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3 years ago

You kick a ball with a speed of 14 m/s at an angle of 51°. How far away does the ball land?

Physics

2 answers:

Ksivusya [100]3 years ago

8 0

Answer:

19.6 m

Explanation:

Apex ; good luck to all :)

In-s [12.5K]3 years ago

4 0

-- The vertical component of the ball's velocity is 14 sin(51°) = 10.88 m/s

-- The acceleration of gravity is 9.8 m/s².

-- The ball rises for 10.88/9.8 seconds, then stops rising, and drops for the
same amount of time before it hits the ground.

-- Altogether, the ball is in the air for (2 x 10.88)/(9.8) = 2.22 seconds
==================================

-- The horizontal component of the ball's velocity is 14 cos(51°) = 8.81 m/s

-- At this speed, it covers a horizontal distance of (8.81) x (2.22) = 19.56 meters
before it hits the ground.

As usual when we're discussing this stuff, we completely ignore air resistance.

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The velocity of a bob on a simple pendulum at the lowest position is 10.56 m/s. What is the maximum vertical height it is able t

kondaur [170]

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Explanation:

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7 0

3 years ago

[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r

blsea [12.9K]

Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b) it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity $V_{f}$ for an object traveling distance h taking the initial vertical component of velocity as $V_{i}$ the kinematics equation is written as

$V_{f}^{2}=V_{i}^{2}+2ah$ where a is acceleration

Substituting g for a where g is gravitational force value taken as 9.81

$V_{f}^{2}=V_{i}^{2}+2gh$

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h

$V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}$ = 20.44275

Therefore, the divers enter with a speed of 20.4 m/s

The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)

The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

$V_{f}^{2}=V_{i}^{2}+2gh$

Since we have final velocity of 25 m/s

$V_{i}^{2}=2gh-V_{f}^{2}$

$V_{i}=\sqrt{(V_{f}^{2}-2gh)}$

$V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}$ = 14.390761 m/s

Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s

In conclusion, the upward initial velocity can’t be physically attained

3 0

3 years ago

How would you describe a fossil that was discovered in a rock at the base of a cliff to a fossil in a rock found at the top of a

ArbitrLikvidat [17]

The one at the base would be much older due to the law of super position, and the rock at the top would be much newer,again, due to the law of super position.

7 0

3 years ago

Choose the correct statement regarding a compound microscope being used to look at a mite. The eyepiece lens forms a virtual ima

Nata [24]

Answer:

a) True. The image of the mite is virtual

e) True. The image must be within the focal length of the eyepiece len

Explanation:

Let's review the general characteristics of compound microscopes

Formed by two converging lenses

Magnification is

M = -L/fo 0.25/fe

Where fo is the focal length of the objective lens and fe is the focal length of the ocular lens, L is the tube length

Let's review the claims

a) True. The image of the mite is virtual

b) False. The effect is the opposite of the magnification equation

c) False. The objective lens forms a real image

d) False. As the seal distance increases the magnification decreases

e) True. The image must be within the focal length of the eyepiece len

4 0

3 years ago

In a photoelectric experiment, a metal is irradiated with light of energy 3.56 eV. If a stopping potential of 1.10 V is required

konstantin123 [22]

Answer:

2.46 eV

Explanation:

It is given that,

The energy of light that fall on the metal = 3.56 eV

The stopping potential of the metal = 1.1 V

We need to find the work function of the metal. It is given by the relation as follows :

W = E-KE ...(1)

Where KE is the kinetic energy of the ejected electron and it is given by :

KE = V×e

= 1.1 eV

Put all the values in formula (1)

W = 3.56 eV - 1.1 eV

= 2.46 eV

Hence, the work function of the metal is 2.46 eV. Hence, the correct option is (c).

5 0

3 years ago