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Akimi4 [234]
3 years ago
9

A train is traveling at a velocity of 90 m/s and it has to travel a distance of 2500 meters to reach the train station. How long

until the train reaches the train station?
Physics
1 answer:
Vlada [557]3 years ago
8 0

Answer:

27.7 or 28 seconds

Explanation:

2500/90=27.7

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A tuning fork labeled 392 Hz has the tip of each of its two prongsvibrating with an amplitude of 0.600mma) What is the maximum s
DanielleElmas [232]

a) 1.48 m/s

The tuning fork is moving by simple harmonic motion: so, the maximum speed of the tip of the prong is related to the frequency and the amplitude by

v_{max}=\omega A

where

v_{max} is the maximum speed

\omega is the angular frequency

A is the amplitude

For the tuning fork in the problem, we have

\omega=2\pi f=2 \pi(392 Hz)=2462 rad/s, where f is the frequency

A=0.600 mm=6\cdot 10^{-4} m is the amplitude

Therefore, the maximum speed is

v_{max}=(2462 rad/s)(6\cdot 10^{-4}m)=1.48 m/s

b)  3.0\cdot 10^{-5} J

The fly's maximum kinetic energy is given by

K=\frac{1}{2}mv_{max}^2

where

m=0.0270 g=2.7\cdot 10^{-5} kg is the mass of the fly

v_{max}=1.48 m/s is the maximum speed

Substituting into the equation, we find

K=\frac{1}{2}(2.7\cdot 10^{-5}kg)(1.48 m/s)^2=3.0\cdot 10^{-5} J

7 0
3 years ago
In the United States, which type of industry is often considered part of an oligopoly?
dezoksy [38]

Answer:

In the United States,cell carrier services are often considered to be a part of an oligopoly

Explanation:

Cell carrier services are a brand value and imagine

6 0
3 years ago
Read 2 more answers
A box of mass 64 kg is at rest on a horizontal frictionless surface. A constant horizontal force F~ then acts on the box and acc
iren2701 [21]

Answer:

The correct answer is;

The magnitude of the force is 35.12 N

Explanation:

To solve the question, we note that the friction is zero and the force causes motion of a stationary mass

One of the equations of motion is required such as

v² = u² + 2× a× s

Where

v = Final velocity = 5.93 m/s

u = Initial velocity = 0 m/s , object at rest

a = acceleration

s = distance moved = 32 meters

But v = Distance/Time = 32 m /5.4 s = 5.93 m/s

Therefore

5.93² = 2×a×32

or a = 35.12/ 64 = 0.55 m/s²

Therefore Force F = Mass m × Acceleration a

Where mass m = 64 kg

Therefore F = 64 kg×0.55 m/s² = 35.12 N

3 0
3 years ago
I need help. which interaction does not take place due to field forces
Advocard [28]

Answer:

Explanation:

Answer D. Interaction between a pen and paper while you write

7 0
3 years ago
A positively charged particle Q1 = +15nC is held fixed at the origin. A second charge Q2 of mass m = 9.5g is floating a distance
Paladinen [302]

Answer:

Q_2=4.4293\times 10^{22}\ nC

Explanation:

Given

charge on the fixed particle, Q_1=15\times 10^{-9}\ C

mass of the second charge, m_2=9.5\times 10^{-3}\ kg

distance between the fixed charge and the floating charge on the top, d=0.25\ m

  • According to the question the second charge is floating just above the fixed positive charge despite of gravity this means that the floating charge is also positive in nature and hence feels the repulsion from the fixed charge which is equal in magnitude to the gravitational force on the charge.

m_2.g=\frac{1}{4\pi\epsilon_0} \times \frac{Q_1.Q_2}{d^2}

where:

\epsilon_0= permittivity of free space

g = acceleration due to gravity

9.5\times 10^{-3}\times 9.8=8.85\times 10^{-9}\times \frac{15\times 10^{-9}\times Q_2}{0.25^2}

Q_2=4.4293\times 10^{22}\ nC

7 0
4 years ago
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