Halfway through her run when the music stopped playing, was there energy in Mila’s phone? Why or why not?

How long does it take anya to cover the distance of 5.00 miles ?

sammy [17]

Here is the full information about the question. Ilya and Anya each can run at a speed of 8.50mph and walk at a speed of 3.50 mph . They set off together on a route of length 5.00 miles . Anya walks half of the distance and runs the other half, while Ilya walks half of the time and runs the other half.
the calculation would be:

t = d / s 
t = 2.5 (half of the total distance) / 8.5 (speed of running) 
This is .294 hours which is about 1058s... 

for the walking part... 

t = d / s 
t = 2.5 / 3.5 
t = 5/7hours = 2571 s.

5 0

3 years ago

A 300 gg ball on a 70-cmcm-long string is swung in a vertical circle about a point 200 cmcm above the floor. The string suddenly

Ratling [72]

Answer:

the tension in the string an instant before it broke = 34 N

Explanation:

Given that :

mass of the ball m = 300 g = 0.300 kg

length of the string r = 70 cm = 0.7 m

At highest point, law of conservation of energy can be expressed as :

$\frac{1}{2} mv^2 = mgh\\\\v = \sqrt{2gh}\\\\v = \sqrt{2*(9.8 \ m/s^2)*(6.00 \ m - 2.00 \ m)}\\\\$

$v = 8.854 \ m/s$

The tension in the string is:

$T = \frac{mv^2}{r}\\\\T = \frac{(0.300 \ kg)*(8.854 \ m/s^2)}{0.70 \ m}\\\\T = 33.59 N\\\\T = 34 \ N$

Thus, the tension in the string an instant before it broke = 34 N

6 0

3 years ago

Monochromatic light with a wavelength of 384 nm passes through a single slit and falls on a screen 86 cm away. If the distance o

valkas [14]

This is a Fraunhofer single slit experiment, where the light passing through the slit produces an interference pattern on the screen, and where the dark bands (minima of diffraction) are located at a distance of
$y= \frac{m\lambda D}{a}$
from the center of the pattern. In the formula, m is the order of the minimum, $\lambda$ the wavelenght, $D$ the distance of the screen from the slit and $a$ the width of the slit.

In our problem, the distance of the first-order band (m=1) is $y=0.22 cm$ . The distance of the screen is D=86 cm while the wavelength is $\lambda = 384 nm=384 \cdot 10^{-7}cm$ . Using these data and re-arranging the formula, we can find a, the width of the slit:
$a= \frac{m \lambda D}{y}= \frac{1 \cdot 384 \cdot 10^{-7}cm \cdot 86 cm}{0.22 cm}=0.015 cm$

3 0

3 years ago

One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction

Sophie [7]

Answer:

F = 156.3 N

Explanation:

Let's start with the top block, apply Newton's second law

F - fr = 0

F = fr

fr = 52.1 N

Now we can work with the bottom block

In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal

we apply Newton's second law

Y axis

N - W₁ -W₂ = 0

N = W₁ + W₂

as the two blocks are identical

N = 2W

X axis

F - fr₁ - fr₂ = 0

F = fr₁ + fr₂

indicates that the lower block is moving below block 1, therefore the upper friction force is

fr₁ = 52.1 N

fr₁ = μ N

a

s the normal in the lower block of twice the friction force is

fr₂ = μ 2N

fr₂ = 2 μ N

fr₂ = 2 fr₁

we substitute

F = fr₁ + 2 fr₁

F = 3 fr₁

F = 3 52.1

F = 156.3 N

7 0

3 years ago

A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s

gulaghasi [49]

(a) 0.96 m/s

The period of the wave corresponds to the time taken for one complete oscillation of the boat, from the highest point to the highest point again. Since the time between the highest point and the lowest point is 2.5 s, the period is twice this time:

$T=2\cdot 2.5 s=5.0 s$

The frequency of the waves is the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{5.0 s}=0.20 Hz$

The wavelength instead is just the distance between two consecutive crests, so

$\lambda=4.8 m$

And the wave speed is given by:

$v=\lambda f=(4.8 m)(0.20 Hz)=0.96 m/s$

(b) 0.265 m

The total distance between the highest point of the wave and its lowest point is

d = 0.53 m

The amplitude is just the maximum displacement of the wave from the equilibrium position, so it is equal to half of this distance. So, the amplitude is

$A=\frac{d}{2}=\frac{0.53 m}{2}=0.265 m$

(c) Amplitude: 0.15 m, wave speed: same as before

In this case, the amplitude of the wave would be lower. In fact,

d = 0.30 m

So the amplitude would be

$A=\frac{d}{2}=\frac{0.30 m}{2}=0.15 m$

Instead, the wave speed would not change, since neither the frequency nor the wavelength of the wave have changed.

8 0

3 years ago