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sashaice [31]
3 years ago
10

during a car ride Sandra travel led directly east for 20 miles and then directly south for 15 miles what was her displacement. 2

5 miles at 59.0 degrees south of east, 25 miles at 25.9 degrees east of south, 25 miles at 36.9 degrees south of east, or 25 miles at 42.5 degrees east of south?
Physics
1 answer:
damaskus [11]3 years ago
8 0
25 miles at 42.5 degrees east of south. (20^2+15^2 =C^2...You get 25 miles...East of South because she traveled a longer distance to the East. Then you find the angle between the original and ending point.) Hope it helps. You may ask me any other questions.
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Societal demands are not a legitimate driving force of scientific endeavor.
UNO [17]
True or false in assuming so false I think
7 0
3 years ago
Transverse waves are sent along a 5.00-m-long string with a speed of 30.00 m/s. The string is under a tension of 10.00 N. What i
Kipish [7]

Answer:

0.055 kg

Explanation:

Given that

Length of the string, l = 5 m

Speed of the wave, v = 30 m/s

Tension on the string, F(t) = 10N

From the formula written in the attachment, we have

v = velocity of the wave, in m/s

F(t) = Tension on the string, in N

U = Mass per length of the string, in kg/m

m = Mass of the string, in kg

l = Length of the string, in m

See attachment for the calculation

3 0
3 years ago
In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.
Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement of the object

t is the time elapsed

\omega is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

And the time taken is

t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

\omega=\frac{2\pi}{5700}=0.0011 rad/s

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

v=\omega r

where

v is the linear velocity

\omega is the angular velocity

r is the radius of the circular orbit

In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m

Therefore, the linear velocity of the telescope is:

v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s

4 0
3 years ago
Pls heplp 70 points!!!!!
Rasek [7]

Answer

the answer is d for sure

5 0
3 years ago
Dolphin echolocation is similar to ultrasound. Reflected sound waves
s344n2d4d5 [400]

Answer:

Waves with high frequencies have shorter wavelengths that work better  than low frequency waves for successful echolocation.

Explanation:

To understand why high-frequency waves work better  than low frequency waves for successful echolocation, first we have to understand the relation between frequency and wavelength.

The relation between frequency and wavelength is given by

λ = c/f

Where λ is wavelength, c is the speed of light and f is the frequency.

Since the speed of light is constant, the wavelength and frequency are inversely related.

So that means high frequency waves have shorter wavelengths, which is the very reason for the successful echolocation because waves having shorter wavelength are more likely to reach and hit the target and then reflect back to the dolphin to form an image of the object.

Thus, waves with high frequencies have shorter wavelengths that work better  than low frequency waves for successful echolocation.

3 0
3 years ago
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