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3 years ago

A 2500-kg car falls over a cliff and converts 1000,000J of PE into KE.

Physics

1 answer:

MArishka [77]3 years ago

7 0

B is correct makes more sense

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2 Points

juin [17]

Answer: c

Explanation:

Analytical methods

3 0

3 years ago

A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm

I am Lyosha [343]

Answer:

Rate of change of magnetic field is $3.466\times 10^3T/sec$

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop $r=\frac{0.13}{2}=0.065m$

Length of the circular loop $L=2\pi r=2\times 3.14\times 0.065=0.4082m$

Wire is made up of diameter of 2.6 mm

So radius $r=\frac{2.6}{2}=1.3mm=0.0013m$

Cross sectional area of wire $A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2$

Resistivity of wire $\rho =2.18\times 10^{-8}m$

Resistance of wire $R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm$

Current is given i = 11 A

So emf $e=11\times 1.67\times 10^{-3}=0.0183volt$

Emf induced in the coil is $e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}$

$0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}$

$\frac{dB}{dt}=3.466\times 10^3=T/sec$

8 0

3 years ago

Which action would be most likely to cause hearing loss?

Archy [21]

If you dont wear protection while shooting a rifle it will damge your hear

4 0

3 years ago

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A father fashions a swing for his children out of a long rope that he fastens to the limb of a tall tree. As one of the children

trasher [3.6K]

Answer:

The centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

Explanation:

The centripetal acceleration is given by:

$a_{c} = \frac{v^{2}}{r}$

Where:

$v^{2}$ : is the tangential speed = 9.50 m/s

r: is the distance = 6.00 m

Hence, the centripetal acceleration is:

$a_{c} = \frac{v^{2}}{r} = \frac{(9.50 m/s)^{2}}{6.00 m} = 15.04 m/s^{2}$

Therefore, the centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

I hope it helps you!

3 0

3 years ago

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how to A golf ball of mass 0.045 kg is hit off the tee at a speed of 45 m/s. The golf club was in contact with the ball for 3.5

Alexandra [31]

Answer: a) 12857.1 m/s/s b) 578.6 N

Explanation:

Impulse = change in momentum

Ft = mV2 - mV1

V = AT, 45 / .0035 = 12857.1 m/s/s

(b) .045 x 12857.1 = 578.6 N

4 0

3 years ago

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