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3 years ago

7

if a ball was dropped from 120 meters. if it takes 2 seconds for the ball to fall 60 meters, how long will it take for the ball

to reach the ground

1 answer:

Sever21 [200]3 years ago

7 0

For the ball to hit the ground it will take 4 seconds since 120 divided by 60 is 2 two times two=4

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Ah electron is a negatively charged particle that has a charge of magnitude, e - 1.60 x 10-19 C. Which one of the following stat

Advocard [28]

Answer:

The correct statement is "The electric field is directed toward the electron and has a magnitude of $\rm \dfrac{ke}{r^2}$ ".

Explanation:

According to Coulomb's law, the magnitude of the electric field due to a static point charge q at a point r distance away from it is given by

$\rm E = \dfrac{k|q|}{r^2}.$

k is the Coulmob's constant.

The direction of the electric field along the line joining the charge and the point where electric field is to be found and it is directed from positive charge to negative charge.

Conventionally, we assume a positive test charge placed at the point where electric field is to be found, the test charge has very small charge such that its charge does not affect the electric field due to the given charge.

The charge on the electron = -e.

The electric field due to an electron is given by

$\rm E = \dfrac{k|-e|}{r^2}=\dfrac{ke}{r^2}.$

The direction of this electric field is from positive test charge, placed at the point where electric field is to be found, towards the electron along the line joining the two.

Thus, the correct statement is "The electric field is directed toward the electron and has a magnitude of $\rm \dfrac{ke}{r^2}$ ".

5 0

2 years ago

A 1kg box is pushed on a flat surface that is 250m long. The box is initially at rest and then pushed with a constant Net force

sasho [114]

Answer:

C) 50 m/s

Explanation:

With the given information we can calculate the acceleration using the force and mass of the box.

Newton's 2nd Law: F = ma

5 N = 1 kg * a
a = 5 m/s²

List out known variables:

v₀ = 0 m/s
a = 5 m/s²
v = ?
Δx = 250 m

Looking at the constant acceleration kinematic equations, we see that this one contains all four variables:

v² = v₀² + 2aΔx

Substitute known values into the equation and solve for v.

v² = (0)² + 2(5)(250)
v² = 2500
v = 50 m/s

The final velocity of the box is C) 50 m/s.

7 0

3 years ago

A resistor of 5 Ω is placed in a circuit. The voltage drop across the resistor is 12 V. What is the current through the resistor

Kaylis [27]

<span>Data:

</span>R = 5 Ω
U = 12 V
i = ?
<span>
Formula:

</span> $U = R*i$ → $i = \frac{U}{R}$ <span>

Solving:

</span> $i = \frac{U}{R}$
$i = \frac{12}{5}$
$\boxed{i = 2.4A}$

Answer:

<span>The current through the resistor is 2.4 Amperes</span><span>

</span>

5 0

3 years ago

A ball is thrown from a rooftop with an initial downward velocity of magnitude vo = 2.9 m/s. The rooftop is a distance above the

Step2247 [10]

Answer:

a) The velocity of the ball when it hits the ground is -20.5 m/s.

b) To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

Explanation:

I´ve found the complete question on the web:

<em />

<em>A ball is thrown from a rooftop with an initial downward velocity of magnitude v0=2.9 m/s. The rooftop is a distance above the ground, h= 21 m. In this problem use a coordinate system in which upwards is positive.</em>

<em>(a) Find the vertical component of the velocity with which the ball hits the ground.</em>

<em>(b) If we wanted the ball's final speed to be exactly 27, 3 m/s from what height, h (in meters), would we need to throw it with the same initial velocity?</em>

<em />

The equation of the height and velocity of the ball at any time "t" are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the ball at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity of the ball at a time "t".

First, let´s find the time it takes the ball to reach the ground (the time at which h = 0)

h = h0 + v0 · t + 1/2 · g · t²

0 = 21 m - 2.9 m/s · t - 1/2 · 9.8 m/s² · t²

Solving the quadratic equation using the quadratic formula:

t = 1.8 s ( the other solution of the quadratic equation is rejected because it is negative).

Now, using the equation of velocity, let´s find the velocity of the ball at

t = 1.8 s:

v = v0 + g · t

v = -2.9 m/s - 9.8 m/s² · 1.8 s

v = -20.5 m/s

The velocity of the ball when it hits the ground is -20.5 m/s.

b) Now we have the final velocity and have to find the initial height. Using the equation of velocity we can obtain the time it takes the ball to acquire that velocity:

v = v0 + g · t

-27.3 m/s = -2.9 m/s - 9.8 m/s² · t

(-27.3 m/s + 2.9 m/s) / (-9.8 m/s²) = t

t = 2.5 s

The ball has to reach the ground in 2.5 s to acquire a velocity of 27.3 m/s.

Using the equation of height, we can obtain the initial height:

h = h0 + v0 · t + 1/2 · g · t²

0 = h0 -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

-h0 = -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

h0 = 38 m

To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

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3 years ago

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Komok [63]

Answer:

c

Explanation:

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3 years ago

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