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shtirl [24]
3 years ago
7

A car of mass 750kg accelerates away from traffic lights. At the end of the first 100m it has reached a speed of 12 m/s. During

this time, its engine provides an average forward force of 780 N, and the average force of friction on the car is 240 N. Calculate the increase in the car's kinetic energy at the end of the first 100 m
Physics
2 answers:
xenn [34]3 years ago
5 0

Answer:

the answer according to me is 54000j

Explanation:

im not sure why the force and friction are given inthe question because the the formula of ke is 1/mv^2

aliina [53]3 years ago
3 0

Answer:

Hruhriehrvdi8d8edeuebueufbrhddvuszu

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A race car traveling at 10. meters per second accelerates at the rate of 1.5 meters per seconds while traveling a distance of 7,
Arturiano [62]

The final speed of the car is 2) 150 m/s

Explanation:

Since the motion of the car is a uniformly accelerated motion, we can solve the problem by using the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the car in this problem, we have

u = 10 m/s

a=1.5 m/s^2

s = 7,467 m

Solving for v, we find the final velocity (and speed) of the car:

v=\sqrt{u^2+2as}=\sqrt{10^2+2(1.5)(7,467)}=150 m/s

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4 0
3 years ago
Read 2 more answers
A bullet with a mass ????b=13.5mb=13.5 g is fired into a block of wood at velocity ????b=253vb=253 m/s. The block is attached to
Lorico [155]

Answer:

450 grams

Explanation:

Given:

mass of the bullet, m = 13.5 g = 0.0135 kg

velocity of the bullet, v = 253 m/s

spring constant of the spring, k = 205 N/m

Compression of the spring, x = 35.0 cm = 0.35 m

Now, the kinetic energy of the moving system (bullet + board) will tend to move the spring

thus,

let velocity of the system, V

now, applying the concept of conservation of momentum, we have

mv = (M + m)V

where,

M is the mass of the block

thus,

V = mv/(M + m)

now,

the kinetic energy of the system = (1/2)(M + m)V²

or

the kinetic energy of the system = (1/2)(M + m)(mv/(m + M))²

Energy gained by the spring = (1/2)kx²

now,

equating both the energies, we get

(1/2)(M + m)(mv/(m + M))² = (1/2)kx²

or

(mv)²/(m + M) = kx²

on substituting the values, we get

(0.0135 × 253)²/(0.0135 + M) = 205 × (0.35)²

or

11.66/(0.0135 + M) = 25.1125

or

M = 0.450 kg = 450 grams

8 0
3 years ago
Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.
Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

Explanation:

Given that,

Positive charge = 24.00  μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})

\theta=\sin^{-1}(\dfrac{1}{4})

\theta=14.47^{\circ}

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

E=\dfrac{2k\lambda}{r}\times2\cos\theat

Put the value into the formula

E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}

E=408094.00\ N/C

E=4.08\times10^{5}\ N/C

Hence, The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

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3 years ago
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katen-ka-za [31]

Answer:

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3 years ago
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A skier is pulled by a towrope up a frictionless ski slope that makes an angle of 12 degrees with the horizontal. The rope moves
MArishka [77]

Answer:

Explanation:

Given,

  • Work done by the rope 900 m/s.
  • Angle of inclination of the slope = \theta\ =\ 12^o
  • Initial speed of the skier = v = 1.0 m/s
  • Length of the inclined surface = d = 8.0 m

part (a)

The rope is doing the work against the gravity on the skier to uplift up to the inclined surface. Therefore the work done by the rope is equal to the work done on the skier due to the gravity

\therefore W_r\ =\ W_g\ =\ 900\ J

In both cases the height attained by the skier is equal. and the work done by gravity does not depend upon the speed of the skier.

part (b)

  • Initial speed of the skier = v = 1.0 m/s.

Rate of the work done by the rope is power of the rope.

Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 1.0}{8.0}\\\Rightarrow P\ =\ 112.5\ Watt

Part (c)

  • Initial speed of the skier = v = 2.0 m/s.

Rate of the work done by the rope is power of the rope.

Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 2.0}{8.0}\\\Rightarrow P\ =\ 225\ Watt

4 0
3 years ago
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