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den301095 [7]
3 years ago
7

Can a body have constant speed and still be acclerating ? Give an Example

Physics
1 answer:
Allushta [10]3 years ago
3 0






Hi Pupil Here's Your answer :::





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An object moving with constant speed can be accelerated if direction of motion changes. For example, an object moving with a constant speed in a circular path has an acceleration because its direction of motion changes continuously.





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Hope this Helps . . . . . . . . .
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A truck initially traveling at a speed of 22 m/s increases at a constant rate of 2.4 m/s^2 for 3.2s. What is the total distance
FinnZ [79.3K]

Answer:

82.7 m

Explanation:

u= 22m/s

a= 2.4 m/s^2.

t= 3.2 secs

Therefore the distance travelled can be calculated as follows

S= ut + 1/2at^2

= 22 × 3.2 + 1/2 × 2.4 × 3.2^2

= 70.4 + 1/2×24.58

= 70.4 + 12.29

= 82.7 m

Hence the distance travelled by the truck is 82.7 m

6 0
3 years ago
Why is the Mid-Atlantic Ridge an ideal place for SWARM to collect electrical conductivity data?
ElenaW [278]

Answer: Mid-ocean ridges are geologically important because they occur along the kind of plate boundary where new ocean floor is created as the plates spread apart. Thus the mid-ocean ridge is also known as a "spreading center" or a "divergent plate boundary." The plates spread apart at rates of 1 cm to 20 cm per year.

3 0
3 years ago
Read 2 more answers
A cyclist accelerates from a velocity of 10 miles/hour east until reaching a velocity of 20 miles/hour east in 5 seconds. What w
Sliva [168]

Answer:

a = 0.894\ m/s^2

Explanation:

<u>Motion with Constant Acceleration</u>

A body moves with constant acceleration when the speed changes uniformly in time. The equation used to find the final speed vf is

v_f=v_o+at

Where vo is the initial speed, a is the acceleration, and t is the time.

The cyclist has an initial speed of vo=10 miles/hour and ends up at vf=20 miles/hour in t=5 seconds.

Both speeds are given in miles/hour and we must convert it to m/s:

1 mile/hour = 0.44704 m/s

10 mile/hour = 4.47 m/s

20 mile/hour = 8.94 m/s

The acceleration is calculated by solving for a:

\displaystyle a=\frac{v_f-v_o}{t}

\displaystyle a=\frac{8.94-4.47}{5}

a = 0.894\ m/s^2

3 0
2 years ago
While jumping on a trampoline you calculate that at the highest peak of your jump you have 900 joules of gravitational potential
BabaBlast [244]

Jumping on a trampoline is a classic example of conservation of energy, from potential into kinetic. It also shows Hooke's laws and the spring constant. Furthermore, it verifies and illustrates each of Newton's three laws of motion.

<u>Explanation</u>

When we jump on a trampoline, our body has kinetic energy that changes over time. Our kinetic energy is greatest, just before we hit the trampoline on the way down and when you leave the trampoline surface on the way up. Our kinetic energy is 0 when you reach the height of your jump and begin to descend and when are on the trampoline, about to propel upwards.

Potential energy changes along with kinetic energy. At any time, your total energy is equal to your potential energy plus your kinetic energy. As we go up, the kinetic energy converts into potential energy.

Hooke's law is another form of potential energy. Just as the trampoline is about to propel us up, your kinetic energy is 0 but your potential energy is maximized, even though we are at a minimum height. This is because our potential energy is related to the spring constant and Hooke's Law.

8 0
3 years ago
A football player kicks the ball at a 45 degree angle. Without an effect from the wind, the ball would travel 60.0m horizontally
Ronch [10]
B when the ball is at its maxium height 
5 0
3 years ago
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