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BigorU [14]
3 years ago
10

It's for #8....I know the answer is 293.2 ft I just don't know how to get it

Physics
1 answer:
Oksi-84 [34.3K]3 years ago
7 0
Kinectic Energy=1/2(mass)(velocity)^2 so 1.2=1/2(.0012)(position/2) so it travels 4000 m. Not sure how it is 293.2 ft
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Is there change in the force of gravity between two objects when their masses are doubled and the distance between them is also
belka [17]

No. The forces of gravity between them would remain unchanged after those three changes.

3 0
3 years ago
A particle that has mass m and charge q enters a uniform magnetic field that has magnitude B and is directed along the x axis. T
Usimov [2.4K]

Answer:

a) The trajectory will be a helical path.

b) θ = 2*π rad

Explanation:

a) Since the initial velocity of the particle has a component parallel (x-component) to the magnetic  field B , the trajectory will be a helical path.

b) Given

t = 2*π*m/(q*B)

We can use the equation

θ = ω*Δt

where

θ is the angular displacement

ω is the angular speed, which is obtained as follows:

ω = q*B/m

then we have

θ = (q*B/m)*2*π*m/(q*B)

⇒  θ = 2*π rad

6 0
3 years ago
Salmon often jump waterfalls to reach their
PilotLPTM [1.2K]

The minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

The given parameters;

  • height of the waterfall, h = 0.432 m
  • distance of the Salmon from the waterfall, s = 3.17 m
  • angle of projection of the Salmon, = 30.8º

The time of motion to fall from 0.432 m is calculated as;

h = v_0_y + \frac{1}{2} gt^2\\\\0.432 = 0 + (0.5\times 9.8)t^2\\\\0.432 = 4.9t^2\\\\t^2 = \frac{0.432}{4.9} \\\\t^2 = 0.088\\\\t = \sqrt{0.088} \\\\t = 0.3 \ s

The minimum velocity of the Salmon jumping at the given angle is calculated as;

X = v_0_x t\\\\3.17 = (v_0\times cos(30.8)) \times 0.3\\\\10.567 = v_0\times cos(30.8)\\\\v_0 = \frac{10.567}{cos(30.8)} \\\\v_0 = 12.3 \ m/s

Thus, the minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

Learn more here: brainly.com/question/20064545

8 0
2 years ago
A runner is moving at a constant speed of 8.00 m/s around a circular track. If the distance from the runner to the center of the
Genrish500 [490]

Answer: Last option

2.27 m/s2

Explanation:

As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.

If we call a_c to the centripetal acceleration then, by definition

a_c =w^2r = \frac{v^2}{r}

in this case we know the speed of the runner

v =8.00\ m/s

The radius "r" will be the distance from the runner to the center of the track

r = 28.2\ m

a_c = \frac{8^2}{28.2}\ m/s^2

a_c = 2.27\ m/s^2

The answer is the last option

3 0
4 years ago
Scenario 3Starting at rest, a 3 kg ball is dropped from the side of a bridge and strikes the ground below at 35m/s. What is the
Nonamiya [84]

   The ball's gravitational potential energy is converted into kinetic energy as it falls toward the ground.

<h3>How can the height of a dropped ball be determined?</h3>

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    The ball's parabolic motion causes it to move at a speed of 26.3 m/s right before it strikes the ground, which is faster than its straight downhill motion, which has a speed of 17.1 m/s. Take note of the rising positive y direction in the above graphic.

To Learn more About potential energy, Refer:

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#SPJ10

4 0
2 years ago
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