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Zielflug [23.3K]
3 years ago
14

The distance traveled, in feet, of a ball dropped from a tall building is modeled by the equation d(t) = 16t2 where d equals the

distance traveled at time t seconds and t equals the time in seconds. What does the average rate of change of d(t) from t = 2 to t = 5 represent?
A. The ball travels an average distance of 112 feet from 2 seconds to 5 seconds.

B. The ball falls down with an average speed of 48 feet per second from 2 seconds to 5 seconds.

C. The ball falls down with an average speed of 112 feet per second from 2 seconds to 5 seconds.

D. The ball travels an average distance of 48 feet from 2 seconds to 5 seconds.
Physics
1 answer:
Ainat [17]3 years ago
6 0
When  t=2, the ball has fallen     d(2) = 16 (2²) = 64 feet .

When  t=5, the ball has fallen     d(5) = 16 (5²) = 400 feet .

Distance fallen from  t=2  until  t=5  is  (400 - 64) = 336 feet.

Time period between  t=2  until  t=5  is  (5 - 2) = 3 seconds.

Average speed of the ball from  t=2  until  t=5  is

                 (distance covered) / (time to cover the distance)

             =            336 feet        /        3 seconds       =  112 feet per second.

That's what choice-C says.        
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5 0
3 years ago
. 30
schepotkina [342]

Answer:

Explanation:

Length if the bar is 1m=100cm

The tip of the bar serves as fulcrum

A force of 20N (upward) is applied at the tip of the other end. Then, the force is 100cm from the fulcrum

The crate lid is 2cm from the fulcrum, let the force (downward) acting on the crate be F.

Using moment

Sum of the moments of all forces about any point in the plane must be zero.

Let take moment about the fulcrum

100×20-F×2=0

2000-2F=0

2F=2000

Then, F=1000N

The force acting in the crate lid is 1000N

Option D is correct

7 0
3 years ago
The tires of a car make 75 revolutions as the car reduces its speed uniformly from 95 km/hr to 55 km.hr. the tires have a diamet
sveta [45]
<span>95 km/h = 26.39 m/s (95000m/3600 secs) 55 km/h = 15.28 m/s (55000m/3600 secs) 75 revolutions = 75 x 2pi = 471.23 radians radius = 0.80/2 = 0.40m v/r = omega (rad/s) 26.39/0.40 = 65.97 rad/s 15.28/0.40 = 38.20 rad/s s/((vi + vf)/2) = t 471.23 /((65.97 + 38.20)/2) = 9.04 secs (vf - vi)/t = a (38.20 - 65.97)/9.04 = -3.0719 The angular acceleration of the tires = -3.0719 rad/s^2 Time is required for it to stop (0 - 38.20)/ -3.0719 = 12.43 secs How far does it go? 65.97 - 38.20 = 27.77 M</span>
7 0
3 years ago
A position vector in the first quadrant has an x-component of 18 m and a magnitude of 30 m. What is the value of its y-component
Snezhnost [94]

Answer:

The value is 24meters

Explanation:

Using

r= xi+yj

To get the magnitude of vector x

We say

/r/= √x²+y²

So

30²= √18² + y²

y= √576

Y= 24m

7 0
3 years ago
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
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