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3 years ago

All waves change speed when they enter a new medium, but they don't always bend. When does bending occur?

1 answer:

5 0

Bending occurs when one side of the wave enters the new medium before the other side of the wave. ... The bending occurs because the two sides of the wave are traveling at different speeds.

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A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

3 years ago

7. Two bikes travelling in the same direction move at a speed of 30 km/hr. The bikes are separated by a distance of 5 km. What w

Fantom [35]

Answer:

Explanation:

Call the bike on the right A

Call the bike on the left B

The car begins it's time when it passes A

4 minutes later, it passes B.

But B has moved in 4 minutes and that is the key to the problem.

How far has B moved.

t = 4 minutes = 4/60 hours = 1/15 of an hour.

d = ?

rate = 30 km / hr

d = r * t

d = 30 km/hr * 1/15 hours = 2 km

The distance between the bikes is 5 km.

So the car has traveled 5 - 2 = 3 km

d = 3 km

r = ?

t = 4 minutes = 1/15 hour

r = d/t = 3/(1/15)= 3 / 0.066666666 = 45 km/hr.

6 0

3 years ago

Find the current if 55 C of charge pass a particular point in a circuit in 5 seconds.

uranmaximum [27]

Answer:

<em>The current is 11 Amperes</em>

Explanation:

<u>Electric Current</u>

The electric current is defined as a stream of charged particles that move through a conductive path.

The current intensity can be calculated as:

$\displaystyle I=\frac{Q}{t}$

Where:

Q = Electric charge

t = Time taken by the charge to move through the conductor

The current intensity is often measured in Amperes.

The charge passing through a point in a circuit is Q= 55 c during t=5 seconds, thus the current intensity is:

$\displaystyle I=\frac{55}{5}$

I = 11 Amp

The current is 11 Amperes

4 0

3 years ago

This. please help me.

kramer

The transfer of heat from the Sun is in the form of thermal radiation (also called infrared radiation)
The transfer of heat at the person’s feet is by thermal conduction. During the day the sand will be hot so if you step on it with bare feet there will be conduction of heat to the feet.

3 0

2 years ago

A 5.00 kg crate is on a 21.0 degree hill. Using X-Y axes tilted down the plane, what is the y-component of the normal force?

Rama09 [41]

Answer:

The y-component of the normal force is 45.74 N.

Explanation:

Given that,

Mass of the crate, m = 5 kg

Angle with hill, $\theta=21^{\circ}$

We need to find the y component of the normal force. We know that the y component of the normal force is given by :

$F_y=F\ \cos\theta\\\\F_y=mg\ \cos\theta\\\\F_y=5\times 9.8\ \cos(21)\\\\F_y=45.74\ N$

So, the y-component of the normal force is 45.74 N. Hence, this is the required solution.

7 0

2 years ago