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Levart [38]
3 years ago
11

A 60 kg pupil runs for 600m in 1 minute uniformly calculate kinetic energy​

Physics
1 answer:
AURORKA [14]3 years ago
8 0

velocity = traveled distance ÷ time of the traveled distance is seconds

velocity = 600 ÷ 60

velocity = 10 m/s

_________________________________

Kinetic Energy = 1/2 × mass × ( velocity )^2

KE = 1/2 × 60 × ( 10 )^2

KE = 30 × 100

KE = 3000 j

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when trying to stop a car on icy pavement in as short a distance as possible, is it better to slam on the brakes and skid to a s
madreJ [45]

Tapping the brakes is almost definitely going to be better than not tapping the brakes.

Using the brakes causes the static friction between your tires and the road to increase; which slows down the car more.

The issue is that when the grip between the brakes and the tire is too strong, the coefficient of friction between the tires isn't high enough to maintain that static contact. This is when the wheels "lock up" and begin to skid. As you mention, kinetic friction is less than static friction, so wheel lock and skidding should be avoided.

This means that tapping the brakes is generally the best braking method in these conditions, as compared to not braking at all.

When considering modern vehicles, your friend missed the mark even more. Most vehicles sold now (as far as I know) have an ABS, an Anti-lock Braking System which actually disengages the brakes when they reach the threshold when you start to lose traction. Basically, you can push down on the brake pedal as hard as you want, and you can feel the brakes start "tapping" on their own, to prevent skidding, increase traction, and allow you to brake in the lowest space possible. Tapping is the method most cars automatically employ on icy surfaces, and if you try it for yourself (in a controlled environment), you will find that tapping the brakes should slow you down a lot faster than not pressing it at all; especially if you avoid skidding.

Hence,

Tapping the brakes is almost definitely going to be better than not tapping the brakes.

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5 0
2 years ago
A 80.0-kg airplane pilot pulls out of a dive by following, at a constant speed of 180km/hr, the arc of a circle whose radius is
wolverine [178]

Answer: 8.33m/s²

Explanation:

Mass of the airplane pilot = 80kg

The speed of plane v = 180km/h = 50m/s

The radius of circle r = 300 m

The acceleration of the plane will be calculated as:

a = v²/r

a = 50²/300

a = 2500/300

a = 8.33m/s²

Note that 180km/h was converted to m/s by

= (180 × 1000)/(60 × 60)

= 180000/3600

= 50m/s

1000 meters = 1 kilometer

60 minutes = 1 hour

60 seconds = 1 minute

3600 seconds = 1 hour

6 0
3 years ago
A substance sample has a Mass of 45.0 g, and the Data Safety
marishachu [46]

The volume of the sample is 48.39 cm³.

<h3 /><h3>What is volume?</h3>

The term volume refers to the amount of three-dimensional space taken up by an item or a closed surface

To calculate the volume of the sample, we use the formula below

Formula:

  • D = m/v
  • v = m/D................ Equation 1

Where:

  • m = Mass of the smaple
  • D = Density of the sample
  • v = Volume of the sample.

From the question,

Given:

  • m = 45 g
  • D = 0.93 g/cm³

Substitute these values into equation 1

  • v = 45/0.93
  • v = 48.39 cm³

Hence, the volume of the sample is 48.39 cm³.

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8 0
2 years ago
Can someone tell me the answer to this?
larisa86 [58]

Answer:

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3 0
3 years ago
Read 2 more answers
- A cannon of 2000 kg fires a shell of 10 kg at
fgiga [73]

1) -0.5 m/s

We can solve the first part of the problem by using the law of conservation of momentum. In fact, the total momentum of the cannon - shell system must be conserved.

Before the shot, both the cannon and the shell are at rest, so the total momentum is zero:

p=0

After the shot, the momentum is:

p=MV+mv

where

M = 2000 kg is the mass of the cannon

m = 10 kg is the mass of the shell

v = 100 m/s is the velocity of the shell (we take as positive the direction of motion of the shell)

V = ? is the velocity of the cannon

Since momentum is conserved, we can write

0=MV+mv

And solving for V, we find the velocity of the cannon:

V=-\frac{mv}{M}=-\frac{(10)(100)}{2000}=-0.5 m/s

where the negative sign indicates that the cannon moves in the direction opposite to the shell.

2) 0.5 m

The motion of the cannon is a uniformly accelerated motion, so we can solve this part by using suvat equation:

v^2-u^2=2as

where

v is the final velocity of the cannon

u = 0.5 m/s is the initial velocity of the cannon (now we take as positive the initial direction of motion of the cannon)

a=-0.25 m/s^2 is the deceleration of the cannon

s is the distance travelled by the cannon

The cannon will stop when v = 0; substituting and solving the equation for s, we find the minimum safe distance required to stop the cannon:

s=\frac{v^2-u^2}{2a}=\frac{0-0.5^2}{2(-0.25)}=0.5 m

7 0
4 years ago
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