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levacccp [35]
2 years ago
11

Use the formula h = −16t2 + v0t. (if an answer does not exist, enter dne.) a ball is thrown straight upward at an initial speed

of v0 = 56 ft/s. (a) when does the ball initially reach a height of 40 ft?
Physics
1 answer:
makkiz [27]2 years ago
3 0
Using the given formula with v0=56 ft/s and h=40 ft 
h = -16t2 + v0t  
40 = -16t2 + 56t 
16t2 - 56t + 40 = 0  
Solving the quadratic equation:  
t= (-b+/-(b^2-4ac)^1/2)/2a = (56+/-((-56)^2-4*16*40)^1/2)/2*16 = (56 +/- 24) / 32 
 We have two possible solutions  
t1 = (56+24)/32 = 2.5 
t2 = (56-24)/32 = 1  
So initially the ball reach a height of 40 ft in 1 second.
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A thin beam of light enters a thick plastic sheet from air at an angle of 32.0° with the normal and continues in the sheet at an
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Answer:

1.36

Explanation:

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i = angle of incidence in air = 32.0° deg

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Using Snell's law

n_{air} Sini = n_{plastic} Sinr

(1) SIn32.0 = n_{plastic} Sin23.0

n_{plastic} = 1.36

5 0
2 years ago
A coworker did not clean his work area before going home this could cause an accident so you quickly clean up the next day you s
defon

Answer:

THE FIRST ONE YOU SHOULD TELL HIM AND THE LAST ONE YOU SHOUDENT DO BECAUSE HE WILL DO IT AGAIN AND EXPECT OTHERS TO CLEAN UP AFTER HIM

Explanation:

5 0
1 year ago
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According to the article, which pattern of brain waves are most conducive to studying new information?
sashaice [31]
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7 0
1 year ago
A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceler
aleksandrvk [35]

Answer:

Part a)

t_1 = \frac{\omega_1}{\alpha}

Part b)

\theta = \frac{1}{2}\alpha t_1^2

Part c)

t = \frac{t_1}{5}

Explanation:

Part a)

As we know that it is having constant torque so here the time taken by it to accelerate is given as

\omega_f = \omega_i + \alpha t

\omega_1 = 0 + \alpha t_1

t_1 = \frac{\omega_1}{\alpha}

Part b)

angular displacement is given as

\theta = \omega_i t_1 + \frac{1}{2}(\alpha) t_1^2

\theta = 0 + \frac{1}{2}\alpha t_1^2

\theta = \frac{1}{2}\alpha t_1^2

Part c)

As we know that the angular deceleration produced by the brakes is given as

\alpha_d = - 5\alpha

now we have

\omega_f = \omega _i + \alpha t

0 = \omega_1 - 5 \alpha_1 t

t = \frac{\omega_1}{5 \alpha}

As we know that

t_1 = \frac{\omega_1}{\alpha}

so we have

t = \frac{t_1}{5}

6 0
1 year ago
A ball of mass 5.0kg is lifted off the floor a distance of 1.7m. 1. What is the change in the gravitational potential energy of
tangare [24]

Answer:

Explanation:

Change in gravitational energy of the ball = mgh

5 mutiply 10 multiply 1.7 = 85J

Potential energy at height = Kinetic energy at bottom

KE= 85J

Velocity

v=5.83m/s

4 0
1 year ago
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