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Nitella [24]
3 years ago
10

Josh has a toy car of mass 3 kg tied to a string of length 2 m. He ties the string to a pole and has the toy car drive in a circ

le around the pole at a speed of 3 m/s.
a. What is the centripetal acceleration of the car?

b. If the tension in the string exceeds 50 N, the string will break. How fast can he make the car go without breaking the string?
Physics
2 answers:
gulaghasi [49]3 years ago
5 0

Part a)

Centripetal acceleration is defined as

a_c = \frac{v^2}{R}

now here we know that

v = 3m/s

R = 2 m

m = 3 kg

now from above formula we have

a_c = \frac{3^2}{2}

a_c = 4.5 m/s^2

Part b)

Maximum possible tension in the string is given as

T = 50 N

now by force equation we have

F = ma

50 = 3 a

a = \frac{50}{3} m/s^2

now again by above formula

\frac{v^2}{R} = \frac{50}{3}

v = \sqrt{\frac{50 \times 2}{3}}

v = 5.77 m/s

aliina [53]3 years ago
5 0

a) a=v^2/R=4.5 m/s/s

b) F=ma=mv^2/R, so

v=\sqrt{FR/m}=5.77 m/s

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8 0
3 years ago
A body falls from the top of the tower and during the last second of its fall it fall through 23mvfind height of tower.
DanielleElmas [232]

Answer:

39.7 m

Explanation:

First, we conside only the last second of fall of the body. We can apply the following suvat equation:

s=ut+\frac{1}{2}at^2

where, taking downward as positive direction:

s = 23 m is the displacement of the body

t = 1 s is the time interval considered

a=g=9.8 m/s^2 is the acceleration

u is the velocity of the body at the beginning of that second

Solving for u, we find:

ut=s-\frac{1}{2}at^2\\u=\frac{s}{t}-\frac{1}{2}at=\frac{23}{1}-\frac{1}{2}(9.8)(1)=18.1 m/s

Now we can call this velocity that we found v,

v = 18 m/s

And we can now consider the first part of the fall, where we can apply the following suvat equation:

v^2-u^2 = 2as'

where

v = 18 m/s

u = 0 (the body falls from rest)

s' is the displacement of the body before the last second

Solving for s',

s'=\frac{v^2-u^2}{2a}=\frac{18.1^2-0}{2(9.8)}=16.7 m

Therefore, the total heigth of the building is the sum of s and s':

h = s + s' = 23 m + 16.7 m = 39.7 m

7 0
3 years ago
Calculate the pressure on a man’s foot when a woman who weighs 520 N steps on his foot with her heel which has an area of 0.001
Vinvika [58]

Answer:

520000  or 520000 pa

Force = 520N

Area of contact = 0.001

Pressure: 520000 or 520000

6 0
2 years ago
A 77.0−kg short-track ice skater is racing at a speed of 12.6 m/s when he falls down and slides across the ice into a padded wal
sukhopar [10]

Answer:

-6112.26  J

Explanation:

The initial kinetic energy, KE_i is given by

KE_i=0.5mv_1^{2} where m is the mass of a body and v_i is the initial velocity

The final kinetic energy, KE_f is given by

KE_f=0.5mv_f^{2} where v_f is the final velocity

Change in kinetic energy, \triangle KE is given by

\triangle KE=KE_f-KE_i=0.5mv_f^{2}-0.5mv_1^{2}=0.5m(v_f^{2}-v_i^{2})

Since the skater finally comes to rest, the final velocity is zero. Substituting 0 for v_f and 12.6 m/s for v_i and 77 Kg for m we obtain

\triangle KE=0.5*77*0^{2}-0.5*77*(0^{2}-12.6^{2})=-6112.26 J

From work energy theorem, work done by a force is equal to the change in kinetic energy hence for this case work done equals <u>-6112.26  J</u>

3 0
2 years ago
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