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balandron [24]
2 years ago
6

Multiply. (2x + 4)(x - 4)

Physics
1 answer:
sergij07 [2.7K]2 years ago
7 0

Answer:

(2x + 4)(x - 4)=2x^2-4x-16

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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.

6 0
3 years ago
Most of the mass of the milky way exists in the form of.
Andrews [41]

Answer: Dark matter.

Explanation: Hope it helps :)

7 0
2 years ago
Read 2 more answers
The temperature of the cooling water as it leaves the hot engine of an automobile is 240 °F. After it passes through the radiato
djverab [1.8K]

Answer:571.09 kJ

Explanation:

Given

Temperature of cooling water from engine exit=240^{\circ} F\approx 115.55^{\circ}C

After Passing through the radiator its temperature decreases to 175^{\circ}F\approx 79.44^{\circ}F

specific heat of water=4.184 J/g^{\circ}C

Volume of water = 1 gallon\approx 3.78 L

density of water \rho =1 gm/mL

Thus mass of water=\rho \times V=3.78\times 1=3.78 kg

Heat transferred to the surrounding is equal to heat absorbed by cooling water

Q=m\cdot c\cdot \Delta T

Q=3.78\times 4.184\times 1000\times (115.55-79.44)

Q=3.78\times 4.184\times 1000\times (36.11)

Q=571.09 kJ

4 0
3 years ago
How do scientific tests help determine the properties of substances
tino4ka555 [31]
It can help determine substances that appear similar but react differently under the same circumstances.
7 0
3 years ago
In which case would electrical potential energy be built up and stored in the electric field? a) A positive charge is moved towa
mr_godi [17]

Answer:

The correct option is B

Explanation:

Although, it is common knowledge that in an electric field, unlike charges attract and like charges repel. However, to build up an electric potential, a positive charge is brought close to another positive charge - this causes work done to be changed to electric potential energy and stored in the electric field.

It should however be noted that when a negative charge is moved away from a positive charge, the negative charge gains electric potential energy.

5 0
2 years ago
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