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3 years ago

Until a train is a safe distance from the station, it must travel at 5 m/s. Once the train is on open track, it can speec

Physics

1 answer:

nadezda [96]3 years ago

4 0

Answer:

The acceleration of the train is 5 m/s².

Explanation:

Given:

let the initial velocity of a train = 5 m/s and

final velocity of a train = 45 m/s

time taken = 8 s

To find:

acceleration: ?

Solution:

We define acceleration as change in velocity per unit time that is the difference between the final velocity and initial velocity divided by time.

$Acceleration = \frac{\textrm{final velocity} - \textrm{initial velocity}}{time} \\$

On substituting the above values we get the required acceleration

$Acceleration = \frac{45 - 5}{8}\\ =\frac{40}{8}\\ =5\ m/s^{2}$

Therefore,the acceleration of the train is 5 m/s².

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A certain capacitor, in series with a resistor, is being charged. At the end of 10 ms its charge is half the final value. The ti

vichka [17]

To solve this problem we will apply the expression of charge per unit of time in a capacitor with a given resistance. Mathematically said expression is given as

$q(t) = e^{-\frac{t}{(R*C)}}$

Here,

q = Charge

t = Time

R = Resistance

C = Capacitance

When the charge reach its half value it has passed 10ms, then the equation is,

$\frac{1}{2}*q_{final} = e^{-\frac{0.01}{(R*C)}}$

$Ln(\frac{1}{2}) = -\frac{0.01}{RC}$

$- RC = \frac{0.01}{Ln(1/2)}$

$RC = 0.014s$

We know that RC is equal to the time constant, then

$T = RC = 0.014s = 14ms$

Therefore the time constant for the process is about 14ms

5 0

3 years ago

b) Assume the rod is 0.60 m long and has a mass of 0.50 kg, and the clay blob has a mass of 0.20 kg and moves at an initial velo

SOVA2 [1]

Answer:

The correct answer is "6.96 rad/s".

Explanation:

The given values are:

Length,

L = 0.6 m

Mass,

m₁ = 0.5 kg

m₂ = 0.2 kg

Initial velocity,

V = 8 m/s

Now,

The final angular velocity will be:

⇒ $\omega =\frac{6m_1V}{(4m_1+3m_2)L}$

By substituting the values, we get

⇒ $=\frac{6\times 0.2\times 8}{(4\times 0.2+3\times 0.5)0.6}$

⇒ $=\frac{9.6}{1.38}$

⇒ $=6.96 \ rad/s$

4 0

3 years ago

What types of crust are colliding between the South American Plate and the Nazca Plate?

Neko [114]

Answer:

Oceanic Crust/Continental Crust

Explanation:

It is a destructive plate boundary between the oceanic crust of the Nazca plate and the continental crust of the South American plate.

3 0

3 years ago

2. What would be the acceleration of the clown at 5 s? (A) 1.6 m/s2 (B) 8.0 m/s2 (C) 2.0 m/s2 (D) 3.4 m/s2 3. After 12 seconds,

Neko [114]

Is there an image that goes with this question?

7 0

2 years ago

A stone is dropped from the upper observation deck of a tower, 250 m above the ground. (Assume g = 9.8 m/s2.) (a) Find the dista

Vitek1552 [10]

(a) $y(t)=250 - 4.9 t^2$

For an object in free-fall, the vertical position at time t is given by:

$y(t) = h + ut - \frac{1}{2}gt^2$

where

h is the initial vertical position

u is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

In this problem,

h = 250 m

u = 0 (the stone starts from rest)

So, the vertical position of the stone is given by

$y(t) = 250 - \frac{1}{2}(9.8) t^2 = 250 - 4.9 t^2$

(b) 7.14 s

The time it takes for the stone to reach the ground is the time t at which the vertical position of the stone becomes zero:

y(t) = 0

Which means

$y(t) = h - \frac{1}{2}gt^2=0$

So for the stone in the problem, we have

$250 - 4.9 t^2 = 0$

Solving for t, we find:

$t=\sqrt{\frac{250}{4.9}}=7.14 s$

(c) -70.0 m/s (downward)

The velocity of an object in free fall is given by the equation

$v(t) = u - gt$

where

u is the initial velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Here we have

u = 0

So if we substitute t = 7.14 s, we find the velocity of the stone at the time it reaches the ground:

$v=0-(9.8 m/s^2)(7.14 s)=-70.0 m/s$

The negative sign means the direction of the velocity is downward.

(d) 6.94 s

In this situation, the stone is thrown downward with an initial speed of 2 m/s, so its initial velocity is

u = -2 m/s

So the equation of the vertical position of the stone in this case is

$y(t) = h + ut - \frac{1}{2}gt^2=250 - 2t - 4.9 t^2$

By solving the equation, we find the time t at which the stone reaches the ground.

We find two solutions:

t = -7.35 s

t = 6.94 s

The first solution is negative, so it has no physical meaning, therefore we discard it. So, the time it takes for the stone to reach the ground is:

t = 6.94 s

6 0

3 years ago