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Marina CMI [18]
3 years ago
11

So, what is the answer of this assignment?

Physics
1 answer:
Ket [755]3 years ago
6 0
What assignment, you have to work with me here i don't know what were working on
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*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude
kvv77 [185]

Answer:

Approximately 3.86\; {\rm N} (given that the magnitude of this charge is -7.35\; {\rm \mu C}.)

Explanation:

If a charge of magnitude q is placed in an electric field of magnitude E, the magnitude of the electrostatic force on that charge would be F = E\, q.

The magnitude of this charge is q = 7.35\; {\rm \mu C}. Apply the unit conversion 1\; {\rm \mu C} = 10^{-6}\; {\rm C}:

\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}.

An electric field of magnitude E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}} would exert on this charge a force with a magnitude of:

\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}.

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0
2 years ago
A car accelerates at a rate of 3 m/s^2. If its original speed is 8 m/s, how many seconds will it take the car to reach a final s
kkurt [141]

Answer:

\Large \boxed{\mathrm{5.67 \ seconds }}

Explanation:

\displaystyle \mathrm{acceleration \ = \ \frac{final \ velocity - initial \ velocity }{elapsed \ time}}

\displaystyle A \ = \ \frac{V_f - V_i }{t}

\displaystyle 3 \ = \ \frac{25 - 8 }{t}

\displaystyle 3 \ = \ \frac{17 }{t}

\displaystyle t \ = \ \frac{17 }{3} \approx 5.67

4 0
3 years ago
A gas cylinder is filled with 4.00 moles of oxygen gas at 300.0 k. the piston is compressed to yield a pressure of 400.0 kpa. wh
Radda [10]
To answer this item, it is assumed that the gas in the cylinder is ideal such that it follows the equation,
                                        PV = nRT
when V is to be calculated,
                                              V = nRT/P
        V = (4)(0.0821 L.atm/molK)(300 K) / (400 kPa/101.325 kPa/atm)  
                                                V = 24.95 L
Thus, the volume of gas in the cylinder is 24.95 L. 
8 0
3 years ago
Please help on this one?
nikdorinn [45]

a is the right answer

4 0
3 years ago
Read 2 more answers
A golf ball has a mass of 120g. Calculate its increase in mass when it is travelling at 40ms1. What is this as a percentage of i
kicyunya [14]

Answer:

Percentage=8.889\times 10^{-13}%

Explanation:

From special theory of relativity the dynamic mass m is related with the rest mass m_{0}of the body as

m=\frac{m_{0} }{\sqrt{1-\frac{v^{2} }{c^{2} } } }

Here, c is the speed of light and v is the velocity of object.

Given mass of the golf ball is 120 g.

m=\frac{120 }{\sqrt{1-\frac{(40)^{2} }{(3\times 10^{8} )^{2} } } }\\m=120(1-\frac{(40)^{2} }{(3\times 10^{8}) ^{2} })^{-\frac{1}{2} } \\

Now applying the binomial theorem and solve the above equation.

m=(1+\frac{1}{2}(\frac{40}{3\times 10^{8} }) ^{2} )\\m=120(1+8.889\times 10^{-15})

Therefore, increase in mass is,

\Delta m=120\times 8.889\times 10^{-15} \\\Delta m=10.6668\times 10^{-13} g

Now percentage of increase in mass with rest mass is,

Percentage=\frac{10.6668\times 10^{-13} g}{120g} \times 100\\Percentage=8.889\times 10^{-13}

Therefore, the percentage of increase in mass with rest mass is Percentage=8.889\times 10^{-13}.

4 0
4 years ago
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