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harina [27]
3 years ago
11

After soccer practice, Coach Miller goes to the roof of the school to retrieve the errant soccer balls. The height of the school

is 3.5 m 3.5 m. He kicks a soccer ball which leaves the roof with a horizontal velocity, but no vertical velocity. Ignoring air resistance, which of the following values indicates how much horizontal velocity is needed for the ball to reach the soccer field located 22 m 22 m from the school?
Physics
1 answer:
blsea [12.9K]3 years ago
3 0
With gravitational acceleration at 9.8, initial height at 3.5m and distance at 22m the initial horizontal velocity is 26.03 ms and the flight time is .845 seconds
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The gravitational force between the charged constituents of the atom is negligible compared with the electric force between them
katrin2010 [14]

Answer: The initial force is reduced a factor 1/4 when the separation between charge is doubled

Explanation: As it well known the electric force between two charges is given by:

Finitial=k*q1*q2/d^2 where d is the distance between charges and k is a constant

if the distance is doubled this means 2*dinitial thus the new force is equal to F initial* 1/4

6 0
3 years ago
Plz help..During takeoff a plane accelerates at 4m/s^2 and takes 40s to reach takeoff speed.
matrenka [14]

Answer:

The velocity of the plane at take off is 160 m/s.

The distance travel by the plane in that time is 3200 meter.

Explanation:

Given:

Acceleration, a = 4 m/s²

Time, t = 40 s

u = 0 i .e initial velocity

To Find:

velocity , v = ?

distance , s =?

Solution:

we have first Kinematic equation

v = u + at

∴ v = 0 + 4×40

∴ v = 160 m/s

Now by Third Kinematic equation

s = ut + \frac{1}{2}at^{2}

∴ s = 0 + 0.5 × 4× 40²

∴ s = 3200 meter

3 0
3 years ago
A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha
Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss} (1)

From the bottom to the top of the second hill

\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss} (2)

Where:

m - Mass of the roller coaster car, in kilograms.

v_{1} - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

g - Gravitational acceleration, in meters per square second.

h_{1} - Height of the first top of the hill with respect to the bottom, in meters.

W_{1, loss} - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

v_{2} - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

h_{2} - Height of the second top of the hill with respect to the bottom, in meters.

W_{2, loss} - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss} (3)

Where W_{loss} is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that m = 175\,kg, g = 9.807\,\frac{m}{s^{2}}, h_{1} = 18\,m, h_{2} = 8\,m and v_{2} = 11\,\frac{m}{s}, then the work done by non-conservative force is:

W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]

W_{loss} = 6574.75\,J

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0
2 years ago
Sally takes two bar magnets and randomly tapes one end of each bar magnet. she labels the magnets A and B. She brings the taped
vlabodo [156]
Since the two taped poles of the magnets labeled A and B attracts one to each other, we know that the two taped poles are oppsosite.

So, you can predict with total certainty that when she brings the taped end of the third magnet (magnet C) near each of the first two magntes, in one case they will attract each other and in the other case they will repele mutually.

You are certain of that because, since the taped poles of the first two magnets are opposite, the pole of the third magnet has to be equal to one of the two first taped poles and opposite to the other of the two firest taped poles.
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