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3 years ago

A body moving with a velocity of 20m/s begins to accelerate at 3 m/s 2 . How far does the body move in 3 seconds

1 answer:

5 0

Given v_in = 20 m/s and a = 3 m/s2, assuming that the body moves at constant acceleration, the motion is modeled by the equation:

s(t) = (v_in)t + (1/2)a(t^2)

where s(t) is the distance traveled

substituting the given,

s(t) = 20t + (3/2)(t^2)

at t = 3

s(t) = 20(3) + (3/2)(3)^2

= 73.5 m

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3 years ago

Identify two factors that determine the state of matter

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Two factors determine whether a substance is a solid, a liquid, or a gas: The kinetic energies of the particles (atoms, molecules, or ions) that make up a substance. Kinetic energy tends to keep the particles moving apart. The attractive intermolecular forces between particles that tend to draw the particles together.

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3 years ago

Read 2 more answers

A piano wire with mass 2.60g and length 84.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplit

likoan [24]

Answer:

Power will be 0.2023 watt

And when amplitude is halved then power will be 0.0505 watt

Explanation:

We have given mass of the Piano wire m = 2.60 gram = 0.0026 kg

Length of wire l = 84 cm = 0.84 m

So mass density $\mu =\frac{m}{l}=\frac{0.0026}{0.84}=0.0031kg/m$

Tension in the wire T = 25 N

Frequency f = 120 Hz

So angular frequency $\omega =2\pi f=2\times 3.14\times 120=753.6rad/sec$

And amplitude A = 1.6 mm = 0.0016 m

We have to find the generated power

Power is given by $P=\frac{1}{2}\sqrt{\mu T}\omega ^2A^2=\frac{1}{2}\times \sqrt{0.0031\times 25}\times 753.6^2\times 0.0016^2=0.2023watt$

From the relation we can see that power $P\ \propto\ A^2$

So if amplitude is halved then power will be $\frac{1}{4}$ times

So power will be equal to $\frac{0.2023}{2}=0.0505watt$

4 0

3 years ago

Two boxes are at rest on a smooth, horizontal surface. The boxes are in contact with one another. If box 1 is pushed with a forc

Maslowich

Answer:

mass of box 1 = 2.20 kg

mass of box 2 = 5.93 kg

Explanation:

Let the mass of box 1 and box 2 is respectively

$m_1$ and $m_2$

so we will have

Force applied on box 1 then acceleration

$a = \frac{F}{m_1 + m_2}$

$1.50 = \frac{12.2}{m_1 + m_2}$

$m_1 + m_2 = 8.13$

Now we know that contact force between them in above case is given as

$F_n = m_2a$

$8.90 = m_2(1.5)$

$m_2 = 5.93 kg$

now we have

$m_1 = 2.20 kg$

8 0

3 years ago

at certain times the demand for electric energy is low and electric energy is used to pump water to a reservoir 45 m above the g

Readme [11.4K]

The mass of water that must be raised is $5.25\cdot 10^7 kg$

Explanation:

Since the process is 70% efficiency, the power in output to the turbine can be written as

$P_{out} = 0.70 P_{in}$

where $P_{in}$ is the power in input.

The power in input can be written as

$P_{in} = \frac{W}{t}$

where

W is the work done in lifting the water

t = 3 h = 10,800 s is the time elapsed

The work done in lifting the water is given by

$W=mgh$

where

m is the mass of water

$g=9.8 m/s^2$ is the acceleration of gravity

h = 45 m is the height at which the water is lifted

Combining the three equations together, we get:

$P_{out} = 0.70 \frac{mgh}{t}$

Where

$P_{out} = 150 MW = 150\cdot 10^6 W$

And solving for m, we find:

$m=\frac{Pt}{0.70gh}=\frac{(1.50\cdot 10^6)(10800)}{(0.70)(9.8)(45)}=5.25\cdot 10^7 kg$

Learn more about power:

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3 0

3 years ago