Ask question

Ask question

All categories

2 years ago

12

Peter left town A at 13:30 and travelled towards town B at an average speed of 40mph. At 13:45 Philip left town A for town B at

an average speed of 30 mph. What was the distance between them at 15:00?

1 answer:

skad [1K]2 years ago

8 0

Answer:

fbddb

Explanation:

dsbse

You might be interested in

which statement tells you where the metalloids are located in the periodic table A.metalloids are the elements to the left of th

IceJOKER [234]

Were is the table at?

5 0

3 years ago

Estimate the radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 6.5 cm from the center of the bul

Deffense [45]

..........................................................

8 0

2 years ago

A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 4.00°C, the resistance of the carbon r

dusya [7]

Answer:

28 degree C

Explanation:

We are given that

$T_1=4.00^{\circ}$

$R_1=217.7 \Ohm$

$R_2=215.1\Ohm$

$\alpha=-5.00\times 10^{-4}C^{-1}$

We have to find the temperature on a spring day when resistance is 215.1 ohm.

We know that

$\alpha(T_2-T_1)=\frac{R_2}{R_2}-1$

Using the formula

$-5.00\times 10^{-4}(T_2-4)=\frac{215.1}{217.7}-1$

$-5\times 10^{-4}(T_2-4)=0.988-1=-0.012$

$T_2-4=\frac{0.012}{5\times 10^{-4}}=24$

$T_2=24+4=28^{\circ}C$

Hence, the temperature on a spring day 28 degree C.

7 0

2 years ago

A boy standing on a bridge above a river throws stone A vertically upward with an initial velocity =15m/s.2 seconds later he dro

Anastaziya [24]

I'm not sure if this is correct but it's what I'll do

This is free-fall problem.
Stone A is thrown upward, at the point it falls down to the place where it was thrown, the velocity is -15m/s.

Now I choose the bridge is the origin. From the bridge, stone A and B fall the same distance which means Ya = Yb ( vertical distance )

Ya = Vo(t + 2) + 1/2a(t+2)^2
= -15(t + 2) + 1/2(9.8)(t^2 + 4t + 4)
= -15t - 30 + 4.5(t^2 + 4t + 4)
= -15t - 30 + 4.5t^2 + 18t + 18
= 4.5t^2 +3t - 12

Yb = Vo(t) + 1/2a(t)^2
= 0 + 4.5t^2

4.5t^2 = 4.5t^2 +3t - 12
0 = 3t - 12
4 = t

Time for Stone B is 4s
Time for Stone A is 6s

7 0

3 years ago

4. Going back to the dog whistle in question 1, what is the minimum riding speed needed to be able to hear the whistle? Remember

jeyben [28]

Answer:

The minimum riding speed relative to the whistle (stationary) to be able to hear the sound at 21.0 kHz frequency is 15.7 m/s

Explanation:

The Doppler shift equation is given as follows;

$f' = \dfrac{v - v_o}{v + v_s} \times f$

Where:

f' = Required observed frequency = 20.0 kHz

f = Real frequency = 21.0 kHz

v = Sound wave velocity = 330 m/s

$v_o$ = Observer velocity = X m/s

$v_s$ = Source velocity = 0 m/s (Assuming the source is stationary)

Which gives;

$20 = \dfrac{330- v_o}{330+0} \times 21$

330 - $v_o$ = (20/21)*330

$v_o$ = 330 - (20/21)*330 = 15.7 m/s

The minimum riding speed relative to the whistle (stationary) to be able to hear the sound at 21.0 kHz frequency = 15.7 m/s.

8 0

2 years ago