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Alex777 [14]
3 years ago
5

a conical pendulum is formed by attaching a 50g mass to a 1.2m string. The mass swings around the circle of radius 25cm (a) calc

ulate the speed of the mass (b) acceleration of the mass (c) tension in the string
Physics
1 answer:
kipiarov [429]3 years ago
4 0
A. 0.5 miles an hour 
B. 0.2 miles an hour
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What is general relativity
juin [17]
General relativity is a theory of space and time. The theory was published by Albert Einstein in 1915. The central idea of general relativity is that space and time are two aspects of spacetime. Spacetime is curved when there is gravity, matter, energy, and momentum.
5 0
3 years ago
What is the weight in Newtons of a ball with a mass of 7.77 kg?
rusak2 [61]

Answer:

76.1N

Explanation:

Given parameters:

Mass of the ball  = 7.77kg

Unknow:

Weight of balloon  = ?

Solution:

Weight is the vertical force applied on a body.

  Weight = mass x acceleration due gravity

So;

  Weight  = mass x acceleration due to gravity

So;

 Weight  = 7.77 x 9.8  = 76.1N

4 0
3 years ago
A 230-km-long high-voltage transmission line 2.0 cm in diameter carries a steady current of 1,100 A. If the conductor is copper
Molodets [167]

Answer:

28.23 years

Explanation:

I = 1100 A

L = 230 km = 230, 000 m

diameter = 2 cm

radius, r = 1 cm = 0.01 m

Area, A = 3.14 x 0.01 x 0.01 = 3.14 x 10^-4 m^2

n = 8.5 x 10^28 per cubic metre

Use the relation

I = n e A vd

vd = I / n e A

vd = 1100 / (8.5 x 10^28 x 1.6 x 10^-19 x 3.14 x 10^-4)

vd = 2.58 x 10^-4 m/s

Let time taken is t.

Distance = velocity x time

t = distance / velocity = L / vd

t = 230000 / (2.58 x 10^-4) = 8.91 x 10^8 second

t = 28.23 years

5 0
3 years ago
Read 2 more answers
You serve a volleyball with a mass of 1.4 kg. The ball leaves with a speed of 13 m/s. Calculate KE
NemiM [27]

Answer:

118.3 J

Explanation:

Givens:

m = 1.4 kg

V = 13 m/s

Formula for kinetic energy:

KE = (1/2)*(m)*(v)^2

KE = .5*(1.4 kg)*(13 m/s)^2

KE 118.3 J

J = Joules

7 0
3 years ago
You are standing a distance of 17.0 meters from the center of a merry-go-round. The merry-go-round takes 9.50 seconds to go comp
GenaCL600 [577]

Answer:

(a)11.24 m/s

(b)7.44 m/s

(c)409 N

(d)539.55\mu

(e) 0

Explanation:

The period for 1 circle 2\pi of the merry go around is 9.5s. It means the angular speed is:

\omega = \theta / t = 2\pi / 9.5 \approx 0.661 rad/s

(a)The speed is

v = \omega * R = 0.661 * 17 = 11.24 m/s

(b) Centripetal acceleration:

a = \frac{v^2}{R} = \frac{11.24^2}{17} = 7.44 m/s^2

(c) Magnitude of the force that keeps you go around at this acceleration

F = ma = 55 * 7.44 = 409 N

(d) let the coefficient of friction by \mu. The frictional force shall be this coefficient multiplied by normal force reverting gravity of the man

F_f = mg\mu = 55*9.81\mu = 539.55\mu

3 0
3 years ago
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