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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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Alan starts from his home and walks 1.3 km east to the library. He walks an additional 0.68 km east to a music store. From there
zepelin [54]

Answer: final Displacement = 0 km, total distance covered =7 km

Explanation:

Given the following :

From home to library = 1.3 km East

Library to music store = 0.68km East

Music store to friend's house = 1.1km North

Friend's house to grocery store = 0.42 km North

Displacement is the net change in distance traveled.

Eastward distance :

To = (1.3 + 0.68)km = 1.98km East

Fro = (0.68 + 1.3)km = 1.98 km East

Δ distance = (1.98 - 1.98) = 0

Northward direction:

To = (1.1 + 0.42)km = 1.52km north

Fro = (0.42 + 1.1)km = 1.52km North

Δ distance = (1.98 - 1.98) = 0

Hence final Displacement = 0

Total distance covered = 2 × (1.3 + 0.68 + 1.1 + 0.42) = 2 × 3.5

= 7km

3 0
3 years ago
A toroid having a square cross section, 5.00 cm on a side, and an inner radius of 15.0 cm has 500 turns and carries a current of
SCORPION-xisa [38]

Answer:

a).β=0.53x10^{-3} T

a).β=0.40 x10^{-4} T

Explanation:

The magnetic field at distance 'r' from the center of toroid is given by:

\beta =\frac{u_{o}*I*N}{2\pi*r}

a).

N=500\\I=0.800A\\r=15cm*\frac{1m}{100cm}=0.15m\\u_{o}=4\pi x10^{-7}\frac{T*m}{A}  \\\beta=\frac{4\pi x10^{-7}\frac{T*m}{A}*0.8A*500}{2\pi*0.15m} \\\beta=0.53x10^{-3}T

b).

The distance is the radius add the cross section so:

r_{1}=15cm+5cm\\r_{1}=20cm

r_{1} =20cm*\frac{1m}{100cm}=0.20m

\beta =\frac{u_{o}*I*N}{2\pi*r1}

\beta =\frac{4\pi x10^{-7}*0.80A*500 }{2\pi*0.20m} \\\beta=0.4x10^{-3} T

3 0
3 years ago
If you remove large amounts of heat from a liquid, what could happen
snow_lady [41]
A liquid becomes a solid when energy is removed. The energy content decreases, and the speed of the particles decrease.
3 0
3 years ago
A 1-kg collar (located at point (2,2) from the origin) is pulled along a vertical, frictionless bar with a force of 10 N applied
faltersainse [42]

Answer:

The acceleration of the collar is 10 m/s²

Explanation:

Given;

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The acceleration of the collar can be calculated by applying Newton's second law of motion;

F = ma

where;

F is the applied force

m is mass of the object

a is the acceleration

a = F / m

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Therefore, the acceleration of the collar is 10 m/s²

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Answer:

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