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Reika [66]
4 years ago
6

The ends of a massless rope are attached to two stationary objects (e.g., two trees or two cars) so that the rope makes a straig

ht line. For this situation, which of the following statements are true?
a. The tension in the rope is everywhere the same.
b. The magnitudes of the forces exerted on the two objects by the rope are the same.
c. The forces exerted on the two objects by the rope must be in opposite directions.
d. The forces exerted on the two objects by the rope must be in the direction of the rope.
Physics
1 answer:
julia-pushkina [17]4 years ago
5 0

Answer: They are all true

a. The tension in the rope is everywhere the same.

b. The magnitudes of the forces exerted on the two objects by the rope are the same.

c. The forces exerted on the two objects by the rope must be in opposite directions.

d. The forces exerted on the two objects by the rope must be in the direction of the rope.

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great rest of Black History Month! :-)  

- Cutiepatutie ☺❀❤

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A uniform cylindrical grinding wheel of mass 50.0 kg and diameter 1.0 m is turned on by an electric motor. The friction in the b
aliya0001 [1]

Answer:

The torque must be applied to the wheel is 15.7 N-m.  

Explanation:

Given that,

Mass of the wheel of cylinder, m = 50 kg

Diameter of the wheel, d = 1 m

Radius, r = 0.5 m      

Initial speed off the wheel is 0

Final angular speed of the wheel, 120 rev/min = 12.56 rad/s

Angular displacement, \theta=20\ rev=125.6\ rad

The torque is given by :

\tau=I\alpha \\\\\tau=\dfrac{mr^2}{2}\times (\dfrac{\omega_f^2}{2\theta})\\\\\tau=\dfrac{50\times1^{2}}{2}\times(\dfrac{12.56^{2}}{2\times125.6})\\\\\tau=15.7\ N-m

So, the torque must be applied to the wheel is 15.7 N-m.

4 0
3 years ago
The gravitational field strength at a distance R from the center of moon is gR. The satellite is moved to a new circular orbit t
3241004551 [841]

Answer:

g'=\frac{g__R}{4}

Explanation:

Given:

  • gravitational field strength of moon at a distance R from its center, g__R
  • Distance of the satellite from the center of the moon, h=2R

<u>Now as we know that the value of gravity of any heavenly body is at height h is given as:</u>

g'=g__{R}} \times \frac{R^2}{(2R)^2}

g'=\frac{g__R}{4}

∴The gravitational field strength will become one-fourth of what it is at the surface of the moon.

6 0
3 years ago
Pulsars are neutron stars that emit X rays and other radiation in such a way that we on Earth receive pulses of radiation from t
stira [4]

Answer:

(a) a_{c} = 5.41\times 10^{9} m/s^{2}

(b) a_{t} = 2.99\times 10^{- 5} m/s^{2}

Given:

Time period of Pulsar, T_{P} = 33.085 ms == 33.085\times 10^{- 3} s

Equatorial radius, R = 15 Km = 15000 m

Spinning time, t_{s} = 9.50\times 10^{10}

Solution:

(a) To calculate the value of the centripetal  acceleration, a_{c} on the surface of the equator, the force acting is given by the centripetal force:

m\times a_{c} = \frac{mv_{c}^{2}}{R}

a_{c} = \frac{v_{c}^{2}}{R}                (1)

where

v_{c} = \frac{distance covered(i.e., circumference)}{ T}

v_{c} = \frac{2\pi R}{Time period, T}           (2)

Now, from (1) and (2):

a_{c} = R\frac({2\pi )^{2}}{T^{2}}

a_{c} = 15000\frac{2\pi )^{2}}{(33.085\times 10^{- 3})^{2}}

a_{c} = 5.41\times 10^{9} m/s^{2}

(b) To calculate the tangential acceleration of the object :

The tangential acceleration of the object  will remain constant and is given by the equation of motion as:

v = u + a_{t}t_{s} = 0

where

u = v_{c}

a_{t} = - \frac{2\pi R}{Tt_{s}}

a_{t} = - \frac{2\pi 15000}{33.085\times 10^{- 3}\times 9.50\times 10^{10}}

a_{t} = 2.99\times 10^{- 5} m/s^{2}

7 0
3 years ago
Plz help me answer this question if u answer them all correctly and respectfuly i will make u brainliest.
Musya8 [376]
Please explain more pls
6 0
3 years ago
A bubble at the bottom of the lake to the surface within 10.0 seconds what is the depth of the lake
stira [4]
I normal bubble starts at the speed of 10mph then speeds up +5 each second. so this would be 600 meters. also, it depends on the density of the water and whats inside the water or liquid in the lake. there are many factors that will change this speed. 
7 0
3 years ago
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