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Anarel [89]
3 years ago
9

Sylvia and Jadon now want to work a problem. Imagine a puck of mass 0.5 kg moving in a circular simulation. Suppose that the ten

sion in the string is 3.5 N, and that the radius of its circular path is 1.0 m. What will Jadon and Sylvia find for the tangential speed of the puck in m/s?
Physics
1 answer:
leva [86]3 years ago
3 0

Answer:

Tangential speed, v = 2.64 m/s

Explanation:

Given that,

Mass of the puck, m = 0.5 kg

Tension acting in the string, T = 3.5 N

Radius of the circular path, r = 1 m

To find,

The tangential speed of the puck.

Solution,

The centripetal force acting in the string is balanced by the tangential speed of the puck. The expression for the centripetal force is given by :

F=\dfrac{mv^2}{r}

v=\sqrt{\dfrac{Fr}{m}}

v=\sqrt{\dfrac{3.5\ N\times 1\ m}{0.5\ kg}}

v = 2.64 m/s

Therefore, the tangential speed of the puck is 2.64 m/s.

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Crescent, gibbous, waxing, and waning.
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A satellite is in circular orbit at an altitude of 1500 km above the surface of a nonrotating planet with an orbital speed of 9.
Ksju [112]

To solve this problem we will use the Newtonian theory about the speed of a body in space for which the speed of a body in the orbit of a planet is summarized as:

v =  \sqrt{\frac{2GM}{R}}

Where,

G = Gravitational Universal Constant

M = Mass of Planet

r = Radius of the planet ('h' would be the orbit from the surface)

The escape velocity is

v = 14.9km/h = 14900m/s

Through this equation we can find the mass of the Planet in function of the distance, therefore

M = \frac{v^2R}{2G}

M = \frac{14900^2R}{2(6.67*10^{-11})}

M = 16.64*10^{17}R

The orbital velocity is

v_o = \sqrt{\frac{GM}{R+h}}

9200^2 = \frac{(6.67*10^{-11})(16.64*10^{17})R}{R+1500*10^3}

11.1*10^7R = (R+15000*10^3)(9200)^2

2.64*10^7R = 12.69*10^{13}

R = 4.81*10^6m

The time period of revolution is,

T = \frac{2\pi(R+h)}{v_o}

T = \frac{2\pi(4.81*10^6+1.5*10^6)}{9200}

T = 4307s

T = 72min = 1hour12min

Therefore the orbital period of the satellite is closes to 1 hour and 12 min

3 0
3 years ago
A sonar emits a sound signal of frequency 40000Hz, towards the bottom of the sea.
kap26 [50]

Answer:

0.000025s

Explanation:

Period it’s. : T(s)= 1/f(Hz)=1/40000Hz=0.000025s

8 0
2 years ago
Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.9 km/h due east. Runner B is initia
Oksanka [162]

Answer:

B meet A 0.01 km east of flagpole

Explanation:

given data

distance A = 5.7 km  west

velocity V1 = 8.9 km/h

distance B = 4.5 km  east

velocity V2 = 7 km/h

to find out

How far runners from the flagpole, when paths cross

solution

we know A and B are 5.7 + 4.5 = 10.2 km apart

and we consider here B will run distance x km for meet

so time will be for B is

time B = distance / velocity

time B = x / 7     ...................1

and

for A distance for meet = ( 10.2 - x ) km

so time A = distance / velocity

time A = ( 10.2 - x )  / 8.9      .............2

now equating equation 1 and 2

time A = time B

x / 7  =   ( 10.2 - x )  / 8.9

x = 4.490

so distance of B run for meet is 4.490 km

so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km

so B meet A 0.01 km east of flagpole

4 0
3 years ago
URGENT!!!!! If a 6 Ω resistor is in a circuit connected to a voltage source, and the current through the circuit is 2 amps, what
Eduardwww [97]

Answer:

Explanation:

according to ohms law we know that

v=IR

given current =2 amps

given resistance =6Ω

so voltage is

v=2*6 =12 V

3 0
3 years ago
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