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deff fn [24]
3 years ago
14

Does the collision between the cart and the brick follow the law of momentum conservation? Make a claim (yes or no) and support

the claim by describing the evidence and reasoning that that supports it
Physics
1 answer:
murzikaleks [220]3 years ago
7 0

Answer:

yes it does

Explanation:

Going by the law of conservation of momentum, when two objects collides, the momentum of one object is transferred to the other object.

This collision can be elastic or inelastic coliision

let the mass of the brick be m1

the mass of the cart be m2

the velocity (initial and final) of the brick  u1 and v1

the velocity (initial and final) of the cart u2 and v2

For elastic collision

m1u1+m2u2= m1v1+m2v2

For inelastic collision the two bodies will move with the same velocity after impact

m1u1+m2u2=(m1+m2)V

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Option A

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An ultra intense laser is one with which intensities greater than 1015 W cm-2 can be achieved.

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This intensity, which was the upper limit of lasers until the invention of the Chirped Pulse Amplification, CPA technique, is the value around which nonlinear effects on the transport of radiation in materials begin to appear.

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3 years ago
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Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.​
Tems11 [23]

Answer:

<em>The range is 35.35 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and g=9.8m/s^2 the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

\displaystyle d={\frac  {v_o^{2}\sin(2\theta )}{g}}

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

\displaystyle d={\frac  {20^{2}\sin(2\cdot 30^\circ )}{9.8}}

\displaystyle d={\frac  {400\sin(60^\circ )}{9.8}}

d=35.35\ m

The range is 35.35 m

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3 years ago
Which of the waves has the smallest amplitude?
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3 years ago
To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel
alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.  

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0
3 years ago
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