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ASHA 777 [7]
4 years ago
6

In which direction does centripetal force act on an object

Physics
2 answers:
Gwar [14]4 years ago
5 0
In a circle or circular motion
KatRina [158]4 years ago
4 0

Toward the center of the circle in which the object is moving.

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A small ball is attached to the lower end of a 0.800-m-long string, and the other end of the string is tied to a horizontal rod.
mestny [16]

Answer:

a = 17.68 m/s²

Explanation:

given,

length of the string, L = 0.8 m

angle made with vertical, θ = 61°

time to complete 1 rev, t = 1.25 s

radial acceleration = ?

first we have to calculate the radius of the circle

 R = L sin θ

 R = 0.8 x sin 61°

 R = 0.7 m

now, calculating at the angular velocity

\omega =\dfrac{2\pi}{T}

\omega =\dfrac{2\pi}{1.25}

  ω = 5.026 rad/s

now, radial acceleration

 a = r ω²

 a = 0.7 x 5.026²

a = 17.68 m/s²

hence, the radial acceleration of the ball is equal to 17.68 rad/s²

7 0
4 years ago
Which other element is most likely to be a nonreactive gas?
krok68 [10]

Answer:

A. Argon

Explanation:

It is a noble gas, a group which is not reactive.

5 0
3 years ago
After traveling for 6.0 seconds, a runner reaches 10m/s. What is the runner's acceleration?
luda_lava [24]

After traveling for 6.0 seconds, a runner reaches 10m/s. What is the runner's acceleration? Answer: 1.67 m/s2

4 0
3 years ago
A sled of mass 2.12 kg has an initial speed of 5.49 m/s across a horizontal surface. The coefficient of kinetic friction between
Darya [45]

Answer:

The speed of the sled is 3.56 m/s

Explanation:

Given that,

Mass = 2.12 kg

Initial speed = 5.49 m/s

Coefficient of kinetic friction = 0.229

Distance = 3.89 m

We need to calculate the acceleration of sled

Using formula of acceleration

a = \dfrac{F}{m}

Where, F = frictional force

m = mass

Put the value into the formula

a=\dfrac{\mu mg}{m}

a=\mu g

a=0.229\times9.8

a=2.244\ m/s^2

We need to calculate the speed of the sled

Using equation of motion

v^2=u^2-2as

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value in the equation

v ^2=(5.49)^2-2\times2.244\times3.89

v=\sqrt{(5.49)^2-2\times2.244\times3.89}

v=3.56\ m/s

Hence, The speed of the sled is 3.56 m/s.

8 0
3 years ago
The escape velocity of any object from Earth is 11.2 km/s. (a) Express this speed in m/s and km/h. (b) At what temperature would
natima [27]

Answer:

a ) 11.1 *10^3 m/s = 39.96 Km/h

b) T_{o2} =1.58*10^5 K

Explanation:

a)v_{es} =v_{rms}= 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h

b)

M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol

gas constant R = 8.31 j/mol.K

v_{rms} = \sqrt{ \frac{3RT}{M}}

So, v_{rms,o2} =\sqrt{ \frac{3RT_{o2}}{M_{o2}}}

multiply each side by M_{o2}, so we have

v_{rms,o2}^2 *M_{o2} =3RT_{o2}

solving for temperature T_{o2}

T_{o2} = \frac{v_{rms,o2}^2 *M_{o2}}{3R}

In the question given,v_{rms} =v_{es}

T_{o2} = \frac{(11.1*10^3)^2 *32.0*10^{-3}}{3*8.31}

T_{o2} =1.58*10^5 K

7 0
4 years ago
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