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A pendulum has 895 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o

LuckyWell [14K]

Newton's law of conservation states that energy of an isolated system remains a constant. It can neither be created nor destroyed but can be transformed from one form to the other.

Implying the above law of conservation of energy in the case of pendulum we can conclude that at the bottom of the swing the entire potential energy gets converted to kinetic energy. Also the potential energy is zero at this point.

Mathematically also potential energy is represented as

Potential energy= mgh

Where m is the mass of the pendulum.

g is the acceleration due to gravity

h is the height from the bottom z the ground.

At the bottom of the swing,the height is zero, hence the potential energy is also zero.

The kinetic energy is represented mathematically as

Kinetic energy= 1/2 mv^2

Where m is the mass of the pendulum

v is the velocity of the pendulum

At the bottom the pendulum has the maximum velocity. Hence the kinetic energy is maximum at the bottom.

Also as it has been mentioned energy can neither be created nor destroyed hence the entire potential energy is converted to kinetic energy at the bottom and would be equivalent to 895 J.

7 0

3 years ago

What is the average de Broglie wavelength of oxygen molecules in air at a temperature of 27°C? Use the results of the kinetic th

asambeis [7]

Answer:

$\lambda = 2.57 \times 10^{-11} m$

Explanation:

Average velocity of oxygen molecule at given temperature is

$v_{rms} = \sqrt{\frac{3RT}{M}}$

now we have

M = 32 g/mol = 0.032 kg/mol

T = 27 degree C = 300 K

now we have

$v_{rms} = \sqrt{\frac{3(8.31)(300)}{0.032}$

$v_{rms} = 483.4 m/s$

now for de Broglie wavelength we know that

$\lambda = \frac{h}{mv}$

$\lambda = \frac{6.6 \times 10^{-34}}{(5.31\times 10^{-26})(483.4)}$

$\lambda = 2.57 \times 10^{-11} m$

7 0

2 years ago

What is the weight of an 1000 gram mass

Reil [10]

That depends on where you weigh it.

-- On Earth, it weighs 9.807 newtons (2.205 pounds).

-- On the moon, it weighs 1.623 newtons (5.84 ounces).

-- On Jupiter, it weighs 24.79 newtons (5.57 pounds).

BTW ... 1,000 grams of mass is called ' one kilogram '.

8 0

3 years ago

Read 2 more answers

A sling is used to give a stone an initial velocity of 20 at an angle of 30 above the horizontal. The stone travels through the

Luba_88 [7]

Answer:

Option E is correct.

There must be a horizontal wind opposite the direction of the stone's motion, because ignoring air resistance when calculating the horizontal range would yield a value greater than 32 m.

Explanation:

Normally, ignoring air resistance, for projectile motion, the range (horizontal distance teavelled) of the motion is given as

R = (u² sin 2θ)/g

where

u = initial velocity of the projectile = 20 m/s

θ = angle above the horizontal at which the projectile was launched = 30°

g = acceleration due to gravity = 9.8 m/s²

R = (30² sin 60°) ÷ 9.8

R = 78.53 m

So, Normally, the stone should travel a horizontal distance of 78.53 m. So, travelling a horizontal distance of 32 m (less than half of what the range should be without air resistance) means that, the motion of the stone was impeded, hence, option E is correct.

There must be a horizontal wind opposite the direction of the stone's motion, because ignoring air resistance when calculating the horizontal range would yield a value greater than 32 m.

Hope this Helps!!!

7 0

3 years ago

Calculate the maximum deceleration (in m/s2) of a car that is heading down a 14° slope (one that makes an angle of 14° with the

lesya [120]

The question is incomplete. Here is the complete question.

Calculate the maximum deceleration of a car that is heading down a 14° slope (one that makes an anlge of 14° with the horizontal) under the following road conditions. You may assum that the weight of the car is evenlydistributed on all four tires and that the sttic coefficient of friction is involved - that is, the tires are not allowed to slip during the deceleration. (Ignore rolling) Calculate for a car: (a) On a dry concrete. (b) On a wet concrete. (c) On ice, assuming that μs = 0.100, the same as for shoes on ice.

Answer: (a) a = - 11.05 m/s²; (b) a = - 10.64 m/s²; (c) a = - 9.84m/s²

Explanation: The image in the attachment describe the forces acting on the car. Observing that, we know that:

$F_{net}$ = - $W_x$ - $f_s$

The $W_x$ is a x-component of force due to gravity (W) and, in this case, is given by: $W_x$ = W.sin(14)

W is described as: W = m.g

Force due to friction ( $f_s$ ) is given by: $f_s$ = μs.N

N is the normal force and, in the system, is equivalent of $W_y$ , so:

$W_y$ = m.g.cos(14)

Therefore, the formula will be:

$F_{net}$ = - $W_x$ - $f_s$

m.a = - (m.g.sin14) - (μs.mg.cos14)

a = - g (sin14 + μscos 14)

a) For dry concrete, μs = 1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 1.cos14)

a = - 11.05 m/s²

b) For wet concrete, μs = 0.7:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin 14 + 0.7.cos14)

a = - 10.64 m/s²

c) For ice, μs = 0.1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 0.1cos14)

a = - 9.84 m/s²

3 0

3 years ago