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3 years ago

How fast must a train accelerates from rest to cover 518 m in the first 7.48 s?

Physics

1 answer:

Rus_ich [418]3 years ago

8 0

Heya!

For this problem, use the formula:

s = Vo * t + (at^2) / 2

Since the initial velocity is zero, the formula simplifies like this:

s = (at^2) / 2

Clear a:

2s = at^2

(2s) / t^2 = a

a = (2s) / t^2

Data:

s = Distance = 518 m

t = Time = 7,48 s

a = Aceleration = ¿?

Replace according formula:

a = (2*518 m) / (7,48 s)^2

Resolving:

a = 1036 m / 55,95 s^2

a = 23,34 m/s^2

The aceleration must be <u>23,34 meters per second squared</u>

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Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.320 mm wide. The diffraction pattern is observed

DanielleElmas [232]

Answer:

$W = 10.28\ mm$

Explanation:

Given,

Red light wavelength = 633 nm

width of slit = 0.320 mm

distance,d = 2.60 m

Condition of first maximum

$a sin \theta_1 = m\lambda$

$\theta_1 =sin^{-1}(\dfrac{m\lambda}{a})$

m = 1

$\theta_1 =sin^{-1}(\dfrac{633\times 10^{-9}}{0.32\times 10^{-3}})$

$\theta_1 = 0.1133^\circ$

Width of the first minima

$y_1 = L tan \theta_1$

$y_1 = 2.60\times tan( 0.11331)$

$y_1 = 5.14 \ mm$

Now, width of the central region

$W = 2 y_1$

$W = 2\times 5.14$

$W = 10.28\ mm$

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3 years ago

5. A body travels in a circle at a constant speed, it

nata0808 [166]

A the same acceleration

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3 years ago

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A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan

mariarad [96]

Answer:

d = 9.69 cm

Explanation:

given,

mass of the block = 1.2 Kg

spring force constant(k) = 730 N/m

spring is compressed = d = ?

rough patch width = 5 cm

μ_k = 0.44

work done by friction = energy lost

$\mu_k mg \times x = \dfrac{1}{2}kd^2-\dfrac{1}{2}mv^2$

$0.44\times 1.2 \times 9.8 \times 0.05 = \dfrac{1}{2}\times 730 \times d^2-\dfrac{1}{2}\times 1.2 \times 2.3^2$

$3.432 = 365 d^2$

$d = \dfrac{3.432}{365}$

d = 0.0969 m

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3 years ago

one arm of a u shaped tube contains water and the other alcohol. if the two fluids meet exactly at the bottom of the U and the a

Tema [17]

Complete Question

One arm of a U-shaped tube (open at both ends) contains water, and the other alcohol.

If the two fluids meet at exactly the bottom of the U, and the alcohol is at a height of 18 cm, at what height will the water be?Assume the density of alcohol is $\rho_a = 790\ kg/m^3$

Answer:

The height of water is $h_w = 0.142 \ m$

Explanation:

From the question we are told that

The height of the alcohol is $h_a =18 \ cm = 0.18 \ m$

The density of the alcohol is $\rho_a = 790\ kg/m^3$

Generally the pressure on both arm of the tube are equal given that they are both open

i,e $P_a = P_w$

Where $P_a$ is pressure of alcohol and $P_w$ is pressure of water

So the pressure on the arm of the tube containing the alcohol is mathematically evaluated as

$P_a = g * h * \rho$

substituting values

$P_a =9.8 * 0.18 * 790$

$P_a = 1394 \ Pa$

Generally the pressure on the arm of the tube containing the water is mathematically evaluated as

$P_w = g * h_w * \rho_w$

where $\rho_w$ is the density of water which has a value $\rho _w = 1000 \ kg/m^3$

$1394 = 9.8 * h_w * 1000$

=> $h_w = \frac{1394}{9800}$

=> $h_w = 0.142 \ m$

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3 years ago

The earth revolves around the sun in its orbit as shown in the figure. Is there any difference in the amount of gravitational fo

Bingel [31]

Answer:

No.

Explanation:

There is no change in gravitational force based on the sizes remaining the same. If there was a change in gravitational force the orbits of all planets would change.

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3 years ago