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2 years ago

Lucy and her bike together have a mass of 120kg. She slows down from 4.5m/s to 3.5m/s. How much kinetic energy does she lose?

1 answer:

7 0

The kinetic energy of a moving object is given by
$K= \frac{1}{2}mv^2$
where m is the object's mass and v its velocity.

In our problem, the initial kinetic energy is:
$K_i = \frac{1}{2} m v_i^2 = \frac{1}{2}(120 kg) (4.5 m/s)^2=1215 J$

while the final kinetic energy is:
$K_f = \frac{1}{2}mv_f^2 = \frac{1}{2}(120 kg)(3.5 m/s)^2= 735 J$

So, the kinetic energy lost by Lucy and her bike is
$\Delta K = K_i - K_f = 1215 J - 735 J = 480 J$

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2 years ago

In a car crash, large accelerations of the head can lead to severe injuries or even death. A driver can probably survive an acce

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Answer:

14.7 m/s

Explanation:

a = acceleration experienced by driver's head = 50 g = 50 x 9.8 m/s² = 490 m/s²

v₀ = initial speed of the driver = 0 m/s

v = final speed of the driver after 30 ms

t = time interval for which the acceleration is experienced = 30 ms = 0.030 s

Using the equation

v = v₀ + a t

Inserting the values

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2 years ago

In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P

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Answer:

Time period for first satellites 24.46 days and for second satellites 37.67 days

Explanation:

Given :

Distance of first satellites $r_{sat1} = 48000 \times 10^{3}$ m

Distance of second satellites $r _{sat2} = 64000 \times 10^{3}$ m

Distance of charon $r_{c} = 19600 \times 10^{3}$ m

Time period of charon $T_{c} = 6.39$ days

From the kepler's third law,

Square of the time period is proportional to the cube of the semi major axis.

$T^{2} = r^{3}$

$\frac{T}{r^{\frac{3}{2} } } = constant$

For first satellites,

$\frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat1} }{r_{sat1} ^{\frac{3}{2} } }$

${T_{sat1} } = 6.39 \times \frac{(48000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}$

$T_{sat1} = 24.46$ days

For second satellites,

$\frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat2} }{r_{sat2} ^{\frac{3}{2} } }$

${T_{sat2} } = 6.39 \times \frac{(64000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}$

$T_{sat2} = 37.67$ days

Therefore, time period for first satellites = 24.46 days and for second satellites 37.67 days

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2 years ago

A metal cube with sides of length a=1cm is moving at velocity v0→=1m/sj^ across a uniform magnetic field B0→=5Tk^. The cube is o

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Answer:

the magnitude of the electric field is 1.25 N/C

Explanation:

The induced emf in the cube ε = LB.v where B = magnitude of electric field = 5 T , L = length of side of cube = 1 cm = 0.01 m and v = velocity of cube = 1 m/s

ε = LB.v = 0.01 m × 5 T × 1 m/s = 0.05 V

Also, induced emf in the cube ε = ∫E.ds around the loop of the cube where E = electric field in metal cube

ε = ∫E.ds

ε = Eds since E is always parallel to the side of the cube

= E∫ds ∫ds = 4L since we have 4 sides

= E(4L)

= 4EL

So,4EL = 0.05 V

E = 0.05 V/4L

= 0.05 V/(4 × 0.01 m)

= 0.05 V/0.04 m

= 1.25 V/m

= 1.25 N/C

So, the magnitude of the electric field is 1.25 N/C

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