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3 years ago

A high speed train travels with an average speed of 227 km/h. The train travels for 2 h.how far does the train travel in meters?

Physics

2 answers:

Sedbober [7]3 years ago

8 0

2 x 227 km/h = 454 m

ella [17]3 years ago

8 0

Answer:

$\boxed {\boxed {\sf 454 \ kilometers}}$

Explanation:

Distance is the product of speed and time.

$d=s*t$

The speed of the train is 227 kilometers per hour. The time is 2 hours.

$s= 227 \ km/hr \\t= 2 \ hr$

Substitute the values into the formula.

$d= 227 \ km/hr * 2 \ hr$

Multiply and note that the hours will cancel each other out.

$d= 227 \ km* 2$

$d=454 \ km$

The train travels <u>454 kilometers</u> in 2 hours.

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Answer:

If density is greater, the object sinks. Saturn is mainly composed of the lightest two gases known, hydrogen and helium. It is the only planet in our solar system whose density is less than water

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2 years ago

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A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi

Dmitriy789 [7]

Answer:

$A = 0.081 \ m$

The value is $u = 0.2569 \ m/s$

Explanation:

From the question we are told that

The mass is $m = 0.750 \ kg$

The spring constant is $k = 17.5 \ N/m$

The instantaneous speed is $v = 39.0 \ cm/s= 0.39 \ m/s$

The position consider is x = 0.750A meters from equilibrium point

Generally from the law of energy conservation we have that

The kinetic energy induced by the hammer = The energy stored in the spring

$\frac{1}{2} * m * v^2 = \frac{1}{2} * k * A^2$

Here a is the amplitude of the subsequent oscillations

=> $A = \sqrt{\frac{m * v^ 2 }{ k} }$

=> $A = \sqrt{\frac{0.750 * 0.39 ^ 2 }{17.5} }$

=> $A = 0.081 \ m$

Generally from the law of energy conservation we have that

The kinetic energy by the hammer = The energy stored in the spring at the point considered + The kinetic energy at the considered point

$\frac{1}{2} * m * v^2 = \frac{1}{2} * k x^2 + \frac{1}{2} * m * u^2$

=> $\frac{1}{2} * 0.750 * 0.39^2 = \frac{1}{2} * 17.5* 0.750(0.081 )^2 + \frac{1}{2} * 0.750 * u^2$

=> $u = 0.2569 \ m/s$

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2 years ago

In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it

bogdanovich [222]

Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{2gh}$

$=\sqrt{2 \times 9.8 \times 0.35}$ (h = 35 cm = 0.35 m)

= 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $$v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}$

$=\frac{0.175}{0.525}+0$

= 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

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3 years ago

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