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4 years ago

2 QUESTIONS 20 POINTS

Physics

2 answers:

irina [24]4 years ago

7 0

I think 1 is x-rays and 2 is ultraviolet light.. I'm not that sure

stich3 [128]4 years ago

3 0

Just finished completitng this kind of info in the course that I am taking! X-rays contain the most energy, and microwaves would have the longest wavelength.

Higher frequency= more energy

If you want to add it to your notes, this is EMR listed in order from highest frequency/highest energy, to lowest frequency/ lowest energy.

Gamme rays, X-rays, UV, Visible light, Infrared, Microwave, and radio waves.

So, Gamme rays have the highest energy and frequency, and radio waves havethe lowest.

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The momentum of an object depends on what two factors help

Mazyrski [523]

The momentum of an object depends on velocity and mass

4 0

3 years ago

The velocity of a 150 kg cart changes from 6.0 m/s to 14.0 m/s. What is the magnitude of the impulse that acted on it?

aleksandrvk [35]

M = 150 kg.

Final velocity, v = 14 m/s

Initial Velocity, u = 6 m/s

Impulse = m(v - u)

= 150*(14 - 6)

= 150*8 = 1200 kgm/s or 1200 Ns

8 0

4 years ago

Read 2 more answers

How many pounds can the rock lift

Shkiper50 [21]

Answer:

450pounds

Explanation:

5 0

3 years ago

A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

svetoff [14.1K]

Answer:

The equation of motion is $x(t)=-$ $\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

= 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

$x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6t}$

The mass is released from the rest x'(0) = 0

$=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

7 0

3 years ago

A 9.0 kg potted plant slides down a 25.0º incline with an acceleration of 2.4 m/s2. What is the coefficient of kinetic friction

mezya [45]

Answer:

0.196

Explanation:

According to Newton's second law

\sumFx = ma

Fm-Ff = ma

Fm is the moving force

Ff is the frictional force

m is the mass = 9kg

a is the acceleration = 2.4m/s²

Fm = Wsin theta

Fm = 9(9.8)sin25°

Fm = 37.28N

Ff is the frictional force = nmgcos theta

Ff = n(9)(9.8)cos25°

Ff = 79.94n

Substitute the given values into the formula

37.28-79.94n = 9(2.4)

-79.94n = 21.6-37.28

-79.94n = -15.68

n = 15.68/79.94

n = 0.196

Hence the coefficient of kinetic friction between the potted plant and the incline is 0.196

7 0

3 years ago