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3 years ago

If a car has an initial velocity of 20 m/s and accelerates at 2.0 m/s² for 100 m, what is its final velocity?

Physics

1 answer:

WINSTONCH [101]3 years ago

8 0

Answer:

28.28 ms-1

Explanation:

v² = u² + 2as

v² = 20² + 2× 2×100

v² = 800

v = 28.28 ms-1

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Work is being done in which of these situations? All motions are at a constant velocity (including velocity = zero).

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Work is done only when there is a displacement of the object on which a force is applied in the direction of the force.

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Space debris that enters earths atmosphere

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3 years ago

A sound wave moving with a speed of 1500 m/s is sent from a submarine to the ocean floor. It reflects off the

Alekssandra [29.7K]

Answer:

the distance between the submarine and the ocean floor is 11,250 m

Explanation:

Given;

speed of the wave, v = 1500 m/s

time of motion of the wave, t = 15 s

The time taken to receive the echo is calculated as;

$time \ of \ motion \ (t) = \frac{total \ distance }{speed \ of \ wave} = \frac{2d}{v} \\\\2d = vt\\\\d = \frac{vt}{2} \\\\d = \frac{1500 \times 15}{2} \\\\d = 11,250 \ m$

Therefore, the distance between the submarine and the ocean floor is 11,250 m

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3 years ago

A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the

BartSMP [9]

Answer:

$r=1.14m$

Explanation:

$\theta$ is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

$F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB$

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

$F_m=F_c\\qvB=F_c(1)$

$F_c$ is the centripetal force and is defined as:

$F_c=m\frac{v^2}{r}$

Here $v$ is the proton's speed and $r$ is the radius of the circular motion. Replacing this in (1) and solving for r:

$qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}$

Recall that 1 J is equal to $6.242*10^{12}MeV$ , so:

$4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J$

We can calculate $v$ from the kinetic energy of the proton:

$K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}$

Finally, we calculate the radius of the proton path:

$r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m$

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4 years ago

two cars go through 2 different crashes. car one 1 experiences a 500N impulse for a duration for 15s, while car2 experiences the

likoan [24]

Answer:

Conservation of momentum.

Momentum is zero after collision, no direction or speed.

Explanation:

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3 years ago