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Ket [755]
4 years ago
7

A 30,000kg train moves at 45m/s How how much kinetic energy with the train have?

Physics
1 answer:
nordsb [41]4 years ago
6 0

Answer:

30,375,000

Explanation:

1/2 x m x v^2

1/2(30000)(45)^2

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How far did a car travel if it was on the road for 55.2s and it traveled at an average speed of 55m/s
pantera1 [17]
The car traveled 1.00363 kilometers in the 55.2s and the speed of 55m/s
3 0
3 years ago
Calculate the net force on your 0.50-cm2 eardrum that air exerts on the inside and the outside after you drive from Denver, Colo
tekilochka [14]

Answer:

1.0563408 N

Explanation:

\rho = Density of air = 0.8 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

h = Altitude

A = Area = 0.5 cm²

Pressure

p=\rho gh\\\Rightarrow p=0.8\times 9.81\times (4301-1609)\\\Rightarrow p=21126.816\ Pa

Pressure

p=\frac{F}{A}\\\Rightarrow F=pa\\\Rightarrow F=21126.816\times 0.5\times 10^{-4}\\\Rightarrow F=1.0563408\ N

The net force on your ear is 1.0563408 N

6 0
3 years ago
A charge of −25µC is distributed uniformly over the surface of a spherical conductor of radius 12.0 cm. Determine the electric f
Bond [772]

Answer:

a) 0 b) 0 c) -9*10⁵ N/C

Explanation:

a)

  • In electrostatic conditions, no electric field can exist inside a conductor.
  • As the distance r=5 cm falls inside the conductor, the electric field is just zero.

b)

  • Same as above, as r=10 cm is still inside the spherical conductor.

c)

  • At r= 50 cm. from the center of the spherical conductor, we can apply Gauss' Law in order to get the value of the electric field.
  • By symmetry, the electric field, at a same distance from the center, must be radial, and constant on a spherical surface concentric with the spherical conductor.
  • So, we can write the following equation for Gauss'Law:

       \int\ {E} \, dA = \frac{Q}{\epsilon_{0} }

  • If E is constant, we can take it out of the integral, and integrate all the closed spherical surface, as follows:

       E* 4*\pi *r^{2}  =\frac{Q}{\epsilon_{0}}

  • So, we can solve for E, as follows:

       E = \frac{Q}{4*\pi*r^{2}*\epsilon_{0}} =  \frac{(-25e-6)C}{4*\pi*(0.5m)^{2}*8.85e-12C2/N*m2}}\\\\   E = -9e5 N/C

  • E = -9*10⁵ N/C (radially inward, taking the outward direction as positive)
8 0
3 years ago
What is the scientific name for a push or a pull?
neonofarm [45]

Answer:

thats a force

Explanation:

4 0
3 years ago
Read 2 more answers
Mr. Turner is driving through downtown Houston at 35 MPH (15.56 m/s). This is above the posted speed limit and there are cars pa
monitta

Answer:

Mr. Turner will not be able to stop in time.

a = - 71.2 m/s²

s₁ = 11.67 m

Explanation:

Since, Mr. Turner is driving at 45.56 m/s and his reaction time for applying brakes is 3/4th of a second. So, the distance covered by car in this time will be:

s₁ = vt

where,

s₁ = distance covered by car before applying brakes = ?

v = speed of car = 15.56 m/s

t = reaction time = 3/4th of second = 0.75 s

Therefore,

s₁ = (15.56 m/s)(0.75 s)

<u>s₁ = 11.67 m</u>

and the distance required for the car to stop after applying brakes is:

s₂ = 1.7 m

So, the total distance traveled by car before stopping will be:

s = s₁ + s₂

s = 11.67 m 1.7 m

s =  13.37 m

since, the driver was 13.3 m ahead.

<u>Therefore, Mr. Turner will not be able to stop in time.</u>

To find the deceleration of Mr. Turner after applying brakes we use 3rd equation of motion:

2as₂ = Vf² - Vi²

where,

a = deceleration = ?

Vf = Final Velocity = 0 m/s

Vi = Initial velocity = 15.56 m/s

Therefore,

2a(1.7 m) = (0 m/s)² - (15.56 m/s)²

a = - (242.1136 m²/s²)/3.4 m

<u>a = - 71.2 m/s²</u>

5 0
3 years ago
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