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kotykmax [81]
3 years ago
14

5) A moving 2 kg object has an initial velocity of 10 m/s. It comes to a stop on a rough

Physics
2 answers:
umka2103 [35]3 years ago
4 0

Answer:

a) kinetic energy=0.5×mass×speed²

kinetic energy=0.5×2×10²=100 joules

Explanation:

im not sure about the rest :(

beks73 [17]3 years ago
4 0

Answer:

a) 100J

b)20N

c) 100J

Explanation:

A) KE = 0.5 x mass x velocity^2

= 0.5 x 2 x 10^2

= 100 J

B) Work done = Force x distance

Since work done = energy transferred

Energy transferred = Force x Distance

Energy transferred/ Distance = Force

100/5 = 20 N

C) Let's assume that all the Kinetic energy of the subject get's transferred into heat energy

Therefore Kinetic energy = Heat energy = 100 J

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A sports car is advertised to be able to stop in a distance of 50.0 m from a speed of 80 km. What is its acceleration and how ma
Flauer [41]

Explanation:

Given that,

Initial speed of the sports car, u = 80 km/h = 22.22 m/s

Final speed of the runner, v = 0

Distance covered by the sports car, d = 80 km = 80000 m

Let a is the acceleration of the sports car.  It can be calculated using third equation of motion as :

v^2-u^2=2ad

a=\dfrac{v^2-u^2}{2d}

a=\dfrac{0-(22.22)^2}{2\times 80000}

a=-0.00308\ m/s^2

Value of g, g=9.8\ m/s^2

a=\dfrac{-0.00308}{9.8}\ m/s^2

a=(-0.000314)\ g\ m/s^2

Hence, this is required solution.

8 0
3 years ago
An object is moving across a surface, but it is not gain or lose speed. Which best describes the objects force?
raketka [301]

Answer:

Gravity. An object is moving across a surface, but it does not gain or lose speed.

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The basic idea. Physicists see gravity as one of the four fundamental forces that govern the universe, alongside electromagnetism and the strong and weak nuclear forces.

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3 0
3 years ago
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Space pilot Mavis zips past Stanley at a constant speed relative to him of 0.800c. Mavis and Stanley start timers at zero when t
inna [77]

Answer:

Explanation:

a. The equation of Lorentz transformations is given by:

x = γ(x' + ut')

x' and t' are the position and time in the moving system of reference, and u is the speed of the space ship. x is related to the observer reference.

x' = 0

t' = 5.00 s

u =0.800 c,

c is the speed of light = 3×10⁸ m/s

Then,

γ = 1 / √ (1 - (u/c)²)

γ = 1 / √ (1 - (0.8c/c)²)

γ = 1 / √ (1 - (0.8)²)

γ = 1 / √ (1 - 0.64)

γ = 1 / √0.36

γ = 1 / 0.6

γ = 1.67

Therefore, x = γ(x' + ut')

x = 1.67(0 + 0.8c×5)

x = 1.67 × (0+4c)

x = 1.67 × 4c

x = 1.67 × 4 × 3×10⁸

x = 2.004 × 10^9 m

x ≈ 2 × 10^9 m

Now, to find t we apply the same analysis:

but as x'=0 we just have:

t = γ(t' + ux'/c²)

t = γ•t'

t = 1.67 × 5

t = 8.35 seconds

b. Mavis reads 5 s on her watch which is the proper time.

Stanley measured the events at a time interval longer than ∆to by γ,

such that

∆t = γ ∆to = (5/3)(5) = 25/3 = 8.3 sec which is the same as part (b)

c. According to Stanley,

dist = u ∆t = 0.8c (8.3) = 2 x 10^9 m

which is the same as in part (a)

7 0
3 years ago
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