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3 years ago

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A coin is tossed upward with an initial velocity of 32 feet per second from a height of 16 feet above the ground. The equation g

iving the object's height h at any time t is h = 16 + 32 t − 16 t 2 . Does the object ever reach a height of 32 feet?

1 answer:

olasank [31]3 years ago

8 0

Answer:

yes.

Explanation:

Suppose height of 32 ft is reached after t seconds . Putting the values in the given equation gives real value of t that means it reaches the height of 32 feet

So 32 = 16 + 32t-16t²

16t²- 32t+16 = 0

t² - 2t + 1 = 0

t = 1 s

So real value of t is obtained . That means it reaches the height of 32 ft.

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2 years ago

Read 2 more answers

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.1 m/s. How long does h

balandron [24]

We need to use the kinematic equation
S=ut+(1/2)at^2
where
S=displacement (+=up, in metres)
u=initial velocity (m/s)
t=time (seconds)
a=acceleration (+=up, in m/s^2)

Substitute values
S=displacement = 1.96-2.27 = -0.31 m (so that shot does not hit his head)
u=11.1
a=-9.81 (acceleration due to gravity)

-0.31=11.1t+(1/2)(-9.81)t^2
Rearrange and solve for t
-4.905t^2+11.1t-0.31=0
t=-0.02756 or t=2.291 seconds
Reject the negative root to give
t=2.29 seconds (to 3 significant figures)

3 0

3 years ago

2200 kg semi truck driving down the highway has lost control. The truck rolls across the median and into oncoming traffic. The t

serious [3.7K]

Answer:

The semi truck travels at an initial speed of 69.545 meters per second downwards.

Explanation:

In this exercise we see a case of an entirely inellastic collision between the semi truck and the car, which can be described by the following equation derived from Principle of Linear Momentum Conservation: (We assume that velocity oriented northwards is positive)

$m_{S}\cdot v_{S}+m_{C}\cdot v_{C} = (m_{S}+m_{C})\cdot v$ (1)

Where:

$m_{S}$ , $m_{C}$ - Masses of the semi truck and the car, measured in kilograms.

$v_{S}$ , $v_{C}$ - Initial velocities of the semi truck and the car, measured in meters per second.

$v$ - Final speed of the system after collision, measured in meters per second.

If we know that $m_{S} = 2200\,kg$ , $m_{C} = 2000\,kg$ , $v_{C} = 45\,\frac{m}{s}$ and $v = -15\,\frac{m}{s}$ , then the initial velocity of the semi truck is:

$m_{S}\cdot v_{S} = (m_{S}+m_{C})\cdot v -m_{C}\cdot v_{C}$

$v_{S} = \frac{(m_{S}+m_{C})\cdot v - m_{C}\cdot v_{C}}{m_{S}}$

$v_{S} = \left(1+\frac{m_{C}}{m_{S}} \right)\cdot v - \frac{m_{C}}{m_{S}} \cdot v_{C}$

$v_{S} = v +\frac{m_{C}}{m_{S}}\cdot (v-v_{C})$

$v_{S} = -15\,\frac{m}{s}+\left(\frac{2000\,kg}{2200\,kg} \right) \cdot \left(-15\,\frac{m}{s}-45\,\frac{m}{s} \right)$

$v_{S} = -69.545\,\frac{m}{s}$

The semi truck travels at an initial speed of 69.545 meters per second downwards.

3 0

2 years ago

A driver of a car enters a new 110 km/h speed zone on the highway. The driver begins to accelerate immediately and reaches 110 k

levacccp [35]

Answer:

30Km/h

Explanation:

acceleration is the change of speed in a given time so when we substract the accelerations we can know how much the car goes per an hour

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2 years ago

Find the angle for the third-order maximum for 591 nm wavelength light falling on a diffraction grating having 1460 lines per ce

Marat540 [252]

Answer:

15.32°

Explanation:

We have given the wavelength $\lambda =591nm=591\times 10^{-9}m$

Diffraction grating is 1460 lines per cm

So $d=\frac{10^{-2}}{1460}=6.71\times 10^{-6}m$ (as 1 m=100 cm )

For maximum diffraction

$dsin\Theta =m\lambda$ here m is order of diffraction

So $6.71\times 10^{-6}sin\Theta =3\times 591\times 10^{-9}$

$sin\Theta =0.264$

$\Theta =15.32^{\circ}$

6 0

2 years ago