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Dahasolnce [82]
3 years ago
7

An airplane starts at rest and accelerates down the runway for 25 s. At the end of the runway, its final velocity is 75 m/s east

. What is its acceleration?
Physics
1 answer:
elena-s [515]3 years ago
7 0

Answer

3m/s²

Explanation

u=o

t=25s

v=75m/s

from Newton's first law of motion,

v=u+at

75=0+a(25)

75=25a

a= 75/25

a=3m/s²

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The train traveled 500 kilometers north to Odessa in 2 hours . What’s the train’s speed ? What is the velocity ?
ElenaW [278]

Explanation:

Speed is distance over time.

500 km / 2 hr = 250 km/hr

Velocity is speed and direction.

250 km/hr north

7 0
3 years ago
A bicycle tire is spinning counterclockwise at 2.70 rad/s. During a time period Δt = 1.50 s, the tire is stopped and spun in the
Andru [333]

Answer:

Δω = -5.4 rad/s

αav = -3.6 rad/s²

Explanation:

<u>Given</u>:

           Initial angular velocity = ωi = 2.70 rad/s

           Final angular velocity = ωf = -2.70 rad/s (negative sign is  

           due to the movement in opposite direction)

           Change in time period = Δt = 1.50 s

<u>Required</u>:

           Change in angular velocity = Δω = ?

           Average angular acceleration = αav = ?

<u>Solution</u>:

          <u>Angular velocity (Δω):</u>

               Δω = ωf - ωi

               Δω = -2.70 - 2.70

               Δω = -5.4 rad/s.

          <u> Average angular acceleration (αav):</u>

               αav = Δω/Δt

               αav = -5.4/1.50

              αav = -3.6 rad/s²

Since, the angular velocity is decreasing from 2.70 rad/s (in counter clockwise direction) to rest and then to -2.70 rad/s (in clockwise direction) so, the change in angular velocity is negative.

7 0
2 years ago
Read 2 more answers
it takes 90 j of work to stretch a spring 0.2 m from its equilibrium position. How muc work is needed to stretch it an additiona
Vinvika [58]

Work needed: 720 J

Explanation:

The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the stretching of the spring from the equilibrium position

In this problem, we have

E = 90 J (work done to stretch the spring)

x = 0.2 m (stretching)

Therefore, the spring constant is

k=\frac{2E}{x^2}=\frac{2(90)}{(0.2)^2}=4500 N/m

Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of

x = 0.2 + 0.4 = 0.6 m

Substituting,

E'=\frac{1}{2}kx^2=\frac{1}{2}(4500)(0.6)^2=810 J

Therefore, the additional work needed is

\Delta E=E'-E=810-90=720 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

7 0
3 years ago
Suppose you have a 113-kg wooden crate resting on a wood floor. (μk = 0.3 and μs = 0.5) (a) What maximum force (in N) can you ex
AURORKA [14]

Answer:

554.27N

Explanation:

(a)  The max frictional force exerted horizontally on the crate and the floor is,

Substitute the values,

μs=0.5

mass=113kg

g=9.81m/s

Ff=μsN

   =μsmg

   =(0.5 x 113 x 9.81)

Ff=554.27N

3 0
2 years ago
Match the following ERA’s Question 4
inna [77]
1. Cenozoic ERA
2. Mesozoic ERA
3. Paleozoic ERA
8 0
2 years ago
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