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monitta
3 years ago
13

You throw a ball straight down from an apartment balcony to the ground below. The ball has an initial velocity of 5.00m/s, direc

ted downward, and it hits the ground 2.00s after it is released. Find the height of the balcony
Please include the steps
Physics
1 answer:
erik [133]3 years ago
5 0
The height of the balcony may be calculated through the equation,
                                h = V₀t + 0.5gt²
where h is the height, V₀ is the initial velocity, g is the gravitational constant and t is time. Substituting the values given above,
                               h = (5 m/s)(2s) + 0.5(10 m/s²)(2 s)²
                                 h  = 30 m
Thus, the height of the balcony is 30 meters. 
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A cannon ball is shot straight upward with a velocity of 72.50 m/s. How high is the cannon ball above the ground 3.30 seconds af
disa [49]

Answer:

Explanation:

Given

Cannon is fired with a velocity of u=72.50\ m/s

Using Equation of motion

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

where

y=displacement

u=initial\ velocity

a=acceleration

t=time

after time t=3.3 s

y=72.50\times 3.3-\frac{1}{2}\times 9.8\times (3.3)^2

y=239.25-53.36

y=185.89\ m

So after 3.3 s cannon ball is at a height of 185.89 m

6 0
3 years ago
What happens when a force exerted on an object cause the object to move?
Charra [1.4K]

Answer:

B. Kinetic energy is created

Explanation:

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2 years ago
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A room with 3.1-m-high ceilings has a metal plate on the floor with V = 0V and a separate metal plate on the ceiling. A 1.1g gla
miss Akunina [59]

Answer:

The ball traveled 0.827 m

Explanation:

Given;

distance between the metal plates of the room, d = 3.1 m

mass of the glass, m = 1.1g

charge on the glass, q = 4.7 nC

speed of the glass ball, v = 4.8 m/s

voltage of the ceiling, V = +3.0 x 10⁶ V

The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;

F = qV/d

|F| = (4.7 x 10⁻⁹ x 3 x  10⁶) / (3.1)

|F| = 4.548 x 10⁻³ N

F = - 4.548 x 10⁻³ N

The net horizontal force experienced by this ball is;

F_{net} = F_c - mg\\\\F_{net} = -4.548 *10^{-3} - (1.1*10^{-3} * 9.8)\\\\F_{net} = -15.328*10^{-3} \ N

The work done between the ends of the plate is equal to product of the  magnitude of net force on the ball and the distance traveled by the ball.

W = F_{net} *h\\\\W = 15.328 *10^{-3} *  h

W = K.E

15.328*10^{-3} *h = \frac{1}{2}mv^2\\\\ 15.328*10^{-3} *h = \frac{1}{2}(1.1*10^{-3})(4.8)^2\\\\ 15.328*10^{-3} *h =0.0127\\\\h = \frac{0.0127}{15.328*10^{-3}}\\\\ h = 0.827 \ m

Therefore, the ball traveled 0.827 m

4 0
3 years ago
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7 0
2 years ago
A uniform magnetic field passes through a horizontal circular wire loop at an angle 19.5 ∘ from the vertical. The magnitude of t
nlexa [21]

To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,

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\epsilon = -(1) \pi (0.220m)^2 cos(19.5\°)(  (3.05T/s)-(13.9T/s)(5.71s))

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Therefore the magnitude of the induced emf is 10.9V

4 0
3 years ago
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