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ANTONII [103]
3 years ago
13

A bus can travel 63 miles in 1.4 hours. If its speed is increased

Physics
1 answer:
Evgesh-ka [11]3 years ago
8 0

Answer:

200

Explanation:

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A local AM radio station broadcasts at a frequency of 696 KHz. Calculate the energy of the frequency at which it is broadcasting
bagirrra123 [75]

Answer:

E=4.61\times 10^{-28}\ J

Explanation:

Given that,

The frequency of local AM radio station, f = 696 KHz = 696000 Hz

We need to find the energy of the frequency at which it is broadcasting.

We know that,

Energy of a wave, E = hf

Where

h is Planck's constant

Put all the values,

E=6.63\times 10^{-34}\times 696000\\\\=4.61\times 10^{-28}\ J

So, the energy of the wave is equal to4.61\times 10^{-28}\ J.

5 0
3 years ago
An electron is moving at a speed of 2.50 ✕ 104 m/s in a circular path of radius of 3.0 cm inside a solenoid. The magnetic field
bogdanovich [222]

Answer:

Explanation:  you would need to divide the two numbers

5 0
2 years ago
What is the magnitude of the force acting on a spring with a spring constant of 275 N/m that is stretched 14.3 cm?
Travka [436]
15 i thank 
if u need more help just ask me

8 0
3 years ago
The table below describes some features of methods used to generate electricity. What is method 3?
blondinia [14]

Answer:

dam/hydro power

Explanation:

estuary is the place where rivers meet the ocean, there are a lot of currents there (waves...) so it likely means thery use a dam or some sort of way of collecting electricity from the motion of the water there. many birds live in estuaries because of being able to find fish to feed from

4 0
2 years ago
A 2.20-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 29.0 N is require
g100num [7]

Answer:

a. 145 N/m b. 1.29 Hz c. 1.62 m/s d.  0 m e. 13.2 m/s² f. ± 0.2 m g. 2.9 J h. 0.54 m/s i. 4.39 m/s²

Explanation:

a. The force constant of the spring

The spring force F = kx and k = F/x where k is the spring constant. F = 29.0 N and x = 0.200 m

k = 29.0 N/0.200 m = 145 N/m

b. The frequency of oscillations, f

f = 1/2π√(k/m)    m = mass = 2.20 kg

f = 1/2π√(145 N/m/2.20 kg) = 1.29 Hz

c. maximum speed of the object

The maximum elastic potential energy of the spring = maximum kinetic  energy of the object

1/2kx² = 1/2mv²

v = (√k/m)x where v is the maximum speed of the object

v = (√145/2.2)0.2 = 1.62 m/s

d Where does the maximum speed occur?

The maximum speed occurs at  0 m

e. The maximum acceleration

a = kx/m = 145 × 0.2/2.2 = 13.2 m/s²

f. The maximum acceleration occurs at x = ± 0.2 m

g. The total energy of the system is the maximum elestic potential energy of the system

E = 1/2kx² = 1/2 × 145 × 0.2² = 2.9 J

h. When x = x₀/3

1/2k(x₀/3)² = 1/2mv²

kx₀²/9 = mv²

v = 1/3(√k/m)x₀ = 1/3(√145/2.2)0.2 = 0.54 m/s

i When x = x₀/3

a = kx₀/3m =  145 × 0.2/(2.2 × 3)= 4.39 m/s²

8 0
3 years ago
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