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valina [46]
2 years ago
6

A box sits on a table. A short arrow labeled F subscript N points up. A short arrow labeled F subscript g points down. A long ar

row labeled F subscript P points right. A short arrow labeled F subscript f points left. Study the force diagram to complete the sentences. The forces acting on the box are . The box will move to the
Physics
1 answer:
Aloiza [94]2 years ago
7 0

Answer:

unbalanced and right

Explanation:

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Karissa is conducting an experiment on the amount of salt that dissolves in water at different temperatures. She repeats her tes
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A student placed an ice cube on a table and observed it for five minutes. He noticed that the ice cube seemed to get smaller and
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the ice cube melted due to the absorption of heat from the surrounding of the ice, the has a heat capacity of zero so it turns it from its solid state to its liquid state which leads to the formation of the puddle

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Lincoln weighs 400 newtons. What’s his mass rounded to the nearest kilogram? Assume that acceleration due to gravity is 9.8 N/kg
Yanka [14]
Weight equals mass times gravitational acceleration=400N, so mass=400/9.8=41kg approx.
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1. Two identical bowling balls of mass M and radius R roll side by side at speed v0 along a flat surface. Ball 1 encounters a ra
UNO [17]

Answer:

1/2

Explanation:

We need to make a couple of considerations but basically the problem is solved through the conservation of energy.  

I attached a diagram for the two surfaces and begin to make the necessary considerations.

Rough Surface,

We know that force is equal to,

F_r = mgsin\theta

F_r = \mu N

F_r = \mu mg cos\theta

Matching the two equation we have,

\mu N = \mu mg cos\theta

\mu = tan\theta

Applying energy conservation,

\frac{1}{2}mv^2_0+\frac{1}{2}I_w^2 = F_r*d+mgh_1

\frac{1}{2}mv^2_0+\frac{2}{5}mR^2\frac{V_0^2}{R^2} = F_r*d+mgh_1

\frac{1}{2}mv^2_0+\frac{mv_0^2}{5} = mgsin\theta \frac{h_1sin\theta}+mgh_1

\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_1+gh_1

h_1 = \frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})

Frictionless surface

\frac{1}{2}mv_0^2+\frac{1}{2}I\omega^2 = mgh_2

\frac{1}{2}m_v^2+\frac{1}{2}\frac{2}{5}mR^2\frac{v_0^2}{R^2} =mgh_2

\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_2

h_2 = \frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})

Given the description we apply energy conservation taking into account the inertia of a sphere. Then the relation between h_1 and h_2 is given by

\frac{h_1}{h_2} = \frac{\frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})}{\frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})}

\frac{h_1}{h_2} = \frac{1}{2}

8 0
2 years ago
A 20 watt lightbulb uses 20 Joules of energy every second.A person expends 50 watts of energy per stair when climbing up stairs.
Pachacha [2.7K]
20W = 20 J/s

Energy expended during climbing stairs = 50 W of energy/stair = 50J/stair

For 20 stairs, Total energy = 50x20 = 1000 J

This can light bulbs for, T= 1000J/20 J/s =50 seconds
6 0
2 years ago
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