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Assoli18 [71]
2 years ago
3

Cindy exerts a force of 40 newtons and moves a chair 6 meters. Her brother Andy pushes a different chair for 6 meters while exer

ting a force of 30 newtons.
Andy does (less or more?) work than Cindy. The difference of their amounts of work is (how many?) Nm.
Physics
2 answers:
zavuch27 [327]2 years ago
5 0
Andy does less and the difference is 60 NM
STALIN [3.7K]2 years ago
4 0
Cindy got more than her brother
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A slender rod is 80.0 cm long and has mass 0.370 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05
nataly862011 [7]

Answer:

1.10 m/s

Explanation:

Linear speed is given by

v=r\omega

Kinetic energy is given by

KE=0.5I\omega^{2}

Potential energy

PE= mgh

From the law of conservation of energy, KE=PE hence

0.5I\omega^{2}=mgh where m is mass, I is moment of inertia, \omega is angular velocity, g is acceleration due to gravity and h is height

Substituting m2-m1 for m and 0.5l for h, \frac {2v}{L} for \omega we obtain

0.5I(\frac {2v}{L})^{2}=0.5Lg(m2-m1)

(\frac {2v}{L})^{2}=\frac {gl(m2-m1)}{I} and making v the subject

v^{2}=\frac {gl^{3}(m2-m1)}{4I}

v=\sqrt {\frac {gl^{3}(m2-m1)}{4I}}

For the rod, moment of inertia I=\frac {ML^{2}}{12} and for sphere I=MR^{2} hence substituting 0.5L for R then I=M(0.5L)^{2}

For the sphere on the left hand side, moment of inertia I

I=m1(0.5L)^{2} while for the sphere on right hand side, I=m2(0.5L)^{2}

The total moment of inertia is therefore given by adding

I=\frac {ML^{2}}{12}+ m1(0.5L)^{2}+ m2(0.5L)^{2}=\frac {L^{2}(M+3m1+3m2)}{12}

Substituting \frac {L^{2}(M+3m1+3m2)}{12} for I in the equation v=\sqrt {\frac {gL^{3}(m2-m1)}{4I}}

Then we obtain

v=\sqrt {\frac {gL^{3}(m2-m1)}{4(\frac {L^{2}(M+3m1+3m2)}{12})}}=\sqrt {\frac {3gL^{3}(m2-m1)}{L^{2}(M+3m1+3m2)}}

This is the expression of linear speed. Substituting values given we get

v=\sqrt {\frac {3*9.81*0.8^{3}(0.05-0.02)}{0.8^{2}(0.39+3(0.02)+3(0.05))}} \approx 1.08 m/s

8 0
2 years ago
A black, totally absorbing piece of cardboard of area A = 1.7 cm2 intercepts light with an intensity of 8.1 W/m2 from a camera s
Furkat [3]

Answer:

2.7x10⁻⁸ N/m²

Explanation:

Since the piece of cardboard absorbs totally the light, the radiation pressure can be found using the following equation:

p_{rad} = \frac{I}{c}

<u>Where:</u>

p_{rad}: is the radiation pressure

I: is the intensity of the light = 8.1 W/m²

c: is the speed of light = 3.00x10⁸ m/s

Hence, the radiation pressure is:

p_{rad} = \frac{I}{c} = \frac{8.1 W/m^{2}}{3.00 \cdot 10^{8} m/s} = 2.7 \cdot 10^{-8} N/m^{2}

Therefore, the radiation pressure that is produced on the cardboard by the light is 2.7x10⁻⁸ N/m².

I hope it helps you!

3 0
2 years ago
Read 2 more answers
Suppose that you lift four boxes individually, each at a constant velocity. The boxes have weights of 3.0 N, 4.0 N, 6.0 N, and 2
Alona [7]

Answer:

The vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

Explanation:

Worked : work can be defined as the product of force and distance.

The S.I unit of work is Joules (J).

Mathematically it can be represented as,

W = F×d.................. Equation 1

d = W/F.............................. Equation 2

where W = work, F = force, d = distance.

<em>Given: W = 12 J</em>

(i) for the 3.0 N weight,

using equation 2

d = 12/3

d= 4 m.

(ii) for the 4.0 N weight,

d = 12/4

d = 3 m.

(iii) for the 6.0 N weight,

d = 12/6

d = 2 m.

(iv) for the 2.0 N weight,

d = 12/2

d = 6 m

Therefore vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

8 0
2 years ago
Capillary waves travel what than long waves
7nadin3 [17]
Faster than. Hope this helps!!!
6 0
2 years ago
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An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid C and 581 km from the center of asteroid
tekilochka [14]

Answer:B

Explanation:

Given

Distance of astronaut From asteroid x is r_x=140 km

Distance of astronaut From asteroid Y is r_y=581 km

Suppose M,M_x,M_y be the masses of Astronaut , asteroid X and Y

If the astronaut is in equilibrium then net gravitational force on it is zero

F_x=F_y

\frac{GMM_x}{r_x^2}=\frac{GMM_y}{r_y^2}

cancel out the common terms we get

\frac{M_x}{r_x^2}=\frac{M_y}{r_y^2}

\frac{M_x}{M_y}=(\frac{r_x}{r_y})^2

\frac{M_x}{M_y}=(\frac{140}{581})^2

\frac{M_x}{M_y}=0.05806\approx 0.0581

4 0
2 years ago
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