Name the net force between objects that allows a car to travel in a circle

The speed of sound in water is 1498 m/s. A sonar signal is sent straight down from a ship at a point just below the water surfac

olga2289 [7]

Answer:

The water is 1310.75 meters deep.

Explanation:

The question gives us the speed of sound and the time the sonar signal takes to travel to the bottom of the water and back.

To find out the distance, we should first divide the time in half, so we only consider the time taken for the sound to reach the bottom of the water body.

This means:

Time = 1.75 / 2 = 0.875 seconds

The distance traveled in this time is:

Distance = Speed * Time

Distance = 1498 * 0.875

Distance = 1310.75 meters

Thus, the water is 1310.75 meters deep.

8 0

4 years ago

A capacitor of cylindrical shape as shown in the red outline, few cm long carries a uniformly distributed charge of 7.2 uC per m

Natali [406]

(a) The magnitude of the electric field at point 5.5m is 2.35 x 10⁴ N/C.

(b) The magnitude of the electric field at point 2.5m is 5.18 x 10⁴ N/C.

<h3>Electric field at a point on the Gaussian surface</h3>

The magnitude of the electric field at a point on the cylindrical Gaussian surface is calculated as follows;

E = λ/2πε₀r

where;

λ is linear charge density
ε₀ is permitivity of free space
r is the position of the charge

<h3>At a distance of 5.5 m</h3>

$E = \frac{\lambda}{2\pi \varepsilon _0 r} \\\\E = \frac{7.2 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 5.5} \\\\E = 2.35 \times 10^4 \ N/C$

<h3>At a distance of 2.5 m</h3>

$E = \frac{\lambda}{2\pi \varepsilon _0 r} \\\\E = \frac{7.2 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 2.5} \\\\E = 5.18 \times 10^4 \ N/C$

Thus, the magnitude of the electric field at points of 5.5m is 2.35 x 10⁴ N/C, and the magnitude of the electric field at points of 2.5m is 5.18 x 10⁴ N/C.

Learn more about electric field here: brainly.com/question/14372859

4 0

2 years ago

A proton with charge 1.602 x 10^-19 C moves at a speed of 300 m/s in a magnetic field at an angle of 65 degrees. If the strength

soldi70 [24.7K]

Answer:

$F_m= 8.28 \times 10^{-16} N$

Explanation:

The magnitude of the force is $F_m=evB\sin\theta\\\implies F_m=1.602 \times 10^{-19}\times 300\times 19\times\sin 65^0= 8.28 \times 10^{-16} N$

3 0

3 years ago

An object that is initially not rotating has a constant torque of 3.6 N⋅m applied to it. The object has a moment of inertia of 6

Korolek [52]

Answer:

0.6

Explanation:

Angular acceleration is equal to Net Torque divided by rotational inertia, which is the rotational equivalent to Newton’s 2nd Law. Therefore, angular acceleration is equal to 3.6/6 which is 0.6. Hope this helped!

3 0

2 years ago

You are standing 2.5m directly in front of one of the two loudspeakers. They are 3.0m apart and both are playing a 686Hz tone in

ahrayia [7]

Answer:

distance from speaker is 17.87 m

Explanation:

given data

distance from loudspeaker = 2.5 m

distance between loudspeaker = 3.0 m

room temperature = 20c

wavelength f = 686Hz

to find out

what distances from the speaker

solution

we know sound velocity c = 331.5 + 0.6 × 20c = 343.5

so wavelength of sound λ = c / f

wavelength = 343.5 / 686 = 0.5 m

when the difference in distance of speaker destructive interference will be

d = λ/2 × (2n-1)

for n = 1, 2 3 4 ..

d = 0.5/2 × (2n-1)

d = 0.250 , 0.75 , 1.25 , 1.750............ for n = 1, 2 3 .............

so

for d = 0.250

side of triangle by hypotenuse of triangle are

$\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x1)$ = 0.250

0.5 x1 = 7.6875

x1 = 15.375 m

for d = 0.75

side of triangle by hypotenuse of triangle are

$\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x2)$ = 0.75

1.5 x2 = 4.6875

x2 = 3.125 m

for d = 1.250

side of triangle by hypotenuse of triangle are

$\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x3)$ = 1.250

2.5 x2 = 1.1875

x3 = 0.475 m

for d = 1.750

x4 will be negative so we stop here

so the distance from speaker here is given below

distance = 2.5 + x

here x = 0.475 , 3.125 and 15.375 so

distance 1 = 2.5 + 0.475 = 2.975 m

distance 2 = 2.5 + 3.125 = 5.625 m

distance 3 = 2.5 + 15.375 = 17.875 m

final distance from speaker is 17.87 m

8 0

3 years ago