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HACTEHA [7]
10 months ago
10

A boy kicks a soccer ball, giving it an initial speed of 7.5m/s at an angle of 27° above the horizontal(=ground). How high will

the ball go? How far will it travel? PLEASE HELP AND SHOW WORKKK
Physics
1 answer:
Gelneren [198K]10 months ago
3 0

When the initial speed given is 7.5m/s at an angle of 27° , ball will go

4.637 meters.

Assume no air opposition to the ball ;

Vertical component of ball is sin 27° =  0.453

0.453* 7.5 = 3.404 meters /sec

Time taken to reach ground is :

3.404 = -3.404+9.8*t

t= 6.808/9.8= 0.694 sec

Horizontal component is 7.5*cos27°= 6.682m/s

Distance = speed * time

=6.682 * 0.694

=4.637 meters

Horizontal distance it can cover in 0.694 sec is 4.637 meters

So range of ball is 4.637 meters.

Form of motion experienced by an object or particle that is projected near surface of the earth and moves along a curve is called Projectile motion. Three types of projectile motion are Horizontal projectile motion. Oblique projectile motion and Projectile motion on an inclined plane.

To know more about projectile motion, refer

brainly.com/question/24216590

#SPJ13

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A bar magnet was placed underneath a sheet of paper where a pile of iron filings sits. In the presence of the energy stored in t
Naya [18.7K]

Answer: 1. The field energy will increase

2. The energy increases, and the lines of force are denser

3. It points toward the field of earths magnetic poles

4. 1 and 2 only

5. 2, 4, 1, 3

Explanation: just took it

4 0
2 years ago
A rubber ball is dropped from a height of 8m. After strikingthe floor, the ball bounces to a height of 5m. a. If the ball had bo
kifflom [539]

Answer:

a) This means the collision between the ball and the floor is elastic.

b) This points to a perfectly inelastic collision between the ball and the floor as they stick together after collision

c) Check Explanation.

Explanation:

Collision of bodies are analysed according to whether both momentum and kinetic energy of the system is conserved, that is, if these two quantities before collision are equal to their values after collision.

In all types of collisions, momentum is usually conserved, but kinetic energy is conserved only in an elastic collision.

A ball dropped from a height of 8 m bounces up back to a height of 5 m.

a. If the ball had bounced to a height of 8m, how would you describe the collision between the ball and the floor?

The ball not bouncing back to a height of 8 m shows energy loss at some point in the total motion of the ball (most likely at the collision). If kinetic energy was conserved, the ball would bounce back up to the height at which it fell from (8 m) after the collision with the floor.

b. If the ball had not bounced at all, how would you describe the collision between the ball and the floor?

If the ball had not bounced at all, this means it lost all of its kinetic energy to the floor, and this points to a perfectly inelastic collision between the ball and the floor as they stick together after collision.

c. What happened to the energy lost by the ball during thecollision?

The energy lost during the collision is converted to another form, most likely responsible for some deformation on the ball & a minute deformation on the floor, converted to some form of heat as a result of the collision or into sound energy, usually, it's a combination of all This!

Hope this Helps!!!

5 0
3 years ago
Could anyone help with this question I do
ra1l [238]
<span>which of those can we not get back once it has been used, the answer is oil or petroleum</span>
8 0
3 years ago
Read 2 more answers
(a) How many kilograms of water must evaporate from a 60.0-kg woman to lower their body temperature by 0.750ºC?
Mrrafil [7]

Answer:

69.69 g

Explanation:

Evaporation of water will take out latent heat of vaporization.  Let the mass of water be m and latent heat of vaporization of water be 2260000 J per kg

Heat taken up by evaporating water

= 2260000 x m J

Heat lost by body

= mass x specific heat of body x drop in temperature

60 x 3500 x .750  ( specific heat of human body is 3.5 kJ/kg.k)

= 157500 J

Heat loss = heat gain

2260000 m= 157500

m = .06969 kg

= 69.69 g

5 0
3 years ago
Use the following equation to help you answer the question. The peak intensity of radiation from a star named Sigma is 2 x 10 6
puteri [66]

Answer:

1449 K

Explanation:

The surface temperature of a star is related to its peak wavelength by Wien's displacement law:

T=\frac{b}{\lambda}

where

T is the surface temperature

b is Wien's displacement constant

\lambda=2\cdot 10^{-6} m

So the surface temperature of the star is

T=\frac{2.898 \cdot 10^{-3} Km}{2\cdot 10^{-6} m}=1449 K

8 0
3 years ago
Read 2 more answers
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