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svlad2 [7]
3 years ago
9

Lacie kicks a football from ground level at a velocity of 13.9 m/s and at an angle of 25.0° to the ground. How long will the bal

l be in the air before it lands? Round your answer to the nearest tenth. s How far will the football travel before it lands? Round your answer to the nearest tenth. m
Physics
2 answers:
RUDIKE [14]3 years ago
6 0

Answer:

1.2s for the first one

15.1m for the second one on time4learning

Explanation:

Step2247 [10]3 years ago
4 0

Answer:

T = 1.2 s

T = 15.1 m = 15 m

Explanation:

This is a case of projectile motion:

TOTAL TIME OF FLIGHT:

The formula for total time of flight in projectile motion is:

T = 2 V₀ Sinθ/g

where,

T = Total Time of Flight = ?

V₀ = Launch Speed = 13.9 m/s

θ = Launch Angle = 25°

g = 9.8 m/s²

Therefore,

T = (2)(13.9 m/s)(Sin 25°)/(9.8 m/s²)

<u>T = 1.2 s</u>

<u></u>

RANGE OF BALL:

The formula for range in projectile motion is:

R = V₀² Sin2θ/g

where,

R = Horizontal Distance Covered by ball = ?

Therefore,

T = (13.9 m/s)²(Sin 2*25°)/(9.8 m/s²)

<u>T = 15.1 m = 15 m</u>

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3 0
3 years ago
HELP ME PLZ
Nadusha1986 [10]

Answer: 1200 000 J = 1.2 MJ

Explanation: Ek = 0.5 mv² = 0.5 · 1500 kg· (40 m/s)²

3 0
3 years ago
Two spherical conductors are separated by a distance much larger than either of their radii. Sphere A has a radius of 11.5 cm an
bonufazy [111]

Explanation:

As the given spheres are connected by a thin wire so, the potential on the spheres are the same.

          \frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}} ......... (1)

Hence, total charge will be as follows.

              q_{1} + q_{2} = Q = -95.5 nC .......... (2)

Using the above two equations, the final equation will be as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

and,    q_{1} = \frac{Qr_{1}}{r_{1} + r_{2}}

Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

                     = \frac{-95.5 \times 74.4 cm}{(11.5 + 74.4) cm}

                     = 82.714 nC

Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.

                       

5 0
3 years ago
If you ride quickly down a hill on a bicycle your eardrums are pushed in before they pop back. Why is this?
Mice21 [21]

Answer:

<em>The difference in pressure between the external air pressure, and the internal air pressure of the middle ear.</em>

Explanation:

First of all, we should note that pressure decreases with height and increases with depth. The air within the middle ear (between the ear drum and the Eustachian tube) adjusts itself to respond to the atmospheric pressure, or when we yawn.  At a high altitude like on the hill, the air pressure in the middle ear, is fairly low (this is to balance the low air pressure at this height). While riding down the hill quickly, there is little time for the air pressure in the ear to readjust itself to the increasing external air pressure, causing the external air to push into the ear drum. Along the way, the air within the middle ear is adjusted by the opening of the Eustachian tube, allowing more air into the space in the middle ear to balance the external air pressure. This readjustment causes the ear to pop.

7 0
3 years ago
A steel ball of mass 50 g is rolled from the left toward a ball of lead of mass 500 g. The steel ball is traveling with a veloci
alina1380 [7]

Answer: It's hard to say without characterizing the collision. But it will be either A if the collision is totally in-elastic, or B if the collision is totally elastic. It could be anywhere in between for partially elastic collisions.

Explanation:

momentum is conserved, so initial system momentum will be left to right.

The velocity of the center of mass is 50(5) / 550 = 0.4545... m/s

In an elastic collision, the lead ball will move off at twice that speed or 0.91 m/s to the right.

The steel ball will bounce back and move away at 0.91 - 5 = -4.1 m/s . The negative sign indicates the steel ball has reversed course and has negative momentum

In a totally in-elastic collision, both balls would move to the right at 0.45 m/s. The steel ball will still have positive momentum.

4 0
3 years ago
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