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3 years ago

a car advertisement States that a certain car can accelerate from rest to70km/h in 7seconds find the car's average acceleration

Physics

1 answer:

Aneli [31]3 years ago

8 0

Answer:

Acceleration is 2.78 m/s²

Explanation:

Since the car is from rest, initial velocity, u is zero.

from first newton's equation of motion:

${ \bf{v = u + at}} \\$

v is final velocity

u is initial velocity

a is acceleration

t is time

70 km/h is (70 × 1000 ÷ 3600) m/s = 19.44 m/s

${ \sf{19.44 = 0 + (a \times 7)}} \\ { \sf{7a = 19.44 }} \\ { \sf{a = 2.78 \: {ms}^{ - 2} }}$

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A sinusoidal wave of angular frequency 1,203 rad/s and amplitude 3.1 mm is sent along a cord with linear density 3.9 g/m and ten

kobusy [5.1K]

Answer:

18.7842493212 W

Explanation:

T = Tension = 1871 N

$\mu$ = Linear density = 3.9 g/m

y = Amplitude = 3.1 mm

$\omega$ = Angular frequency = 1203 rad/s

Average rate of energy transfer is given by

$P=\dfrac{1}{2}\sqrt{T\mu}\omega^2y^2\\\Rightarrow P=\dfrac{1}{2}\sqrt{1871\times 3.9\times 10^{-3}}\times 1203^2\times (3.1\times 10^{-3})^2\\\Rightarrow P=18.7842493212\ W$

The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W

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3 years ago

PLEASE HURRYY!!!!The diagram shows two balls released from a device at the same time. The ball on the left falls freely from res

LenaWriter [7]

Answer:

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Explanation:

4 0

3 years ago

Read 2 more answers

An electron is released from rest in a weak electric field given by E =-2.30 x 10-10 N/Cj. After the electron has traveled a ver

Goshia [24]

Explanation:

It is given that,

An electron is released from rest in a weak electric field of, $E=2.3\times 10^{-10}\ N/C$

Vertical distance covered, $s=1\ \mu m=10^{-6}\ m$

We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

$v^2-u^2=2as$

$v^2=2as$ .............(1)

Electric force is $F_e$ and force of gravity is $F_g$ . As both forces are acting in downward direction. So, total force is:

$F=mg+qE$

$F=9.1\times 10^{-31}\times 9.8+1.6\times 10^{-19}\times 2.3\times 10^{-10}$

$F=4.57\times 10^{-29}\ N$

Acceleration of the electron, $a=\dfrac{F}{m}$

$a=\dfrac{4.57\times 10^{-29}\ N}{9.1\times 10^{-31}\ kg}$

$a=50.21\ m/s^2$

Put the value of a in equation (1) as :

$v=\sqrt{2as}$

$v=\sqrt{2\times 50.21\times 10^{-6}}$

v = 0.010 m/s

So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.

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3 years ago

HELP ME PLZ WILL GIVE BRAINLIEST ALSO 24 POINTS

Ludmilka [50]

Answer:

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Explanation:

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3 years ago

A spring with a spring constant of 120 J/m2 is fixed to a wall, free to oscillate. On the other end, a ball with a mass of 1500

Neporo4naja [7]

Answer:

A. 4.47 m/s

Explanation:

As the ball oscillates, it mechanical energy, aka the total kinetic and elastics energy stays the same. For the ball to be at maximum speed, its elastic energy i 0 and vice versa. When the ball is at rest, its kinetic energy is 0 and its elastic energy is at maximum at 50 cm, or 0.5 m

1500 g = 1.5 kg

$E_e = E_k$