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3 years ago

The diagram below shows a 1.0 x 105newton truck at rest on a hill that makes an angle of 8.0° with the

Physics

1 answer:

Digiron [165]3 years ago

7 0

Answer:

$it \: is \: mg \sin(8.0 \degree) \\ = 1.0 \times 10 {}^{5} \times \sin(8.0 \degree) \\ = 9.9 \times 10 {}^{4} newtons$

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Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .

4 0

3 years ago

___ is the combination of all colors of light

TiliK225 [7]

The color white is what you'd see when every color of light is combined.

5 0

3 years ago

Read 2 more answers

At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81m/s2 . A watermelon has a weight of 58.0N at the sur

kvasek [131]

Answer: A) mass on earth surface = 5.91kg

B) mass on surface of jupiter = 5.91kg

C) weight on surface of jupiter = 10.697N

Explanation:

The relationship between weight (W), mass (m) and acceleration due gravity (g) is given below

W=mg

From the question, g= 9.8m/s² and weight on the surface on the earth is 58N

A) The mass of watermelon on earth is

m = 58/ 9.8 = 5.91kg

B) the mass of the watermelon on jupiter is 5.91kg.

You will notice this is the same as the mass of watermelon on earth and that is so because mass is a scalar quantity that does not depends on the distance away from the center of the earth (unlike weight which is a vector) thus making it constant all through any location.

C) mass of watermelon is 5.91kg, g=9.8m/s² weight of watermelon on jupiter is given below as

W = mg

W = 5.91 x 9.8

= 10.697N.

6 0

3 years ago

A 1,500-kg car accelerates along a flat track to a speed of 20 m/s. how much does it's gravitational potential energy increases

lesya [120]

Look at bitesizes physics section, they have all the information you need to complete this question.

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3 years ago

You’ve made the hypothesis that the stepper the slope , the faster a ball will be rolling when it reaches the bottom .

BigorU [14]

Answer:

B) how steep the slope is

Explanation:

Because you have to know how is the influence of the steep of the slope in the time that a ball reaches the bottom. The steep of the slope is the variable that you would have to change in an experiment.

I hope this is useful for you

regards

4 0

3 years ago