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tigry1 [53]
3 years ago
11

What is the acceleration of a car that goes from 0 MS to 60 MS and six seconds

Physics
1 answer:
Len [333]3 years ago
8 0
A = Δv/Δt
a = v2-v1/t2-t1
a = 60MS - 0MS / 6 seconds
a = 60 MS/ 6 seconds
a = 10 m/s^2
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A 66.0 kg diver is 3.10 m above the water, falling at speed of 6.10 m/s. Calculate her kinetic energy as she hits the water. (Ne
castortr0y [4]

Kinetic energy as she hits the water is 3300 joule.

To find the answer, we need to know about the Newton's equation of motion.

<h3>What's the Newton's equation of motion to determine the final velocity?</h3>
  • The final velocity is determined as

V²=U²+2aS

  • V= final velocity, U= initial velocity, a= acceleration and S= distance
<h3>What's the final velocity of the driver falling from 3.10m with initial velocity of 6.10m/s?</h3>
  • Here, a= 9.8m/s², U= 6.10m/s and S= 3.10m
  • So, V²= 6.1²+2×9.8×3.10= 98
  • V= √98= 10m/s
<h3>What's the kinetic energy of the driver when touches the water?</h3>

Kinetic energy= 1/2×mass×velocity²

= 1/2 × 66 × 10²

= 3300J

Thus, we can conclude that the kinetic energy of the driver is 3300 Joule.

Learn more about the kinetic energy here:

brainly.com/question/25959744

#SPJ4

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2 years ago
1. Amplitude, loudness, volume, dynamics, and intensity are related but distinct terms. How are they different?
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An equiconvex lens has power 4D. what will be the radius OF curvature of each
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How much energy (in joules) is released when 0.06 kilograms of mercury is condensed to a liquid at the same temperature?
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3 years ago
Read 2 more answers
A) A real object 4.0 cm high stands 30.0 cm in front of a converging lens of focal length 23 cm. Find the image distance, the im
vfiekz [6]

Explanation:

Given that,

Height of object = 4.0 cm

Distance of the object u= -30.0 cm

Focal length = 23 cm

We need to calculate the image distance

Using lens's formula

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Where, u = object distance

v = image distance

f = focal length

Put the value into the formula

\dfrac{1}{v}-\dfrac{1}{-30}=\dfrac{1}{23}

\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{30}

\dfrac{1}{v}=\dfrac{7}{690}

v=98.57\ cm

We need to calculate the height of the image

Using formula of height

\dfrac{h'}{h}=\dfrac{-v}{u}

Where, h' = height of image

h = height of object

\dfrac{h'}{4}=\dfrac{-98.57}{-30}

h'=\dfrac{98.57\times4}{30}

h'=13.14\ cm

The image is real and inverted.

(b). Now, object distance u = 13.0 cm

We need to calculate the image distance

Using lens's formula

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Put the value into the formula

\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{13.0}

\dfrac{1}{v}=-\dfrac{10}{299}

v=-29.9\ cm

We need to calculate the height of the image

\dfrac{h'}{4}=\dfrac{-(-29.9)}{-13.0}

h=-\dfrac{29.9\times4}{13.0}

h=-9.2\ cm

The image is virtual and erect.

Hence, This is required solution.

6 0
3 years ago
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