Can somebody pls pls help

What 2 things do you need to make sound? help ASAP

inessss [21]

Answer:

- A vibrating object

- a medium to travel

HOPE IT HELPS :)

PLEASE MARK IT THE BRAINLIEST!

8 0

3 years ago

1-A train travels 100 km to reach town A in one hour and 15 min. The train stops at station A for 45 minutes. Then it travels 15

kirill [66]

Answer 1) : 62.5 km/hour is the average velocity of the train.

2) The final velocity of the car at the end of 75 m is 14.69 m/s

Explanation:

1) Displacement of the train = 100 km + 150 km = 250 km

Total time train took =1 hour 15 min+ 45 min + 2 hours = 240 min = 4 hours

Average velocity= $\frac{Displacement}{time}=\frac{250 km}{4 hour}=62.5 km/h$

62.5 km/hour is the average velocity of the train.

2) The acceleration of the car, a= 1.2 $m/s^2$

Distance covered by the car,s = 75 m

Initial velocity of the car , $v_i$ = 6 m/s

Final velocity of thre car , $v_f$ =?

Using third equation of motion:

$v_{f}^2=v_{i}^2+2as=(6 m/s)^2+2\times 1.2 m/s\times 75 m=216 m^2/s^2$

$v_{f}=14.69 m/s$

The final velocity of the car at the end of 75 m is 14.69 m/s

8 0

3 years ago

At 4.00 l, an expandable vessel contains 0.864 mol of oxygen gas. how many liters of oxygen gas must be added at constant temper

svet-max [94.6K]

Givens
=====
V = 4.00 L
T = 273oK We're assuming the temperature does not change, just the pressure.
n = 0.864 moles
R = 8.314 joules / mole * oK
P = ?????

Formula
======
PV = n*R*T
P = n*R*T/V
P = 0.864 * 8.314 * 273 / 4
P = 490 kpa

You have to add 1.6 – 0.864 = 0.736 moles of gas.

We have to assume that the temperature and pressure remain the same when we add the 0.736 moles of gas. We are now looking for the volume.

PV = n*R*T

V = 0.736 * 8.314 * 273 / 490

V = 3.41 L Remember this is at about 4 atmospheres so we have to convert to Standard Pressure.

Total Volume = 3.41 + 4.00 = 4.41

V1 * P1 = V2 * P2

P1 = 490 kPa

P2 = 101 kPa

V1 = 7.41 L

V2 = ????

7.41* 490 = V2 * 101 V2 = 7.41 * 490 / 101 V2 = 35.94 L

You had 4 L now you need 31.94 more.

6 0

3 years ago

A deer with a mass of 146kg is running head on toward you with a speed of 17 m/s. You are going north. Find the momentum of the

Pavel [41]

Momentum = Mass * Velocity = 146 * 17 = 2482 kgm/s

6 0

3 years ago

Read 2 more answers

After flying for 15 min in a wind blowing 42 km/h at an angle of 19° south of east, an airplane pilot is over a town that is 48

masha68 [24]

Answer:

The speed of the airplane relative to the air is 209.47km/hr

Explanation:

Whenever we are solving a physics problem, it's really useful to start by drawing a diagram of the problem (See picture attached). It will help us visualize the problem better.

Now, we know that the plane flew for an amount of time of 15 minutes. For our dimensions to be the same, we need to turn those 15min to hours, like this:

$15min*\frac{1hr}{60min}=0.25hr$

Once our time is rewritten as hours, we can now calculate the velocity towards north of the plane.

$V=\frac{distance}{time}$

the plane traveled a distance to the north of 48km so the velocity is:

$V=\frac{48km}{0.25hr}$

so

V=192km/hr j

Now, we can calculate the x and y-components of the velocity of the wind. The problem states that the wind is blowing at 42km/hr at an angle of 19° south of east, so the x and y-components of the velocity of the wind are:

$V_{x}=42km/hr*cos(-19^{o} )=39.71 i$

and

$V_{y}=42km/hr*sin(-19^{o} )=-13.67 j$

So the velocity of the wind can be expressed as a vector as:

$V_{wind}=(39.71i - 13.67j)km/hr$

Once we know this, we can find the velocity of the plane with respect of the wind on x and on y:

$V_{plane x}=V_{plane/wind x}+V_{wind x}$

$V_{plane/wind x}=V_{plane x}-V_{wind x}$

$V_{plane/wind x}=(0-39.71 i)km/hr$

$V_{plane/wind x}= -39.71 i km/hr$

and

$V_{plane y}=V_{plane/wind y}+V_{wind y}$

$V_{plane/wind y}=V_{plane y}-V_{wind y}$

$V_{plane/wind y}=192km/hr j - (- 13.67j)km/hr$

$V_{plane/wind x}= 205.67 j km/hr$

So the velocity of the plane with respect to the wind can be rewritten as:

$V_{plane/wind x}= (-39.71i + 205.67 j) km/hr$

Since the problem asks us to find the speed of the plane with respect to the wind, this means that we need to find the magnitude of the velocity, since the speed is a scalar defined to be the magnitude of the velocity.

so:

$speed=\sqrt{(-39.71)^{2}+(205.67)^{2} }$

speed= 209.47 km/hr

Therefore, the speed of the airplane relative to the air is 209.47km/hr

6 0

3 years ago