When would a blackbody curve of a star moving toward earth have its peak shifted?

Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the

leonid [27]

Answer:

$KE_A=33\ J$

$KE_B=99\ J$

Explanation:

Given:

Let mass of the particle B be, $m_B=m$

then the mass of particle A, $m_A=3m$

Energy stored in the compressed spring, $E=132\ J$

Now when the compression of the particles with the spring is released, the spring potential energy must get converted into the kinetic energy of the particles and their momentum must be conserved.

Kinetic energy:

$\frac{1}{2}m_A.v_A^2+\frac{1}{2}m_B.v_B^2=132$

$3m.v_A^2+m.v_B^2=264$ .............................(1)

<u>Using the conservation of linear momentum:</u>

$m_A.v_A+m_B.v_B=0$

$3m.v_A+m.v_B=0$ .............................(2)

Put the value of $v_A$ from eq. (2) into eq. (1)

$3m\times (\frac{-v_B}{3})^2+m.v_B^2=264$

$v_B^2=\frac{198}{m}$ ...........................(3)

<u>Now the kinetic energy of particle B:</u>

$KE_B=\frac{1}{2}\times m_B\times v_B^2$

$KE_B=\frac{1}{2}\times m\times \frac{198}{m}$

$KE_B=99\ J$

Put the value of $v_B^2$ form eq. (3) into eq. (1):

$v_A^2=\frac{22}{m}$

<u>Now the kinetic energy of particle A:</u>

<u /> $KE_B=\frac{1}{2}\times 3m\times \frac{22}{m}$ <u />

$KE_A=33\ J$

6 0

3 years ago

Human perception of the frequency of sound waves is called:

Olin [163]

Answer:

The study of the human perception of sound is called psychoacoustics.

Explanation:

5 0

3 years ago

For what angle of projetion is the range is maximum

VladimirAG [237]

Answer:

46 degrees

Explanation:

8 0

3 years ago

An object is pulled along the ground by exerting a force of 35 pounds a rope that makes a 10° angle with the ground. How much wo

GREYUIT [131]

<h2>Answer:</h2>

Workdone = 601.33 Nm

<h2>Explanation:</h2>

Workdone = Force(F) x horizontal distance(d).

Wordone = F x d

Where F is the horizontal component of the force

Force = 35 pounds

Recall that;

1 pound force = 4.4N

=> 35 pounds = 35 x 4.4N = 154N

=> Force = 154N

But since the rope makes a 10⁰ angle with the ground,

The horizontal component of the force F is Force x cos10⁰

=> F = 154 cos 10⁰

=> F = 154 x 0.9848

=> F = 151.66N

Horizontal distance (d) = 13 feet

Recall that;

1 foot = 0.305m

=> 13 feet = 13 x 0.305m = 3.965m

=> d = 3.965m

=> Substituting the values of F and d into the equation above, we have;

Workdone = F x d

Workdone = 151.66N x 3.965m

Workdone = 601.33Nm

8 0

3 years ago

What is the maximum frictional force in the knee joint of a person who supports 76. 0 kg of her mass on that knee?

Bad White [126]

The maximum frictional force in the knee joint of a person who supports 76. 0 kg of her mass on that knee 119.28N.

Given,

m=76kg, μ=0.16, g=9.81 m/ $s^{2}$

frictional force = mμg= 76*0.16*9.81= 119.28 N

<h3>Frictional force</h3>

An opposing force to the relative motion of two bodies in contact is known as frictional force. Always acting in the opposite direction from the direction of motion, frictional force is applied to a moving body. Because it resists motion, it aids in lowering the moving object's speed. The force is one of touch. Four broad categories can be made for the force of friction depending on the sort of motion that occurs between the two objects. Static friction is the force of friction between an object and the surface it is put on.

What is the maximum frictional force in the knee joint of a person who supports 76. 0 kg of her mass on that knee?

Learn more about frictional force here:

brainly.com/question/13707283

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4 0

2 years ago