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professor190 [17]
3 years ago
11

An object moves with constant acceleration 3.10 m/s2 and over a time interval reaches a final velocity of 12.4 m/s. If its initi

al velocity is −6.20 m/s, what is its displacement during this interval?
Physics
2 answers:
kap26 [50]3 years ago
7 0

Answer:

25.05 m

Explanation:

Using newton's equation of motion.

v² = u²+2as ................. Equation 1

Where v = final velocity of the object, u = initial velocity of the object, a = acceleration of the object, s = displacement of the object.

make s the subject of the equation

s = (v²-u²)/2a............. Equation 2

Given: v = 12.4 m/s, u = -6.20 m/s, a = 3.10 m/s²

Substitute into equation 2

s = (12.4²-(-6.2)²)/(2×3.1)

s = (153.76-38.44)/6.2

s = 155.32/6.2

s = 25.05 m.

Hence the displacement = 25.05 m

Harrizon [31]3 years ago
3 0

Answer:

Explanation:

Given:

a = 3.10 m/s^2

vf = 12.4 m/s

vi = -6.2 m/s

t = (vf - vi)/a

= (12.4 + 6.2)/3.1

= 6 s

displacement = (vf - vi)*t

= (12.4 + 6.2) * 6

= 111.6 m.

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3 years ago
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A car with a mass of 1,500 kg requires a centripetal force of 640 N to safely follow a circular curve in the road at 13.4 m/s. W
Tamiku [17]
<h2>Answer</h2>

Radius is 421 m.

<u>Explanation </u>

A car with a mass of 1,500 kg requires a centripetal force of 640 N to safely follow a circular curve in the road at 13.4 m/s. Therefore, for radius we use formula which is Fc = mv ^ 2 / r,

As mass = m = 1500kg,

Centripetal force = Fc = 640N,

Velocity = v = 13.4 m / s

By putting values, Fc = mv ^ 2 / r,

r = mv ^ 2 / Fc,

=> r = ( 1500kg ) . ( 13.4 m / s) ^ 2 / 640,

=> r = ( 1500kg ) . ( 179.56 ) / 640,

r = 269340 / 640,

=> r = 420.84 m.

Radius is 421 m.

8 0
3 years ago
Read 2 more answers
A bicyclist is riding to the left with a velocity of 14 \,\dfrac{\text m}{\text s}14 s m ​ 14, start fraction, start text, m, en
Gnesinka [82]

Answer:

-2.0 m/s²

Explanation:Acceleration is the rate of change of velocity.

\begin{aligned}a&=\dfrac{\text{Change in velocity}}{\text{Change in time}}\\ \\ &=\dfrac{v_f-v_i}{\Delta t} \end{aligned}

a

​

 

=

Change in time

Change in velocity

​

=

Δt

v

f

​

−v

i

​

​

​

Hint #22 / 3

We can calculate the bicyclist's acceleration from the final velocity v_fv

f

​

v, start subscript, f, end subscript, initial velocity v_iv

i

​

v, start subscript, i, end subscript, and time interval \Delta tΔtdelta, t.

\begin{aligned}a&=\dfrac{v_f-v_i}{\Delta t}\\ \\ &=\dfrac{-21\,\dfrac{\text m}{\text s}-(-14\,\dfrac{\text m}{\text s})}{3.5\,\text s}\\ \\ &=-2.0\,\dfrac{\text m}{\text s^2}\end{aligned}

a

​

 

=

Δt

v

f

​

−v

i

​

​

=

3.5s

−21

s

m

​

−(−14

s

m

​

)

​

=−2.0

s

2

m

​

​

Hint #33 / 3

The acceleration of the bicyclist is -2.0\,\dfrac{\text m}{\text s^2}−2.0

s

2

m

​

minus, 2, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction.

5 0
3 years ago
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