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AURORKA [14]
3 years ago
7

A metal bar has a frictionless axle going through its center of mass. You notice that the bar is not level (flat), but that it i

s tilted at a 30 degree angle (the right end is below the horizontal and the left end is above the horizontal) and that the bar is not rotating away from this orientation. You can say that: a. The net force isn't zero and the net torque is counter-clockwise on the bar b. The net force is zero but the net torque is counter-clockwise in the bar c. The net force is zero bot the net torque is clockwise on the bar d. The net force isn't zero and the net torque is clockwise on the bar e. The net force is zero and the net torque is zero on the bar
Physics
1 answer:
ANTONII [103]3 years ago
5 0

Answer:

yep

Explanation:

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The mass of the sun is 1.99x10^30 kg . Jupiter is 7.79x10^8 km away from the sun and the mass of 1.90x10^27 kg
maks197457 [2]

Question is missing:

"What is the gravitational force between the Sun and Jupiter?"

Answer:

4.16\cdot 10^{23} N

Explanation:

The gravitational force between two objects is given by

F=G\frac{m_1m_2}{r^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

In this problem, we have

m_1 = 1.99\cdot 10^{30} kg is the mass of the sun

m_2 = 1.90\cdot 10^{27} kg is the mass of Jupiter

r=7.79\cdot 10^8 km = 7.79\cdot 10^{11} m is their separation

Solving the equation, we find

F=(6.67\cdot 10^{-11})\frac{(1.99\cdot 10^{30})(1.90\cdot 10^{27})}{(7.79\cdot 10^{11})^2}=4.16\cdot 10^{23} N

6 0
3 years ago
Consider the system shown in fig. 6-26. the rope and pulley have negligible mass, and the pulley is frictionless. the coefficien
Anestetic [448]
Need and answer choice if you have one
4 0
3 years ago
THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THANKS!!!
poizon [28]

Answer:

391.67Hz

Explanation:

The fundamental frequency formula in string is expressed as;

Fo = V/2L

V is the velocity of the wave = 329m/s

L is the length of the string = 42cm = 0.42m

Substitute

Fo = 329/2(0.42)

Fo = 329/0.84

Fo = 391.67Hertz

Hence the  fundamental frequency of a mandolin string is 391.67Hz

4 0
3 years ago
Calculate the current flowing if a charge of 36 kilocoulombs flows in 1 hour.
kupik [55]

Answer:

2hrs and some mins

Explanation:

bc 2×36= 17 =)

3 0
3 years ago
Rank the following objects by their accelerations down an incline (assume each object rolls without slipping) from least to grea
Alexxx [7]

Answer:

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

Starting point. Highest point

          Em₀ = U = m g h

Final point. To get off the ramp

          Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

         

energy is conserved

        Em₀ = Em_f

        mgh = ½ m v² +1/2 I w²

angular and linear velocity are related

         v = w r

         w = v / r

we substitute

          mg h = ½ v² (m + I / r²)

          v² = 2 gh   \frac{m}{m+ \frac{I}{r^2} }

          v² = 2gh    \frac{1}{1 + \frac{I}{m r^2} }

this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

         v² = v₀² + 2 a L

where L is the length of the plane

         v² = 2 a L

         a = v² / 2L

we substitute

         a = g \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

let's use trigonometry

         sin θ = h / L

         

we substitute

         a = g sin θ   \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

      I = m r²

we look for the acerleracion

      a₁ = g sin θ    \frac{1}{1 + \frac{mr^2 }{m r^2 } }1/1 + mr² / mr² =

      a₁ = g sin θ    ½

b) solid cylinder

       I = ½ m r²

       a₂ = g sin θ  \frac{1}{1 + \frac{1}{2}  \frac{mr^2}{mr^2} } = g sin θ   \frac{1}{1+ \frac{1}{2} }

       a₂ = g sin θ   ⅔

c) hollow sphere

     I = 2/3 m r²

     a₃ = g sin θ   \frac{1}{1 + \frac{2}{3} }

     a₃ = g sin θ \frac{3}{5}

d) solid sphere

     I = 2/5 m r²

     a₄ = g sin θ  \frac{1 }{1 + \frac{2}{5} }

     a₄ = g sin θ  \frac{5}{7}

We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)

a) a₁ = g sin θ ½ = g sin θ      \frac{105}{210}

b) a₂ = g sinθ ⅔ = g sin θ     \frac{140}{210}

c) a₃ = g sin θ \frac{3}{5}= g sin θ       \frac{126}{210}

d) a₄ = g sin θ \frac{5}{7} = g sin θ      \frac{150}{210}

the order of acceleration from lower to higher is

   

     a₁ <a₃ <a₂ <a₄

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

8 0
3 years ago
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