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3 years ago

Peter left Town A at 13:30 and travelled towards Town B at an

Physics

1 answer:

pochemuha3 years ago

4 0

Answer:

Explanation: From 13:30 to 15:00, it past: 1 h 30 mins = 1.5

Then, the distance covered by Peter: 40x1.5= 60 miles

From 13:45 to 15:00, it pasts; 1 h 15min =1.25

Then, the distance covered by Philip. 30 x 1.25 = 37.5 miles

Lastly, the distance between them: 60-37.5= 22.5 miles

So the answer is 22.5

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A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 50 cm/s. Find the rate at which th

Drupady [299]

The radius of the circle, in cm, after t seconds would be 50t
The area, A, of the circle after t seconds is expressed in the equation: A = pi * r^2
A = pi * (50T)^2 = pi*2500*t^2
The change of area per unit time is obtained by differentiating the equation
A' = pi*2500*2*t A' = pi*5000*t
when t = 3 secondsA' = pi*5000*3 = 47122 cm2/s

6 0

3 years ago

What is the basis for rutherford's planetary model?

Olenka [21]

The basis for Rutherford's Planetary model, was the results he got from experiments.

He observed that most of the alpha particles he fired at a gold foil, passed through the foil, but only few were deflected back. So he concluded that most of the Atom would be empty space, with a positive entity at the center.

3 0

3 years ago

sing a rope that will snap if the tension in it exceeds 361 N, you need to lower a bundle of old roofing material weighing 455 N

yaroslaw [1]

Answer:

a)-2m/s^2

b)27.2m/s

Explanation:

Hello! The first step to solve this problem is to find the mass of the block remembering that the definition of weight force is mass by gravity (g=9.8m / s ^ 2)

W=455N=weight

W=mg

W=455N=weight

$m=\frac{W}{g} =\frac{455}{9.81}=46.38kg$

The second step is to draw the free body diagram of the body (see attached image) and use Newton's second law that states that the sum of the forces is equal to mass by acceleration

$a=\frac{T-W}{m} =\frac{361-455}{46.38kg} =-2m/s^2$

for point b we use the equations of motion with constant acceleration to find the velocity

$Vf=\sqrt{X(2)(a)+Vo^2}$

Where

Vf = final speed

Vo = Initial speed =0

A = acceleration =2m/s

X = displacement =6.8m

Solving

$Vf=\sqrt{(6.8)(2)(2)+0^2}=27.2m/s$

4 0

3 years ago

The components of a 10.8-meters-per-second velocity at an angle of 34.° above the horizontal are

Kisachek [45]

Answer:

6.0 m/s vertical and 9.0 m/s horizontal

Explanation:

For the vertical component, we use the formula:

Sin(34°) = y / 10.8

Then we solve for y:

0.559 = y / 10.8

y = 6.0

And for the horizontal component, we use the formula:

Cos(34°) = x / 10.8

Then we solve for x:

0.829 = x / 10.8

y = 9.0

So the answer is " 6.0 m/s vertical and 9.0 m/s horizontal".

3 0

3 years ago

How long will it take a plane to fly 1256km if it travels 500km/hr?

expeople1 [14]

Answer:

<h3>The answer is 2.51 s</h3>

Explanation:

The time taken can be found by using the formula

$t = \frac{d}{v} \\$

d is the distance

v is the velocity

From the question we have

$t = \frac{1256}{500} \\ = 2.512$

We have the final answer as

Hope this helps you

4 0

2 years ago