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Annette [7]
3 years ago
13

An ant starts at one edge of a long strip of paper that is 34.2 cm wide. She travels at 0.4 cm/s at an angle of 52◦ with the lon

g edge. How long will it take her to get across? Answer in units of s.
Physics
1 answer:
nordsb [41]3 years ago
5 0

Answer:

t=108.50s

Explanation:

The ant is under an uniform motion. So, we get the time taken by the ant to cross the strip from the definition of speed:

v=\frac{x}{t}\\t=\frac{x}{v}

In this case, we use the vertical speed of the ant (v_y=vsin\theta), since the ant crosses the strip that is 34.2 cm wide (vertically). So:

t=\frac{y}{v_y}\\t=\frac{y}{vsin\theta}\\t=\frac{34.2cm}{0.4\frac{cm}{s}sin(52^\circ)}\\t=108.50s

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The current through a 10 ohm resistor connected to a 120 volt power supply is
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Answer:I=12 A

Explanation:

Given

Resistance R=10 \Omega

Voltage V=120 V

According to ohm's law current through a conductor is directly proportional to the voltage applied.

V\propto I

V=IR

where V=Voltage

I=Current

R=resistance

I=\frac{V}{R}

I=\frac{120}{10}

I=12 A

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3 years ago
When an electron in a certain excited energy level in a one-dimensional box of length 2.00 Å makes a transition to the ground st
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Calculate the wavelength associated with an electron with energy 2000 eV.

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3 years ago
An object is NOT accelerating if it is moving
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At a constant speed

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What two measurements are needed to determine density?
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7 0
3 years ago
Read 2 more answers
You pull a solid nickel ball with a density of 8.91 g/cm3 and a radius of 1.40 cm upward through a fluid at a constant speed of
Sunny_sXe [5.5K]

Answer:

P = 1.090\,N

Explanation:

The constant speed means that ball is not experimenting acceleration. This elements is modelled by using the following equation of equilibrium:

\Sigma F = P - W + F_{D}

\Sigma F = P - \rho \cdot V \cdot g + c\cdot v = 0

Now, the exerted force is:

P = \rho \cdot V \cdot g - c\cdot v

The volume of a sphere is:

V = \frac{4\cdot \pi}{3}\cdot R^{3}

V = \frac{4\cdot \pi}{3}\cdot (0.014\,m)^{3}

V = 1.149\times 10^{-5}\,m^{3}

Lastly, the force is calculated:

P = (8910\,\frac{kg}{m^{3}} )\cdot (1.149\cdot 10^{-5}\,m^{3})\cdot (9.81\,\frac{m}{s^{2}} )+(0.950\,\frac{kg}{s})\cdot (0.09\,\frac{m}{s} )

P = 1.090\,N

5 0
3 years ago
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