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kifflom [539]
1 year ago
13

A ball is tossed in the air and released. It moves up, reverses direction, falls back down again, and is caught at the same heig

ht it was released. Considering the time interval after the ball is released and before it is caught, when does the gravitational potential energy of the ball have its maximum value?
Physics
2 answers:
PolarNik [594]1 year ago
8 0

Answer:

The potential energy has a  maximum when the ball is a time that is half of the time for total travel

Explanation:

Generally potential energy is a the varies directly with the height according to this formula

            PE =mgh

and the ball attains a maximum height when the time is equal to half of the total time taken to travel  

Anni [7]1 year ago
3 0

Answer:

The potential energy is highest at it point of Max height...

Explanation:

From the formula potential energy = mgh...

The greater the height, the greater the potential energy.

So considering the time interval it's at the time it takes the ball to reach the reversal point ( its Max height)

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Two trucks with equal mass are attracted to each other with a gravitational force of 5.3 x 10 -4 N. The trucks are separated by
HACTEHA [7]

Answer:

7327 kg or 7.3 tons

Explanation:

We have Newton formula for attraction force between 2 objects with mass and a distance between them:

F_G = G\frac{M_1M_2}{R^2}

where G =6.67408*10^{-11} m^3/kgs^2 is the gravitational constant on Earth. M = M_1 = M_2 is the masses of the 2 objects. and R = 2.6m is the distance between them.

F_G = 5.3*10^{-4}N is the attraction force.

5.3*10^{-4} = 6.67408*10^{-11}\frac{M^2}{2.6^2}

7941265 = \frac{M^2}{2.6^2}

M^2 = 53682948.76

M = \sqrt{53682948.76} \approx 7327 kg or 7.3 tons

3 0
2 years ago
In a closed system, the loss of momentum of one object_____ the gain in momentum of another object.
densk [106]

In a closed system, the loss of momentum of one object  is same as________ the gain in momentum of another object

according to law of conservation of momentum, total momentum before and after collision in a closed system in absence of any net external force, remains conserved . that is

total momentum before collision = total momentum after collision

P₁ + P₂ = P'₁ + P'₂

where P₁ and P₂ are momentum before collision for object 1 and object 2 respectively.

P'₁ - P₁  = - (P'₂ -  P₂)

so clearly gain in momentum of one object is same as the loss of momentum of other object

6 0
2 years ago
You’ve been given the challenge of balancing a uniform, rigid meter-stick with mass M = 95 g on a pivot. Stacked on the 0-cm end
Mariulka [41]

Answer: d = 4750n/3.1+95n

Explanation:

Using the principle of moment to solve the question.

Sum of clockwise moments = sum of anti clockwise moments

Since there are n identical coins with mass 3.1g placed at point 0cm, 1 coin will have mass of 3.1/n grams

Taking moment about the pivot,

Mass 3.1/n grams will move anti-clockwisely while the mass 95g will move in the clockwise direction.

Since its a meter rule (100cm) the distance from the center mass(95g) to the pivot will be 50-d (check attachment for diagram).

To get 'd'

We have 3.1/n × d = 95 × (50-d)

3.1d/n = 4750-95d

3.1d = 4750n-95dn

3.1d+95dn=4750n

d(3.1+95n) = 4750n

d = 4750n/3.1+95n

6 0
2 years ago
You are designing a generator with a maximum emf 8.0 V. If the generator coil has 200 turns and a cross-sectional area of 0.030
shutvik [7]

Answer:

7.1 Hz

Explanation:

In a generator, the maximum induced emf is given by

\epsilon= 2\pi NAB f

where

N is the number of turns in the coil

A is the area of the coil

B is the magnetic field strength

f is the frequency

In this problem, we have

N = 200

A=0.030 m^2

\epsilon=8.0 V

B = 0.030 T

So we can re-arrange the equation to find the frequency of the generator:

f=\frac{\epsilon}{2\pi NAB}=\frac{8.0 V}{2\pi (200)(0.030 m^2)(0.030 T)}=7.1 Hz

4 0
1 year ago
Read 2 more answers
What is the total kinetic energy of a 0.15 kg hockey puck sliding at 0.5 m/s and rotating about its center at 8.4 rad/s? The dia
ycow [4]
The mass of the puck is
m = 0.15 kg.
The diameter of the puck is 0.076 m, therefore its radius  is
r = 0.076/2 = 0.038 m
The sliding speed is
v = 0.5 m/s
The angular velocity is
ω = 8.4 rad/s

The rotational moment of inertia of the puck is
I = (mr²)/2
  = 0.5*(0.15 kg)*(0.038 m)²
  = 1.083 x 10⁻⁴ kg-m²

The kinetic energy of the puck is the sum of the translational and rotational kinetic energy.
The translational KE is
KE₁ = (1/2)*m*v²
       = 0.5*(0.15 kg)*(0.5 m/s)²
       = 0.0187 j

The rotational KE is
KE₂ = (1/2)*I*ω²
       = 0.5*(1.083 x 10⁻⁴ kg-m²)*(8.4 rad/s)²
       = 0.0038 J

The total KE is
KE = 0.0187 + 0.0038 = 0.0226 J

Answer: 0.0226 J


4 0
1 year ago
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