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Hoochie [10]
3 years ago
7

A car moving in a straight line starts at x=0 at t=0. it passes the point x=25.0m with a speed of 11.0m/s at t=3.0s. it passesth

e point x=385m with a speed of 45.0m/s at t=20.0s. find the average velocity and the average acceleration, between t=3.0s and t=20.0s
Physics
1 answer:
Ksenya-84 [330]3 years ago
5 0

Between time 3 second to 20 seconds the time difference = 17 seconds

Distance traveled = 385 - 25 = 360 m

So average velocity = 360/17 = 21.18 m/s

Change in velocity at t =3 seconds and t = 20 seconds is given by 45 - 11 = 34 m/s

So average acceleration = 34/17 = 2  m/s

         

You might be interested in
A water wheel rotates with a period of 2.26 s. If the water wheel has a radius of l.94 m,
Dafna11 [192]

Answer:

The correct answer is B)

Explanation:

When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel.  So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.

The formula for calculating the velocity of a point on the edge of the wheel is given as

V_{r} = 2π r / T

Where

π is Pi which mathematically is approximately 3.14159

T is period of time

Vr is Velocity of the point on the edge of the wheel

The answer is left in Meters/Seconds so we will work with our information as is given in the question.

Vr = (2 x 3.14159 x 1.94m)/2.26

Vr = 12.1893692/2.26

Vr = 5.39352619469

Which is approximately 5.39

Cheers!

7 0
3 years ago
Which of the following represents an element?
Digiron [165]
H2 is the correct answer
7 0
3 years ago
Which situation will change the direction of the bicycle?
Elodia [21]

Answer:

Dont know if this is right but i say C

Explanation:

6 0
3 years ago
An airplane accelerates from a speed of 88m/s to a speed of 132 m/s during a 15 second time interval. How far did the airplane t
Gelneren [198K]

Answer:

1650\:\mathrm{m}

Explanation:

We can use the following kinematics equations to solve this problem:

v_f=v_i+at,\\{v_f}^2={v_i}^2+2a\Delta x.

Using the first one to solve for acceleration:

132=88+a(15),\\15a=44,\\a=\frac{44}{15}=2.9\bar{3}\:\mathrm{m/s^2}.

Now we can use the second equation to solve for the distance travelled by the airplane:

132^2=88^2+2\cdot2.9\bar{3}\cdot \Delta x,\\\Delta x= \frac{9680}{2\cdot2.9\bar{3}},\\\Delta x =\fbox{$ 1650\:\mathrm{m}$}(three significant figures).

6 0
3 years ago
It is claimed that if a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all its kinet
Amiraneli [1.4K]

Answer:

354.72 m/s

Explanation:

m = mass of lead bullet

c = specific heat of lead = 128 J/(kg °C)

L = Latent heat of fusion of lead = 24500 J/kg

T_{i} = initial temperature = 27.4 °C

T_{f} = final temperature = melting point of lead =  327.5 °C

v = Speed of lead bullet

Using conservation of energy

Kinetic energy of bullet = Heat required for change of temperature + Heat of melting

(0.5) m v^{2} = m c (T_{f} - T_{i}) + m L\\(0.5) v^{2} = c (T_{f} - T_{i}) + L\\(0.5) v^{2} = (128) (327.5 - 27.4) + 24500\\(0.5) v^{2} = 62912.8\\v = 354.72 ms^{-1}

3 0
3 years ago
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