈ </span><span>0.45N. </span>

6 0

2 years ago