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RSB [31]
3 years ago
15

A 2500-N net force acting on a 880-kg car accelerates it at a rate of ______ m/s/s

Physics
1 answer:
Bond [772]3 years ago
8 0

Answer:

a = 2.84 m/s²

Explanation:

Given that,

Net force, F = 2500 N

Mass of the car, m = 880 kg

We need to find the acceleration of the car. Net force is given by :

F = ma

a=\dfrac{F}{m}\\\\a=\dfrac{2500\ N}{880\ m/s^2}\\\\a=2.84\ m/s^2

So, the acceleration of the car is 2.84 m/s².

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11. A disk 8.00 cm in radiu: rotates at a constant rate of 1200 revinin about its central axis Determine (a) its angular speed i
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A disk 8.00 cm in radiu: rotates at a constant rate of 1200 revinin about its central axis Determine (a) its angular speed in zadians per second, (b) the tangential speed at a point 3.00 cm from its center, (c) the radial aceleration of a point on the rim, and (d) the total distance a point our tke rim noves ign-2.00 s (E) The moment of inertia if it's mass is 2Kg? is the answer

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2 years ago
Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 168 cm , but its circumference is decreasi
DedPeter [7]

Answer:

103.1 V

Explanation:

We are given that

Initial circumference=C=168 cm

\frac{dC}{dt}=-15cm/s

Magnetic field,B=0.9 T

We have to find the magnitude of the  emf induced in the loop after exactly time 8 s has passed since the circumference of the loop started to decrease.

Magnetic flux=\phi=BA=B(\pi r^2)

Circumference,C=2\pi r

r=\frac{C}{2\pi}

r=\frac{168}{2\pi} cm

\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}(-15)=-\frac{15}{2\pi} cm/s

\int dr=-\int \frac{15}{2\pi}dt

r=-\frac{15}{2\pi}t+C

When t=0

r=\frac{168}{2\pi}

\frac{168}{2\pi}=C

r=-\frac{15}{2\pi}t+\frac{168}{2\pi}

E=-\frac{d\phi}{dt}=-\frac{d(B\pi r^2)}{dt}=-2\pi rB\frac{dr}{dt}

E=-2\pi(-\frac{5}{2\pi}t+\frac{168}{2\pi})B\times -\frac{15}{2\pi}

t=8 s

B=0.9

E=2\pi\times \frac{15}{2\pi}\times 0.9(-\frac{15}{2\pi}(8)+\frac{168}{2\pi})

E=103.1 V

6 0
3 years ago
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