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sukhopar [10]
3 years ago
12

A box is against a wall. A person pushes on the box, but the box does not move. Is this situation an example of a force

Physics
2 answers:
yuradex [85]3 years ago
8 0

Situations are not forces.

But there are four forces in the scenario you described:

1). the force with which the person pushes the box

2). the normal force ... the wall pushes on the box in the opposite direction

3). gravity pulls the box downward

4). the force of static friction pushes the box upward.

#1 and #2 are balanced horizontal forces.

#3 and #4 are balanced vertical forces.

The net force on the box is zero.

So the box does not accelerate.

Leya [2.2K]3 years ago
7 0

Answer: I don’t think so, because the box is ALREADY against the wall so you can’t move it further into the wall

Explanation:

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The amount of heat needed to raise the temperature of a substance by \Delta T is given by
Q=m C_s \Delta T
where
m is the mass of the substance
Cs is its specific heat capacity
\Delta T is the increase in temperature

For oxygen, the specific heat capacity is approximately 
C_s = 0.92 J/(g K)
The variation of temperature for the sample in our problem is 
\Delta T= -15^{\circ}C-(-30^{\circ} C)=+15^{\circ}C=15 K
while the mass is m=150 g, so the amount of heat needed is
Q=m C_s \Delta T=(150 g)(0.92 J/g K)(15 K)=2070 J
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3 years ago
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if you use the compound pulley, you will need to pull twice the distance but with less force. the force you need is equal to one
algol13

Answer:

F= 25/2 = 12.5N

Explanation:

When you use a compound pulley the force required depends on the mechanical advantage of the compound pulley. This is known as rate of loss of distance or the ratio of the force to the load.

M.A = Effort distance /Load distance. OR M.A = Load/Effort

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4 years ago
A resistor, inductor, and a battery are arranged in a circuit. The circuit has an inductance of L = 1 H and a resistance of 1.4
shtirl [24]

Answer:

\tau \approx 7.14 \times 10^{-4}s \approx0.714ms

Explanation:

In a LC circuit The time constant τ is the time necessary for 60% of the total current (maximum current), pass through the inductor after a direct voltage source has been connected to it. The time constant can be calculated as follows:

\tau =\frac{L}{R}

Therefore, the time needed for the current to reach a fraction f = 0.6(60%) of its maximum value is:

\tau =\frac{1}{1.4\times 10^{3}} =7.142857143 \times 10^{-4} \approx7.14 \times 10^{-4}s

8 0
4 years ago
Problem 1: Three beads are placed along a thin rod. The first bead, of mass m1 = 24 g, is placed a distance d1 = 1.1 cm from the
Svet_ta [14]

Answer:

b)  x_{cm} = 4.88 cm , c) x_{cm}’= 1/M  (m₁ d₁ + m₃ d₃) and d)

x_{cm}’= 1.88 cm

Explanation:

The definition of mass center is

    x_{cm} = 1/M ∑ xi mi

Where mi, xi are the mass and distance from an origin for each mass and M is the total mass of the object.

Part b

Apply this equation to our case.

Body 1

They give us the mass (m₁ = 24 g) and the distance (d₁ = 1.1 cm) from the origin at the far left

Body 2

They give us the mass (m₂ = 12.g) and the distance relative to the distance of the body 1, let's look for the distance from the left end (origin)

    D₂ = d₁ + d₂

    D₂ = 1.1 + 1.9

    D₂ = 3.0 cm

Body 3

Give the mass (m₃ = 56 g) and the position relative to body 2, let's find the distance relative to the origin

    D₃ = D₂ + d₂

    D₃ = 3.0 + 3.9

    D₃ = 6.9 cm

With this data we substitute and calculate the center of mass

    M = m₁ + m₂ + m₃

    M = 24 + 12 + 56

    M = 92 g

    x_{cm} = 1/92 (1.1 24 + 3.0 12 + 6.9 56)

    x_{cm} = 1/92 (448.8)

    x_{cm} = 4,878 cm

    x_{cm} = 4.88 cm

This distance is from the left end of the bar

Par c)

In this case we are asked for the same calculation, but the reference system is in the center marble, we have to rewrite the distance with the reference system in this marble.

Body 1

It is at   d1 = -1.9 cm

It is negative for being on the left and the value is the relative distance of 1 to 2

Body 2

d2 = 0 cm

The reference system for her

Body 3

d3 = 3.9 cm

Positive because that is to the left of the reference system and is the relative distance between 2 and 3

Let's write the new center of mass (xcm')

    x_{cm} ’= 1/M  (m₁ d₁ + m₂ d₂ + m₃ d₃)

   

   x_{cm}’= 1/M  (m₁ d₁ + m₃ d₃)

Part d) Let's calculate the value of the center of mass

    x_{cm}’= 1/92 ((24 (-1.9) +56 3.9)

    x_{cm}’= 1/92 (172.8)

    x_{cm}’= 1.88 cm

This distance is to the right of the central marble

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-Dominant- [34]
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