All categories

3 years ago

Physics A-Level, both of these questions

Physics

1 answer:

podryga [215]3 years ago

8 0

I can only help you with question 7
B is the right answer.
A is wrong because when the angle is critical angle the light leaves medium 1 parallel to the surface.
C is wrong because light reflects only when travelling from a denser to a rarer/less dense medium.speed of light is lower in denser medium.
D is wrong because the frequency of the light doesn't change when travelling from one medium to another.

You might be interested in

Your heart pumps blood at a pressure of 100 mmHg and flow speed of 60 cm/s. At your brain, the blood enters capillaries with suc

Savatey [412]

Answer:

1.28 m

Explanation:

Generally, pressure of fluid is given by

$P=\rho g h$ where g is acceleration due to gravity, h is the height and $\rho$ is the density

Considering that the pressure for mercury is same as for blood only that the height and density of fluid are different then

$\rho_b g h_b= \rho_m g h_m$

Since g is constant, then

$\rho_b h_b= \rho_m h_m$

Making $h_b$ the subject of the formula then

$h_b=\frac {\rho_m h_m}{\rho_b}$

Where subscripts m and b denote mercury and blood respectively

Assuming density of blood is 1060 Kg/m3, density of mercury as 13600 Kg/m3 and substituting height of mercury for 0.1 m then

$h_b=\frac {13600*0.1}{1060}=1.283018868 m \approx 1.28 m$

7 0

3 years ago

A 5.0 kg block hangs from the ceiling by a mass-less rope. A Second block with a mass of 10.0 kg is attached to the first block

gayaneshka [121]

The tension in the first and second rope are; 147 Newton and 98 Newton respectively.

Given the data in the question

Mass of first block; $m_1 = 5.0kg$

Mass of second block, $m_2 =10kg$

Tension on first rope; $T_1 =\ ?$

Tension on second rope; $T_2 =\ ?$

To find the Tension in each of the ropes, we make use of the equation from Newton's Second Laws of Motion:

$F = m\ *\ a$

Where F is the force, m is the mass of the object and a is the acceleration ( In this case the block is under gravity. Hence ''a" becomes acceleration due to gravity $g = 9.8m/s^2$ )

For the First Rope

Total mass hanging on it; $m_T = m_1 + m_2 = 5.0kg + 10.0kg = 15.0kg$

So Tension of the rope;

$F = m\ * \ g\\\\F = 15.0kg \ * 9.8m/s^2\\\\F = 147 kg.m/s^2\\\\F = 147N$

Therefore, the tension in the first rope is 147 Newton

For the Second Rope

Since only the block of mass 10kg is hang from the second, the tension in the second rope will be;

$F = m\ * \ g\\\\F = 10.0kg \ * 9.8m/s^2\\\\F = 98 kg.m/s^2\\\\F = 98N$

Therefore, the tension in the second rope is 98 Newton

Learn More, brainly.com/question/18288215

4 0

2 years ago

Read 2 more answers

E14. A ball rolls off a table with a horizontal velocity of 5 m/s. If

Shkiper50 [21]

a) Vertical velocity: 5.9 m/s

b) Horizontal velocity: 5 m/s

Explanation:

The motion of the ball is the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction)

- A uniformly accelerated motion (constant acceleration) along the vertical direction

Here we want to find the vertical component of the ball's velocity. This can be done by using the suvat equation for the vertical motion:

$v_y = u_y +gt$

where:

$v_y$ is the vertical velocity at time t

$u_y=0$ is the initial vertical velocity (zero because the ball has been thrown horizontally)

$g=10 m/s^2$ is the acceleration of gravity (here we take downward as positive direction)

Substituting t = 0.6 s, which is the total time of flight, we find the vertical velocity of the ball just before it hits the ground:

$v_y=0+(9.8)(0.6)=5.9 m/s$

The motion along the vertical direction is an accelerated motion, because there is a force (the force of gravity) acting on the ball and that it causes an acceleration in the ball.

However, there are no forces acting in the horizontal direction on the ball (if we neglect the air resistance): this means that the acceleration of the ball in the horizontal direction is zero.

As a consequence, this also means that the horizontal component of the ball's velocity is constant during the motion.

Since the ball was thrown from the table with an initial horizontal velocity of 5 m/s, this means that the horizontal velocity of the ball just before it hits the floor is still

$v_x = 5 m/s$

8 0

4 years ago

Sam shared with his class that he lives on Oak Street. Sam is likely how old?

Lyrx [107]

D FOUR years old hope this helps

mark brainliest please

4 0

3 years ago

Read 2 more answers

an engine has been design to work between source and the sink at temperature 177 degree Celsius and 27 degree Celsius respective

irina [24]

Given data

Source temperature (T₁) = 177°C = 177+273 = 450 K

Sink temperature (T₂) = 27°C = 27+273 = 300 K

Energy input (Q₁) = 3600 J ,

Work done = ?

We know that, efficiency (η) = Net work done ÷ Heat supplied

η = W ÷ Q₁

W = η × Q₁

First determine the efficiency ( η ) = ?

Also, we know that ( η ) = (T₁ - T₂) ÷ (T₁)

= 33.3% = 0.333

Now, Work done is W = η × Q₁

= 0.33 × 3600

W = 1188 J

Work done by the engine is 1188 J

4 0

3 years ago