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1 year ago

The acceleration of a train moving from rest and its speed reaches 36m/sec in 9 sec

Physics

1 answer:

jenyasd209 [6]1 year ago

4 0

Answer:

4 m/s^2

Explanation:

The acceleration is defined as: Δv/Δt (the difference of the velocity over a time period in which happens that difference).

Remember that a difference is calculated by subtracting the initial value of a physical quantity from its final value.

In our case:

Δv = Vfinal - Vinitial = 36m/s - 0 m/s = 36m/s

Δt = 9s

a = Δv/Δt = 36m/s / 9s = 4m/s^2

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What single property was the most important in Jesseca’s material?

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Answer:

Jesseca wanted to create a material that reflected most of the light that fell on it.

Explanation:

Plato Answer

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2 years ago

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In the steady state 1.2 ✕ 1018 electrons per second enter bulb 1. There are 6.3 ✕ 1028 mobile electrons per cubic meter in tungs

bekas [8.4K]

Answer:

E=12.2V/m

Explanation:

To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

The equation is given by,

$V=\frac{I}{nAq}$

Where,

V= Drift Velocity

I= Flow of current

n= number of electrons

q = charge of electron

A = cross-section area.

For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is

$\frac{I}{q} = 1.2*10^{18}$

$A= 1.3*10^{-8}m^2$

$n=6.3*10^{28} e/m^3$

$\omicron{O} = 1.2*10^{-4}(m/s)(N/c)$ Mobility

We can find the drift velocity replacing,

$V = \frac{1.2*10^{18}}{(1.3*10^{-8})(6.3*10^{28})}$

$V= 1.465*10^-3m/s$

The electric field is given by,

$E= \frac{V}{\omicron{O}}$

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$E=12.2V/m$

7 0

3 years ago

Here on Earth you hang a mass from a vertical spring and start it oscillating with amplitude 1.9 cm. You observe that it takes 3

Vlada [557]

Answer:

T = 3.23 s

Explanation:

In the simple harmonic movement of a spring with a mass the angular velocity is given by

w = √ K / m

With the initial data let's look for the ratio k / m

The angular velocity is related to the frequency and period

w = 2π f = 2π / T

2π / T = √ k / m

k₀ / m₀ = (2π / T)²

k₀ / m₀ = (2π / 3.0)²

k₀ / m₀ = 4.3865

The period on the new planet is

2π / T = √ k / m

T = 2π √ m / k

In this case the amounts are

m = 6 m₀

k = 10 k₀

We replace

T = 2π√6m₀ / 10k₀

T = 2π √6/10 √m₀ / k₀

T = 2π √ 0.6 √1 / 4.3865

T = 3.23 s

5 0

3 years ago

Your answer should be precise to 0.1 m/s. Use a gravitational acceleration of 10 m/s/s. At it lowest point, a pendulum is moving

saw5 [17]

Explanation:

It is given that,

Speed, v₁ = 7.7 m/s

We need to find the velocity after it has risen 1 meter above the lowest point. Let it is given by v₂. Using the conservation of energy as :

$\dfrac{1}{2}mv_1^2=\dfrac{1}{2}mv_2^2+mgh$

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$v_2^2=(7.7)^2-2\times 10\times 1$

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So, the velocity after it has risen 1 meter above the lowest point is 6.26 m/s. Hence, this is the required solution.

4 0

3 years ago

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I need help please, thank you!!!

stiks02 [169]

Answer:

135 mph

because 7+9=16

8 is left

if 7=40

9=50

then 8=45

add 40+50+45mph

=135

4 0

2 years ago

The acceleration of a train moving from rest and its speed reaches 36m/sec in 9 sec​

The acceleration of a train moving from rest and its speed reaches 36m/sec in 9 sec