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3 years ago

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You apply a very small force, say 0.001 newtons, to a very large truck, with a mass of 2000 kilograms. What can you say for sure

about what will happen to the truck?
The truck will accelerate (move from rest to motion) as long as that tiny force is larger than any force of friction that opposes it.

The truck will not move because small forces cannot move large objects.

The truck will accelerate (move from rest to motion) as long as the small applied force is large enough to overcome the inertia of the truck.

An extremely small force can never accelerate such a large truck. The truck will not move under any circumstance.

1 answer:

Fofino [41]3 years ago

5 0

Answer:

it will stay still. unless its in space.

Explanation:

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Find the kinetic energy of a ball of mass 200 grams moving at a speed of 20 m/s.

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Answer:

40KJ

Explanation:

m=200g

V=20m/s

Ek=?

Ek=(m*V²)/2

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you have dissolved 10g sodium oxide in 200 ml water .calculate the concentration of the solution .physical

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Answer:

0.05

Explanation:

Divide the mass of the solute by the total volume of the solution. Write out the equation C = m/V, where m is the mass of the solute and V is the total volume of the solution. Plug in the values you found for the mass and volume, and divide them to find the concentration of your solution.

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A student plucks a fixed-end string, creating a standing wave with 6.00 nodes (including any nodes at the ends). The string is t

Vesna [10]

1) 2.5 wavelengths

2) 0.208 m

3) 1731 Hz

Explanation:

1)

Standing waves are waves that do not propagate, but instead the particles of the medium just oscillate around a fixed position. Examples of standing waves are the waves produced on a string with fixed ends.

The points of a standing wave in which the amplitude of the oscillation is always zero are called nodes.

The two fixed ends of the string are two nodes. In this problem, we have a total of 6 nodes along the string: this means that there are 4 additional nodes apart from the two ends of the string.

Therefore, this also means that the string oscillate in 5 different segments.

One wavelength is equal to 2 segments of the oscillation: therefore, since here there are 5 segments, this means that the number of wavelengths that we have in this string is

$n=\frac{5}{2}=2.5$

2)

The wavelength of a wave is the distance between two consecutive crests (or throughs) of the wave.

The wavelength of a standing wave can be also measured as the distance between the nth-node and the (n+2)-th node: so, basically, the wavelength in a standing wave is twice the distance between two nodes:

$\lambda = 2 d$

where

$\lambda$ is the wavelength

d is the distance between two nodes

Here the length of the string is

L = 0.520 m

And since it oscillates in 5 segments, the distance between two nodes is

$d=\frac{L}{5}=\frac{0.520}{5}=0.104 m$

And therefore, the wavelength is

$\lambda=2d=2(0.104)=0.208 m$

3)

The frequency of a wave is the number of complete oscillations of the wave per second.

The frequency of a wave is related to its speed and wavelength by the wave equation:

$v=f\lambda$

where

v is the speed

f is the frequency

$\lambda$ is the wavelength

In this problem:

v = 360 m/s is the speed of the wave

$\lambda=0.208 m$ is the wavelength

Therefore, the frequency is

$f=\frac{v}{\lambda}=\frac{360}{0.208}=1731 Hz$

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3 years ago

What is the equivalent capacitance of the three capacitors in the figure (figure 1)?

vekshin1

The equivalent capacitance of the combination is $\dfrac{1}{C} = \dfrac{1}{C_1}+\dfrac{1}{C_2}$ where C1 and C2 are the capacitance of both capacitors in series.

<h3>What is equivalent capacitor?</h3>

Let the capacitance of both capacitors be C1 and C2. For a series connected capacitors, same charge flows through the capacitors but different voltage flows through them.

Let Q be the amount of charge in each capacitors,

V be the voltage across each capacitors

C be the capacitance of the capacitor.

Using the formula Q = CV where V = Q/C... (1)

For the large capacitor with capacitance of the capacitor C1,

$Q = C_1V_1;$

$V_1 = \dfrac{Q}{C_1}...(2)$

where $V_1$ is the voltage across $C_1,$

For the small capacitor with capacitance of the capacitor $C_2$ ,

$Q = C_2V_2$ ;

$V_2 = \dfrac{Q}{C_2} ... (3)$

where $V_2$ is the voltage across $C_2$ ,

Total voltage V in the circuit will be;

$V = V_1+V_2... (4)$

Substituting equation 1, 2 and 3 in equation 4, we have;

$\dfrac{Q}{C} = \dfrac{Q}{C_1} +\dfrac{ Q}{C_2}$

$\dfrac{Q}{C} = Q({\dfrac{1}{C_1}+\dfrac{1}{C_2})$

Since change Q is the same for both capacitors since they are in series, they will cancel out to finally have;

$\dfrac{1}{C} = ({\dfrac{1}{C_1}+\dfrac{1}{C_2})$

This gives the equivalent capacitance of the combination.

To know more about equivalence capacitance follow

brainly.com/question/5626146

#SPJ4

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2 years ago