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4 years ago

12

If you toss a ball vertically upward in a uniformly moving train, it returns to its starting place. Will it do the same if the t

rain is accelerating?

1 answer:

gogolik [260]4 years ago

3 0

Answer:No

Explanation:

No

As the train is accelerating so train velocity will be more as compared to the ball and thus will cover more distance as compared to the ball.

When the ball is thrown upward with some velocity, it also possesses the train velocity at the time of throwing but as time passes velocity of train increases due to acceleration of the train. This causes the ball to fall behind the point of launch.

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What is true about the force between charges?

shusha [124]

Answer:

The statements A and D are true

The statements B and C are false

Explanation:

The force between charges can be explained by the Coulomb's Law.

According to Coulomb:

1 - Like charges always repel each other

2- Unlike or opposite charges always attrack each other

3- Force between 2 charges is given by:

$F=k\frac{q_1q_2}{r^2}$

where F is the force between 2 charges and r is the distance between 2 charges.

We can see that Force F is inversely proportional to the distance r

Which means that when F increase, r decreases and when F decreases, r increases

5 0

4 years ago

Read 2 more answers

A cylinder with moment of inertia I1 rotates with angular speed ω0 about a frictionless vertical axle. A second cylinder, with m

MAXImum [283]

Answer:

Part(a): The final angular velocity is $\omega_{f} = \dfrac{I_{1}\omega_{i}}{(I_{1} + I_{2})}$

Part(b): The ratio of the rotational energies is $\dfrac{k_{f}}{k_{i}}& = \dfrac{I_{1}}{(I_{1} + I_{2})}$ ,showing the the energy of th system will decrease.

Explanation:

Part(a):

If ' $I$ ' be the moment of inertia of an object and ' $\omega$ ' be its angular velocity then the angular momentum ' $L$ ' of the object can be written as

$L = I \omega$

If ' $I_{1}$ ' and ' $I_{2}$ ' be the moment of inertia of the two cylinders and ' $\omega_{1}$ ' and ' $\omega_{2}$ ' be the initial angular velocity of the cylinders and ' $\omega_{1}{'}$ ' and ' $\omega_{2}^'}$ ' be their respective final angular velocity, then from conservation of angular momentum,

$I_{1} \omega_{1} + I_{2} \omega_{2} = I_{1} \omega_{1}^{'} + I_{2} \omega_{2}^{'}$

Given, $\omega_{1} = \omega_{i},~\omega_{2} = 0,~\omega_{1}^{'} = \omega_{2}^{'} = \omega_{f}$ . From the above expression

$&& I_{1} \omega_{i} = (I_{1} + I_{2}) \omega_{f}\\&or,& \omega_{f} = \dfrac{I_{1}\omega_{i}}{(I_{1} + I_{2})}$

Part(b):

Initial kinetic energy

$K_{i} = \dfrac{1}{2} I_{1} \omega_{i}^{2}$

and Final kinetic energy

$K_{f} = \dfrac{1}{2}(I_{1} + I_{2}) \omega_{f}^{2}$

Substituting the value of $\omega_{f}$ ,

$&& K_{f} = \dfrac{1}{2}(I_{1} + I_{2})\dfrac{I_{1}^{2}\omega_{i}^{2}}{(I_{1} + I_{2})^{2}} = \dfrac{1}{(I_{1} + I_{2})} \dfrac{1}{2}I_{1}\omega_{i}^{2} = \dfrac{1}{(I_{1} + I_{2})} K_{i}\\&\dfrac{k_{f}}{k_{i}}& = \dfrac{I_{1}}{(I_{1} + I_{2})}$

The above expression shows that the ebergy of the system will decrease.

7 0

3 years ago

The water line from the street to my house is 1 inch diameter and made of PVC (i.e. smooth). The line is roughly 450 ft long. Th

jekas [21]

Answer:

The right solution is "126 Psi".

Explanation:

The given values are:

P₁ = 130 psig

i.e.,

= $130\times 6.894$

= $896.22 \ Kpa$

or,

= $896.22\times 10^3 \ Pa$

Z₂ = 10ft

= 3.05 m

$\delta$ = 1000 kg/m³

According to the question,

Z₁ = 0

V₁ = V₂

As we know,

⇒ $\frac{P_1}{\delta_g} +\frac{V_1^2}{2g} +Z_1=\frac{P_2}{\delta_g} +\frac{V_2^2}{2g} +Z_2$

On substituting the values, we get

⇒ $\frac{P_1}{\delta_g} +0+0=\frac{P_2}{\delta_g} +0+Z_2$

⇒ $\frac{896.22\times 10^3}{1000\times 9.8} =\frac{P_2}{1000\times 9.8} +3.05$

⇒ $P_2=866330 \ P_a$

i.e.,

⇒ $=866330\times 0.000145$

⇒ $=126 \ Psi$

8 0

3 years ago

A bullet is shot from a rifle with a speed of 3,015 feet per second. Assuming the billet moves at a constant velocity what is th

VMariaS [17]

Answer:

Explanation:

Given

final speed v = 3015ft/s

initial speed = 0m/s

Distance S = 4146m/s

Required

Time

Using the equation of motion to get the time;

S = (v+u)/2 * t

since 1m = 3.28084ft'

4146m = 3.28084 * 4146 = 13,602.36264feet

13,602.36264 = 3015+0/2 * t

13,602.36264 = 1,507.5t

t = 13,602.36264/ 1,507.5

t = 9.023 secs

Hence it will take 9.023 seconds the rescuer die the biller to strike a target

3 0

3 years ago

How does the comet's energy change as it moves from point B to point C?

Oksanka [162]

B. Kinetic energy increase, gravitational potential energy decreases

5 0

3 years ago