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lord [1]
4 years ago
7

Which statement is true for a car that first goes around a curve of radius r at a constant speed v, and then goes around the sam

e curve at half the original speed?
The centripetal force is twice as big the second time around.

The centripetal force is four times as big the second time around.

The centripetal force is one-half as big the second time around.

The centripetal force is one-fourth as big the second time around.
Physics
1 answer:
sasho [114]4 years ago
3 0
Centripetal force is equal to (mv^2)/r
The way I use to answer these question is to set every variable to 1
m=1
v=1
r=1
so centripetal force =1
then change the variable we're looking at
and since we're find when it's half we could either change it to 1/2 or 2, but 2 is easier to use
m=1
v=2
r=1
((1)×(2)^2)/1=4
So the velocity in the 1st part is half the velocity in the 2nd part and the centripetal force is 4× less
The answer is the centripetal force is 1/4 as big the second time around
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An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.
Anna35 [415]

Answer

Given,

Energy absorbed, Q_H = 1.69\ kJ

Energy expels,Q_C =  1.25\ kJ

Temperature of cold reservoir, T = 27°C

a) Efficiency of engine

 \eta = \dfrac{Q_H - Q_C}{Q_H}\times 100

 \eta = \dfrac{1.69 - 1.25}{1.69}\times 100

\eta =26.03 %

b) Work done by the engine

 W = Q_H- Q_C

 W =1.69 - 1.25

 W = 0.44\ kJ

c) Power output

     t = 0.296 s

   P = \dfrac{W}{t}

   P = \dfrac{0.44}{0.296}

   P = 1.486\ kW

8 0
3 years ago
During an earthquake, _______ travels through the Earth's interior as _______ waves.
nordsb [41]

Answer:

During an earthquake, seismic waves travels through the Earth's interior as body or p waves.

Explanation:

If neither of the bold words look familiar from your lesson feel free to ignore this answer

4 0
4 years ago
A solid disk of radius 5.50 cm and mass 1.25 kg , which is rolling at a speed of 1.50 m/s , begins rolling without slipping up a
shtirl [24]

Answer:

Explanation:

Deceleration of solid disk = g sin10/1 + k²/r² = g sin 10 / 1 + 1/2 = g sin 10 x 2/3

[ k is radius of gyration of disk which is equal to( 1/√2)x r ]

deceleration a = -1.1345 m/s²

v = u - at , t = u / a = 1.5 / 1.1345 = 1.322 s.

5 0
3 years ago
A block of 200 g is attached to a light spring with a force constant of 5 N / m and freely in a horizontal plane vibrates. The m
Finger [1]

Answer:

m  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 mm  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 m

Explanation:

That is a reason

8 0
3 years ago
A plant box falls from the windowsill 25.0 m above the sidewalk and hits the cement 3.0 s later. What is the box's velocity when
mamaluj [8]

Answer:

22m/s

Explanation:

To find the velocity we employ the equation of free fall: v²=u²+2gh

where u is initial velocity, g is acceleration due to gravity h is the height, v is the velocity the moment it hits the ground, taking the direction towards gravity as positive.

Substituting for the values in the question we get:

v²=2×9.8m/s²×25m

v²=490m²/s²

v=22.14m/s which can be approximated to 22m/s

8 0
3 years ago
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