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Nonamiya [84]
3 years ago
9

Andrew is pushing this block to the right across on the floor. Which force in the free body diagram is pointing in the wrong dir

ection?
Question 11 options:

the applied force


friction


the normal force


the weight

Physics
2 answers:
soldier1979 [14.2K]3 years ago
8 0

Answer:

the normal force

Explanation:

The free-body diagram represents all the forces acting on an object. In this example, there are four forces acting on the box: an applied force, the friction (which always act opposite to the applied force), the weight of the box (which is always downward), and the normal force.

The normal force is the reaction force exerted by the surface on which the box is moving on the box, and this reaction force is always opposite to the force exerted by the box on the surface. Since the latter is downward, it means that the normal force must be upward, so in the diagram it is wrong.

Deffense [45]3 years ago
6 0

Normal force since it's perpendicular to the contact force and should be going up.

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Convert an acceleration of 12m/s^2 to km/h^2​
ozzi

Answer:

43.2

because to convert from m/sec to kmph we need to multiply by 3600/1000

4 0
3 years ago
2. An object with moment of inertia ????1 = 9.7 x 10−4 kg ∙ m2 rotates at a speed of 3.0 rev/s. A 20 g mass with moment of inert
svetoff [14.1K]

Answer:

2.85 rad/s

Explanation:

5 cm = 0.05 m

20 g = 0.02 kg

When dropping the 2nd object at a distance of 0.05 m from the center of mass, its corrected moments of inertia is:

I_2 = 1.32\times10^{−6} + 0.02 * 0.05^2 = 0.513\times10^{-4} kgm^2

So the total moment of inertia of the system of 2 objects after the drop is:

I = I_1 + I_2 = 9.7\times10^{-4} + 0.513\times10^{-4} = 0.0010213 kgm^2

From here we can apply the law of angular momentum conservation to calculate the post angular speed

\omega_1 I_1 = \omega_2 I

\omega_2 = \omega_1 \frac{I_1}{I} = 3 \frac{9.7\times10^{-4}}{0.0010213} = 2.85 rad/s

6 0
3 years ago
A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 20.1 N; when it is completely i
Paha777 [63]

Answer:

A) V = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}

B) d = 4181.49 kg/m^{3} = 4.18 g/cm^{3}

Explanation:

A) Using the Archimedes' force we can find the weight of water displaced:

W_{d} = W_{a} - W_{w}

Where:

W_{a}: is the weight of the block in the air = 20.1 N

W_{w}: is the weight of the block in the water = 15.3 N

W_{d} = W_{a} - W_{w} = 20.1 N - 15.3 N = 4.8 N

Now, the mass of the water displaced is:

m = \frac{W_{d}}{g} = \frac{4.8 N}{9.81 m/s^{2}} = 0.49 kg

The volume of the block can be found using the mass of water displaced and the density of the water:

V = \frac{m}{d} = \frac{0.49 kg}{997 kg/m^{3}} = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}

B) The density of the block can be found as follows:

d = \frac{W_{a}}{g*V} = \frac{20.1 N}{9.81 m/s^{2}*4.92 \cdot 10^{-4} m^{3}} = 4181.49 kg/m^{3} = 4.18 g/cm^{3}

I hope it helps you!            

6 0
3 years ago
In order for work to happen you MUST have?
Nataly_w [17]

Explanation:

it requires energy

hope it helps

7 0
2 years ago
If the voltage across a circuit of constant resistance is doubled, the power dissipated by that circuit will
densk [106]

Answer:

The voltage will quadruple

Explanation:

The power dissipated in a circuit is given by

P=\frac{V^2}{R}

where

V is the voltage

R is the resistance

In this problem, the voltage across the circuit is doubled:

V' = 2V

So the new power dissipated is

P'=\frac{V'^2}{R}=\frac{(2V)^2}{R}=4\frac{V^2}{R}=4 P

so, the power dissipated will quadruple.

6 0
3 years ago
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