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lina2011 [118]
3 years ago
12

13. A catapult launches a boulder with an upward velocity of 148 ft/s. The height of the boulder (h) in feet after t seconds is

given by the function h = –16t² + 148t + 30. How long does it take the boulder to reach its maximum height? What is the boulder’s maximum height? Round to the nearest hundredth, if necessary.
A) 9.25 s; 30 ft
B) 4.63 s; 640.5 ft
C) 4.63 s; 1,056.75 ft
D) 4.63 s; 372.25 ft

16. Solve the equation using the Zero Product Property.
(2x – 4)(2x – 1) = 0
A) 2, –1over2
B) 2, 1over2
C) –2, 2
D) –2, 1over2
Physics
2 answers:
Lostsunrise [7]3 years ago
5 0
32t + 148 = 0148 = 32t4.625 = tt ≈ 4.63h = -16 • (4.625)^2 + 148 • 4.625 + 30 = 372.25
ANSWER: is D. 4.63 sec; 372.25 ft
2.) 2x - 4 = 0 and 2x - 1 = 0x = {4/2, ½}x = {2,½} ANSWER: is B 2, 1 over 2
SVEN [57.7K]3 years ago
4 0

13. Answer: The height will be maximum 372.25 ft at 4.63 s.

Explanation:

Given that,

Upward velocity v = 148 ft/s

Function

h = -16t^2+148t+30.....(I)

On differentiate

\dfrac{dh}{dt}=-32t+148....(II)

for maximum height,

\dfrac{dh}{dt}=0

Put the value of \dfrac{dh}{dt} in equation (II)

-32t+148=0

t = \dfrac{148}{32}

t = 4.63\ s

The maximum height is

Put the value of t in equation(I)

h = -16\times(4.63)^2+148\times4.63+30

h = 372.25 ft

Hence, The height will be maximum 372.25 ft at 4.63 s.

16. Answer: The value of x is 2, \dfrac{1}{2}.

Explanation:

Given that,

(2x-4)(2x-1)=0

Using zero product property

2 x-4=0

x = 2

And,

2x-1=0

x = \dfrac{1}{2}

Hence, The value of x is 2, \dfrac{1}{2}.

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RoseWind [281]

Answer:

a) \Delta U_{g} = 12.945\,J, b) \Delta U_{k} = 12.945\,J, c) k = 2930.059\,\frac{N}{m}

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)

\Delta U_{g} = 12.945\,J

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

\Delta U_{k} = 12.945\,J

c) The spring constant of the gun is:

\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}

k = \frac{2\cdot \Delta U_{k}}{x^{2}}

k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}

k = 2930.059\,\frac{N}{m}

4 0
3 years ago
The MAC is 58 inches, The CG limits are from 26% to 43% MAC. If the CG is found to be
eimsori [14]

By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.

<h3>What are the limits?</h3>

First, we need to find the limits.

We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.

So if 58 in is the 100%, the 26% and 43% of that are:

  • 26% → (26%/100%)*58in = 0.26*58 in = 15.08 in
  • 43% → (43%/100%)*58in = 0.43*58 in = 24.94 in.

But we know that the CG is found to be 45.5% MAC, then it measures:

(45.5%/100%)*58in = 0.455*58in = 26.39 in

We need to compare it with the largest limit, so we get:

26.39 in - 24.94 in = 1.45 in

This means that the CG is 1.45 inches out of limits.

If you want to learn more about percentages, you can read:

brainly.com/question/14345924

6 0
2 years ago
Which ball has the greater average speed during the 1-s interval after release (assuming neither hits the ground during that tim
notka56 [123]

Answer:

The ball thrown downward

Explanation:

When the ball is thrown vertically, the acceleration of it is the gravity acceleration independent if it is thrown downward or upward. However, the acceleration is a vector, so, when the ball is thrown upward, the movement is against the gravity, so the acceleration is negative, and so, the velocity decreases during time; and when the ball is thrown downward, the movement goes to the gravity, so the acceleration is positive, so the velocity increase after time passes.

3 0
3 years ago
A 2.0kg rock is thrown straight up into the air with a speed of 30m/s. Ignore air resistance. What is the net force acting on th
bazaltina [42]

Answer:

19.6N

Explanation:

Given parameters:

Mass of rock = 2kg

Speed  = 30m/s

Unknown:

Net force on the rock  = ?

Solution:

The net force acting on this rock is a function of the acceleration due to gravity acting upon it.

 Net force  = weight  = mass x acceleration due to gravity

 Net force  = 2 x 9.8  = 19.6N downward

6 0
2 years ago
A large truck is moving 22.0 m/s. If it’s momentum is 125,000 kg • m/s, what is the trucks mass
strojnjashka [21]

Answer:

p = mv

m = p/v = 125000/22 = 5682 kg

Explanation:

Direct application of the momentum equation

p = mv

where,

p: momentum

m: mass

v: object velocity

steps:

-------

1) check for units consistency (  SI or Imperial)

2) separate the variable you are looking for.

3) DONE! :DD

6 0
3 years ago
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