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Zarrin [17]
2 years ago
5

A 1 kg flashlight is dropped from rest and falls to the floor without air resistance . At the point during its fall, when it is

0.7 meters above the floor, its potential energy exactly equals its kinetic energy. How fast is the flashlight moving at this point, in m/s 3.7
Physics
1 answer:
alexgriva [62]2 years ago
8 0

Answer:

The speed of the flashlight at that point is 3.7 m/s

Explanation:

When an object of mass M is at a height H above the ground, the potential energy of the object is:

U = M*H*g

Where g is the gravitational acceleration, g = 9.8 m/s^2

And for an object with velocity v, the kinetic energy is:

K = (M/2)*v^2

We know that when the flashlight of mass  1kg is 0.7 meters above the ground, the potential energy is equal to the kinetic energy, then:

M = 1kg

H = 0.7m

g = 9.8 m/s^2

Replacing these in the equations, we get:

U = K

(1kg)*(0.7m)*(9.8 m/s^2) = ((1kg)/2)*v^2

As the mass factor appears in both sides, we can remove it:

(0.7 m)*(9.8 m/s^2) = (v^2)/2

Now we can multiply both sides by 2:

2*(0.7 m)*(9.8 m/s^2) = v^2

Now let's apply the square root to both sides:

√(2*(0.7 m)*(9.8 m/s^2)) = v = 3.7 m/s

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A sled of mass 10 kg slides along the ice. it has an initial speed of 2 m/s but stops because of friction. How much work is done
NeX [460]

Answer: The correct answer is option B.

Explanation:

Mass of the sled = 10 kg

Initial speed of the sled = 2 m/s

Kinetic energy of the sled = \frac{1}{2}mv^2

\frac{1}{2}\times 10 kg\times (2 m/s)^2=20 Joules

Work done by the sled = 20 joules

The work done by the friction will be in opposite direction and equal to the magnitude of the work done of the sled that - 20 J.

Hence, correct answer is option B.

6 0
3 years ago
Read 2 more answers
A 1430 kg car speeds up from 7.50 m/s to 11.0 m/s in 9.30 s. Ignoring friction, how much power did that require?(unit=W)PLEASE H
Aleks04 [339]

Answer:

Kinetic energy = (1/2) (mass) (speed²)

Original KE = (1/2) (1430 kg) (7.5 m/s)²  =  40,218.75 joules

Final KE  =  (1/2) (1430 kg) (11.0 m/s)²  =   86,515 joules

Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules

Power = work/time = 46,296.25 joules / 9.3 sec  =  4,978.1 watts .

Explanation:

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3 0
2 years ago
In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward
Vladimir79 [104]

Answer:

The answer is "1.01 \times 10^{-13}"

Explanation:

Using the law of conservation for energy. Equating the kinetic energy to the potential energy.

KE=U=\frac{kqq'}{r}\\\\

Calculating the closest distance:

\to r=\frac{kqq'}{KE}\\\\

=\frac{k(2e)(79e)}{KE}\\\\=\frac{k(2)(79)e^2}{KE}\\\\=\frac{9.0\times 10^9 \ N \cdot \frac{m^2}{c}(2)(79)(1.6 \times10^{-19} \ C)^2}{(2.25\ meV) (\frac{1.6 \times 10^{-13} \ J}{1 \ MeV})}\\\\

=\frac{9.0\times 10^9 \times 2\times 79\times 1.6 \times10^{-19}\times 1.6 \times10^{-19} }{(2.25 \times 1.6 \times 10^{-13}) }\\\\=\frac{3,640.32\times 10^{-29}}{3.6 \times 10^{-13} }\\\\=\frac{3,640.32}{3.6} \times 10^{-16}\\\\=1011.2 \times 10^{-16}\\\\=1.01 \times 10^{-13}

5 0
3 years ago
A certain computer chip that has dimensions of 3.67 cm and 2.93 cm contains 3.5 million transistors. If the transistors are squa
Evgesh-ka [11]

Answer:

1.56 × 10^-3 cm.

Explanation:

So, we are given the following parameters from the question above;

Length = 3.67 cm, breadth = 2.93 cm, and the number of embedded transistors = 3.5 million.

Step one: find the area of the computer chip.

Therefore, Area = Length × breadth.

Area = 3.67 cm × 2.93 cm.

Area of the computer chip = 10.7531 cm^2. = 10.75 cm^2.

Step two: find the area of one transistor

The area of one transistor is; (area of the computer chip) ÷ (number of embedded transistors).

Hence;

The area of one transistor= 10.7531/4.4 × 10^6.

The area of one transistor= 2.44 × 10^-6 cm^2.

=> Note that We have our transistors as square, therefore;

The maximum dimension = √ (2.44 × 10^-6) cm^2.

The maximum dimension= 1.56 × 10^-3 cm.

7 0
3 years ago
Find the momentum of a 500,000 kg train that is stopped on the tracks?
Aneli [31]

Answer:

The answer should be A) 0m/s

Explanation:

It is stopped on the train tracks therefore it is not moving.

Please tell me if I am wrong because I'm not 100% sure on this. Hope it's right and that it helped you.

5 0
3 years ago
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