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3 years ago

What is the acceleration of a car that is going at a steady speed of 60 mph?

Physics

1 answer:

PilotLPTM [1.2K]3 years ago

8 0

Answer:

Explanation: you are only accelerating if you are slowing down, speeding up, or changing direction

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Of all planets in our solar system jupiter has the greatest gravitational

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3 years ago

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A string is tied between two posts separated by 2.4 m. When the string is driven by an oscillator at frequency 567 Hz, 5 points

Alex787 [66]

Explanation:

The given data is as follows.

Length (l) = 2.4 m

Frequency (f) = 567 Hz

Formula to calculate the speed of a transverse wave is as follows.

f = $\frac{5}{2l} \times v$

Putting the gicven values into the above formula as follows.

f = $\frac{5}{2l} \times v$

567 Hz = $\frac{5}{2 \times 2.4 m} \times v$

v = 544.32 m/s

Thus, we can conclude that the speed (in m/s) of a transverse wave on this string is 544.32 m/s.

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3 years ago

. A solenoid coil consists of a single layer of 250 circular turns of wire with each turn having a 0.02m radius. The axial lengt

Vinil7 [7]

Answer:

a) L = 3.29 10⁻⁴ H, b)U = 5.33 10⁻² J

Explanation:

a) The inductance is a solenoid this given carrier

L = $\frac{N \ \phi_B }{I}$

The magnetic field inside the solenoid is

B = μ₀ $\frac{N}{l} i$

hence the magnetic flux

Ф_B = B. A = μ₀ $\frac{N \ A}{l \ i}$

we substitute in the expression of inductance

L = N² μ₀ A /l

let's find the area of each turn

A = π r²

A = π 0.02²

A = 1.2566 10⁻³ m²

let's calculate

L = 250² 4π 10⁻⁷ 1.2566 10⁻² / 0.3

L = 3.29 10⁻⁴ H

b) The stored energy is

U = ½ L i²

let's calculate

U = ½ 3.29 10⁻⁴ 18²

U = 5.33 10⁻² J

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3 years ago

A boat leaves the dock at t = 0.00 s and, starting from rest, maintains a constant acceleration of (0.461 m/s2)i relative to the

liberstina [14]

Answer:

At t=4.82 s, the boat is moving at 3.464 m/s.

At t=4.82 s, the boat is 13.112 m from the dock.

Explanation:

The speed of the boat in j'th direction remains constant for all times (vj=2.16 m/s), however, the speed in i'th direction is changing due to the constant acceleration (0.461 m/s^2)i.

In order to find the velocity of the boat a t=4.82 s, first we need to compute the velocity of the boat relative to the water in the i direction (vi_b) at t=4.82 s:

vi_b = a*t = (0.461 m/s^2)*(4.82 s) = 2.222 m/s

Now, we add this velocity to the velocity of the water in the i direction:

vi = vi_b + vi_w = 2.222 m/s + 0.486 m/s = 2.708 m/s

Therefore, the speed of the boat at t = 4.82 s is: v = (vi, vj) = (2.708, 2.16) m/s. Finally, to find its speed, we just calculate the magnitude of v and obtain that the speed is: 3.464 m/s.

For the second question, first we will find the distance that the boat moved in the i'th direction and then in the j'th direction.

The speed in the i'th direction, for all times, is given by:

(0.485 + 0.461*t) and in order to find the distance advanced in the i'th direction (di) during 4.82 s, we need to integrate this velocity:

di = 0.485*t + (0.461*t^2)/2 (evaluated from t=0 to t =4.82) = 0.485*(4.82) + (0.461*(4.82)^2)/2 = 2.337 + 5.634 = 7.971 m

The speed in j'th direction, for all times, is given by:

2.16 and in order to find the distance advanced in the j'th direction (dj) during 4.82 s, we need to integrate this velocity:

dj = 2.16*t (evaluated from t=0 to t =4.82) = (2.16)*(4.82) = 10.411 m

Using Pythagoras' Theorem, we find that the the boat is at 13.112 m from the dock at t = 4.82 s.

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3 years ago

A _______ satellite records reflected wavelengths from Earth's surface.

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3 years ago

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