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VladimirAG [237]
2 years ago
11

A block slides on a table pulled by a string attached to a hanging weight. In case 1 the block slides without friction and in ca

se 2 there is kinetic friction between the sliding block and the table. In which case is the tension in the string the biggest? Please explain.
Physics
2 answers:
ycow [4]2 years ago
7 0

The tension in the string with friction would be the biggest because of the involvement of the force of gravity. This would result in that the friction force that is acting on the system. There is no friction in the frictionless system, and only the force of gravity is relevant.

ankoles [38]2 years ago
7 0

Answer:

In case 2

Explanation:

Thinking process:

The friction is a force that tends to oppose motion.

In case 1, the block slides without friction. This means that the surface is a smooth surface. Thus, the net force needed to pull the string is minimal.

In case 2, there is kinetic friction between the sliding block and the table. Now, the kinetic friction is given by the formula:

F = \muN\mu N

where, \mu = coefficient of friction and N = normal force.

Thus, the net force needed:

F_{net} = F_{pull} - \mu N

Thus, the force is larger in case 2.

Therefore, the tension is larger in case 2.

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Two disks with the same rotational inertia i are spinning about the same frictionless shaft, with the same angular speed ω, but
valentina_108 [34]

Answer:

3. none of these

Explanation:

The rotational kinetic energy of an object is given by:

K=\frac{1}{2}I \omega^2

where

I is the moment of inertia

\omega is the angular speed

In this problem, we have two objects rotating, so the total rotational kinetic energy will be the sum of the rotational energies of each object.

For disk 1:

K_1 = \frac{1}{2}I (\omega)^2 = \frac{1}{2}I\omega^2

For disk 2:

K_2 = \frac{1}{2}I(-\omega)^2 = \frac{1}{2}I\omega^2

so the total energy is

K=K_1 + K_2 = \frac{1}{2}I\omega^2 + \frac{1}{2}I\omega^2 = I\omega^2

So, none of the options is correct.

5 0
2 years ago
50 POINTS! A Boy throws a ball horizontally a distance of 22m downrange from the top of a tower that is 20.0m tall. What is his
DerKrebs [107]

The ball's horizontal and vertical velocities at time t are

v_x=v_{xi}

v_y=v_{yi}-gt

but the ball is thrown horizontally, so v_{yi}=0. Its horizontal and vertical positions at time t are

x=v_{xi}t

y=20.0\,\mathrm m-\dfrac g2t^2

The ball travels 22 m horizontally from where it was thrown, so

22\,\mathrm m=v_{xi}t

from which we find the time it takes for the ball to land on the ground is

t=\dfrac{22\,\rm m}{v_{xi}}

When it lands, y=0 and

0=20.0\,\mathrm m-\dfrac{9.8\frac{\rm m}{\mathrm s^2}}2\left(\dfrac{22\,\rm m}{v_{xi}}\right)^2

\implies v_i=v_{xi}=11\dfrac{\rm m}{\rm s}

7 0
2 years ago
A highly charged piece of metal (with uniform potential throughout) tends to spark at places where the radius of curvature is sm
k0ka [10]

Answer:

look it up

Explanation:

8 0
2 years ago
Question
nataly862011 [7]

The density of a material is constant and it is given by the ratio of the mass to the volume of the material

The mass of the liquid and the full bottle ae

The mass of the liquid is <u>450 g</u>

The mass of the filled bottle is <u>530 grams</u>

<u></u>

The reason the above values are correct are as follows:

The given parameters are;

Volume of the bottle, V = Half litre

Mass of the bottle, m_b = 80 g

The volume of liquid in the bottle when filled, V = 500 cm³

The density of the olive oil with which the bottle is filled, ρ = 0.9 g/cm³

a. Required:

To calculate the mass of oil in the bottle

Solution:

The volume of oil in the bottle when the bottle is filled, V = 500 cm³

Density, \ \rho = \dfrac{Mass}{Volume}  = \dfrac{m}{V}

The mass of the liquid, m = ρ × V

∴ m = 0.9 g/cm³ × 500 cm³ = 450 g

The mass of the liquid, m = <u>450 g</u>

<u></u>

b. The mass of the oil in the bottle, m = grams

The mass of the full bottle, m_{filled} = m + m_b

∴ m_{filled} = 450 g + 80 g = 530 g

The mass of the full bottle, m_{filled} = <u>530 grams</u>

Learn more about density here:

brainly.com/question/18110802

4 0
2 years ago
A large solar panel on a spacecraft in Earth orbit produces 1.0 kW of power when the panel is turned toward the sun. What power
Mandarinka [93]

Answer:

e*P_s = 11 W

Explanation:

Given:

- e*P = 1.0 KW

- r_s = 9.5*r_e

- e is the efficiency of the panels

Find:

What power would the solar cell produce if the spacecraft were in orbit around Saturn

Solution:

- We use the relation between the intensity I and distance of light:

                                  I_1 / I_2 = ( r_2 / r_1 ) ^2

- The intensity of sun light at Saturn's orbit can be expressed as:

                                  I_s = I_e * ( r_e / r_s ) ^2

                                  I_s = ( 1.0 KW / e*a) * ( 1 / 9.5 )^2

                                  I_s = 11 W / e*a

- We know that P = I*a, hence we have:

                                  P_s = I_s*a

                                  P_s = 11 W / e

Hence,                       e*P_s = 11 W

3 0
2 years ago
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