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3 years ago

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For an object to be in static equilibrium, the net force must be equal to zero and and The center of gravity must be above the s

upport point. The moment of inertia must be zero. The normal force must equal the weight. The net torque must be equal to zero.

1 answer:

iren [92.7K]3 years ago

8 0

Answer:

The normal force must equal the weight.

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How to solve these two questions?

hammer [34]

1) See attached figure

The relationship between charge and current is:

$i = \frac{Q}{t}$

where

i is the current

Q is the charge

t is the time

Therefore, the current is the rate of change of the charge passing through a given point over time.

This means that for a graph of charge over time, the current is just equal to the slope of the graph.

For the graph in this problem:

- Between t = 0 and t = 2 s, the slope is

$\frac{50-0}{2-0}=25 C/s$

therefore the current is

i = 25 A

- Between t = 2 s and t = 6 s, the slope is

$\frac{-50-(50)}{6-2}=-25 C/s$

therefore the current is

i = -25 A

- Between t = 6 s and t = 8 s, the slope is

$\frac{0-(-50)}{8-6}=25 C/s$

therefore the current is

i = 25 A

The figure attached show these values plotted on a graph.

2) $15 \mu C$

The previous equation can be rewritten as

$Q = i t$

This equation is valid if the current is constant: if the current is not constant, then the total charge is simply equal to the area under a current vs time graph.

Here we have the current vs time graph, so we gave to find the area under it.

The area of the first triangle is:

$A_1 = \frac{1}{2}(0.001 s)(0.010 A)=5\cdot 10^{-6} C$

While the area of the second square is

$A_2 = (0.002 s - 0.001 s)(0.010 A)=1\cdot 10^{-5}C$

So, the total area (and the total charge) is

$Q=A_1 +A_2 = 5\cdot 10^{-6} + 1\cdot 10^{-5} = 1.5\cdot 10^{-5}C=1.5 \mu C$

3 0

3 years ago

Anybody know this answer?

Ber [7]

It seperates naturally

3 0

2 years ago

What is the force exerted

Norma-Jean [14]

Answer:

I wish i saw the field shown to the right

Explanation:

3 0

3 years ago

The force F shown in Figure 4.30 has a moment of 40 Nm about the pivot. Calculate the magnitude

salantis [7]

$\boxed{\sf \tau=rFsin\theta}$

Put values

$\\ \rm\hookrightarrow 40=2Fsin40$

$\\ \rm\hookrightarrow Fsin40=20$

$\\ \rm\hookrightarrow 0.64F=20$

$\\ \rm\hookrightarrow F=31.25N$

8 0

2 years ago

A proton moving at 8.9 × 106 m/s through a magnetic field of 0.96 T experiences a magnetic force of magnitude 3.8 × 10−13 N. Wha

goblinko [34]

Answer: 15.66 °

Explanation: In order to solve this proble we have to consirer the Loretz force for charge partcles moving inside a magnetic field. Thsi force is given by:

F=q v×B = qvB sin α where α is teh angle between the velocity and magnetic field vectors.

From this expression and using the given values we obtain the following:

F/(q*v*B) = sin α

3.8 * 10^-13/(1.6*10^-19*8.9*10^6* 0.96)= 0.27

then α =15.66°

8 0

2 years ago