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2 years ago

What is the acceleration of an object with a constant velocity

Physics

2 answers:

valina [46]2 years ago

6 0

Zero.

Acceleration is defined as the change in velocity over time.

Since in your case there is no change, there is no acceleration, so it is zero:

Or in formula: a=Δvt

Where a=acceleration, Δv=change in velocity and t=time

Verizon [17]2 years ago

6 0

Acceleration is 0 m/s/s, Net force is zero. An object moving at constant velocity does not accelerate and does not have an unbalanced force acting on it.

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A child rides her bike at a rate of 12.0 km/hr down the street. A squirrel suddenly runs in front of her so she applies the brak

Sedbober [7]

By v = u - at
=>8 = 12 - a x 0.25 
=>a = 4/0.25 km/hr/sec 
=>a = 16km/hr/sec

I hope this helped!

3 0

3 years ago

Read 2 more answers

What is the velocity of an object that has been in free fall for 1.5s?

Crank

Answer:

D. 15 m/s downward

Explanation:

v = at + v₀

v = (-9.8 m/s²) (1.5 s) + (0 m/s)

v = -14.7 m/s

Rounded to two significant figures, the answer is D, 15 m/s downward.

8 0

2 years ago

The greater the amplitude of a wave a. The quieter the sound b. The higher the frequency of the sound c. The dimmer the light in

Stolb23 [73]

Answer:

The greater the amplitude the greater the energy.

(Think of a water wave - which carries greater energy a 1 ft wave or

a 10 ft wave)

8 0

2 years ago

A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s.

SVEN [57.7K]

Answer:

a) Knowns, initial speed $v_{i}=13.0 m/s$ , final speed $v_{f}=0 m/s$ and gravity due it is a constant $g=9.8m/s^{2}$

b) The maximum high reached by the dolphin is $y_{max}=8.62 m$

c) Total time is $t=2.65s$

Explanation:

a) First of all the initial speed is given at the start of the problem, gravity is constant and final speed is known any object thrown straight up reaches its max high at $0m/s$ speed.

b) Second, now that we know final speed we use $v_{f} =v_{i}-gt$ , as we clear for $t=\frac{v_{i}-v_{f} }{g}=\frac{20.0m/s}{9.8m/s^{2} }=1.32 s$ .

Then we use $y=v_{i}t-\frac{1}{2} gt^{2}=(20.0m/s)(1.32s)-\frac{1}{2} (9.8m/s^{2} ) (1.32s)^{2} =8.62m$

c)Third, finally we can use $y=v_{i}t-\frac{1}{2} gt^{2}$ , as we know $y=0m$ when the dolphin fall into the water again and $v_{i} =13.0m/s$ , then we have $0=(13m/s)t-\frac{1}{2} (9.8m/s^{2} )t^{2}$ is a quadratic form $0=t(13.0-4.9t)$ so we have $t_{i}=0s$ and $t_{f}=\frac{13}{4.90} =2.65s$