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horsena [70]
3 years ago
10

PLS HELP. A flatbed car of a train moves 8 m/s to the east. A jogger runs along to top the flatbed car (which is not very safe)

with a velocity of 6 m/s to the west. How fast is the jogger moving?
Physics
1 answer:
jonny [76]3 years ago
5 0

Answer:

2 m/s

Explanation: Given that a flatbed car of a train moves 8 m/s to the east. A jogger runs along to top the flatbed car (which is not very safe) with a velocity of 6 m/s to the west.

Since the jogger is moving in an opposite direction to the direction of the train, and velocity is the distance covered in a specific direction, the jogger will be moving at a velocity relative to the velocity of the train.

Velocity = (8 - 6) m/s

Velocity = 2 m/s

Therefore, the jogger will be moving at the speed of 2 m/s

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What is the symbol F g mean
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Answer:

Disambiguation

Explanation:

Fg is an abbreviation used in physics to describe the amount of force exerted by gravity on an object, normally represented in units such as the newton.

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How many degrees equal π in radians? How many revolutions (turns) equal π radians? ​
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Answer:

It follows that the magnitude in radians of one complete revolution (360 degrees) is the length of the entire circumference divided by the radius, or 2πr / r, or 2π. Thus 2π radians is equal to 360 degrees, meaning that one radian is equal to 180/π ≈ 57.295779513082320876 degrees.

Explanation:

6 0
2 years ago
Recall that photosynthetic rates remain relatively constant in regions near the equator. Imagine that tropical environments pers
solniwko [45]

Answer:E.

a straight line sloping upward (atmospheric CO2 levels would not seasonally oscillate, but would have increased over time)

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8 0
3 years ago
A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con
Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5  x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \  \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm

The magnitude of the electric field is now given as;

E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0
3 years ago
A stone is thrown vertically upward with a speed of 18.0 . (a)How fast is it moving when it reaches a height of 11.0 ? (b)How lo
aliina [53]
For the first part, we are looking for Vf when dy=11.0
Upward is positive, downward is negative. 
So <span>Vf = square root [2(-9.8)(11.0) + (18.0)^2] </span>
<span>Vf = 10.4 m/s your answer is correct. 

For the part b, t is equals to the time took to reach and dy is equals to 11.0
you did, </span>11= 18t m/s-(1/2) 9.8t^2 then <span>-11 + 18t- 9.8t^2. By quadratic formula, for the way down the answer is 2.9 s while on it's way up, the answer is 0.77 s</span><span>
 </span> 
5 0
3 years ago
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