All categories

2 years ago

How do you calculate average speed of the bus between 0 h and 2.34 h using this table?

Physics

1 answer:

Helen [10]2 years ago

8 0

Fallout is awesome, but I didn't care too much for the base building mainly because of the amount of time it takes to make it perfect.

You might be interested in

A 6 kilogram block in outer space is moving at -100 m/s (to the left). It suddenly experiences three forces as shown below (in t

Blizzard [7]

Answer:

x = 1474.9 [m]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces must be equal to the product of mass by acceleration.

We must understand that when forces are applied on the body, they tend to slow the body down to stop it.

So as the body continues to move to the left, it is slowing down. Therefore we must calculate this deceleration value using Newton's second law. We must perform a sum of forces on the x-axis equal to the product of mass by acceleration. With leftward movement as negative and rightward forces as positive.

ΣF = m*a

$10 +12*sin(60)= - 6*a\\a = - 3.39[m/s^{2}]$

Now using the following equation of kinematics, we can calculate the distance of the block, before stopping completely. The initial speed must be 100 [m/s].

$v_{f}^{2} =v_{o}^{2}-2*a*x$

where:

Vf = final velocity = 0 (the block stops)

Vo = initial velocity = 100 [m/s]

a = - 3.39 [m/s²]

x = displacement [m]

$0 = 100^{2}-2*3.39*x\\x=\frac{10000}{2*3.39}\\x=1474.9[m]$

4 0

2 years ago

The average intensity of light emerging from a polarizing sheet is 0.708 W/m2, and that of the horizontally polarized light inci

Pachacha [2.7K]

Answer:

Angle θ = 30.82°

Explanation:

From Malus’s law, since the intensity of a wave is proportional to its amplitude squared, the intensity I of the transmitted wave is related to the incident wave by; I = I_o cos²θ

where;

I_o is the intensity of the polarized wave before passing through the filter.

In this question,

I is 0.708 W/m²

While I_o is 0.960 W/m²

Thus, plugging in these values into the equation, we have;

0.708 W/m² = 0.960 W/m² •cos²θ

Thus, cos²θ = 0.708 W/m²/0.960 W/m²

cos²θ = 0.7375

Cos θ = √0.7375

Cos θ = 0.8588

θ = Cos^(-1)0.8588

θ = 30.82°

4 0

3 years ago

An object of mass 2.0 kg is attached to the top of a vertical spring that is anchored to the floor. The unstressed length of the

poizon [28]

Answer:

The value is $A = 0.014 \ m$

Explanation:

From the question we are told that

The mass of the object is $m = 2.0 \ kg$

The unstressed length of the string is $l = 0.08 \ m$

The length of the spring when it is at equilibrium is $l_e = 5.9 \ cm = 0.059 \ m$

The initial speed (maximum speed)of the spring when given a downward blow $v = 0.30 \ m/s$

Generally the maximum speed of the spring is mathematically represented as

$u = A * w$

Here A is maximum height above the floor (i.e the maximum amplitude)

and $w$ is the angular frequency which is mathematically represented as

$w = \sqrt{\frac{k}{m} }$

$u = A * \sqrt{\frac{k}{m} }$

=> $A = u * \sqrt{\frac{m}{k} }$

Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

$b = l -l_e$