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Leya [2.2K]
4 years ago
7

Suppose an object’s initial velocity is 10 m/s and its final velocity is 4 m/s. Mass is constant. What can best be concluded abo

ut the object based on the work-energy theoremWork is positive, the environment did work on the object, and the energy of the system increases.
Work is positive, the object did work on the environment, and the energy of the system increases.
Work is negative, the object did work on the environment, and the energy of the system decreases.
Work is negative, the environment did work on the object, and the energy of the system decreases.
Physics
2 answers:
grandymaker [24]4 years ago
8 0
Correct answer is:
"Work is negative, the object did work on the environment, and the energy of the system decreases. "

In fact, the work-energy theorem states that the work done is equal to the variation of kinetic energy:
W=\Delta K=K_f - K_i
where W is the work, Kf the final kinetic energy and Ki the initial kinetic energy. Since the kinetic energy depends on the velocity v by:
K= \frac{1}{2} mv^2
and since the final velocity is less than the initial velocity, \Delta K is negative (the kinetic energy of the system is decreased), and the work is negative.This also means that the object did work on the environment: in fact, by doing work, the object gave part of its kinetic energy to the environment, and so its kinetic energy decreased.
serg [7]4 years ago
7 0

C. Work is negative, the object did work on the environment, and the energy of the system decreases

pls give brainliest

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4 years ago
A mass of 4kg suspended by a light string 2m long and at rest is projected horizontally with a velocity of 1.5 m/s. find the ang
Dafna11 [192]

Answer:

19.5°

Explanation:

The energy of the mass must be conserved. The energy is given by:

1) E=\frac{1}{2}mv^2+mgh

where m is the mass, v is the velocity and h is the hight of the mass.

Let the height at the lowest point of the be h=0, the energy of the mass will be:

2) E=\frac{1}{2}mv^2

The energy when the mass comes to a stop will be:

3) E=mgh

Setting equations 2 and 3 equal and solving for height h will give:

4) h=\frac{v^2}{2g}

The angle ∅ of the string with the vertical with the mass at the highest point will be given by:

5) cos\phi=\frac{l-h}{l}

where l is the lenght of the string.

Combining equations 4 and 5 and solving for ∅:

6) \phi={cos}^{-1}(\frac{l-h}{l})={cos}^{-1}(1-\frac{h}{l})={cos}^{-1}(1-\frac{v^2}{2gl})

8 0
4 years ago
A billiard ball moving at 5 m/s strikes a stationary ball of the same mass. After the collision, the original ball moves at a ve
SIZIF [17.4K]

90 degrees - 30 = 60 degrees

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8 0
3 years ago
A particle with a charge of −1.24×10−8c is moving with instantaneous velocity v⃗ = (4.19×104m/s)i^ + (−3.85×104m/s)j^ . part a w
Bezzdna [24]

Answer:

F = 0i (in the x-direction), 0j (in the y-direction),-8.59*10^-4 N k (In the z-direction)

Explanation:

The force given by charged particles moving in a magnetic field is given below (cross is cross product, they don't have that format in the equation tool):

F=qv (cross) B\\

Now we can perform the cross product between v and B

v(cross)B = \left[\begin{array}{cc}4.19*10^{4} &-3.85*10^{4}\\1.8&0&\en[tex]v(cross)B = 69300 (kg*m/(s^2*C))\\d{array}\right][/tex]

Now multiply by Q (charge) to get the force

F = -1.24*10^-8 * 69300\\F = -8.59*10^-4N

F = -8.59*10^-4 N k

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6 0
3 years ago
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Leno4ka [110]

Answer:

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Explanation:

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4 0
4 years ago
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