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gayaneshka [121]
3 years ago
15

8. A point of Charge +3.0 X 10^-7 Coulomb is placed 2.0 X 10^-2 from a second

Physics
1 answer:
Oksi-84 [34.3K]3 years ago
4 0
Use the formula F = (9x10^9 Q * q) / r^2
Message me if you need help.
You might be interested in
0.235 moles of a diatomic gas expands doing 205 J of work while the temperature drops 88 K. Find Q.? (Unit=J)
Harman [31]

The heat released by the gas is -225 J

Explanation:

First of all, we have to calculate the change in internal energy of the gas, which for a diatomic gas is given by

\Delta U = \frac{5}{2}nR\Delta T

where

n = 0.235 mol is the number of moles

R=8.314\cdot J/mol K is the gas constant

\Delta T = -88 K is the change in temperature

Substituting,

\Delta U = \frac{5}{2}(0.235)(8.314)(-88)=-430 J

Now we can us the 1st law of thermodynamics to find the heat absorbed/released by the gas:

\Delta U = Q -W

where

\Delta U = -430 J is the change in internal energy

Q is the heat

W = 205 J is the heat done by the gas

Solving for Q,

Q=\Delta U + W = -430 + 205 =-225 J

Since the sign is negative, it means the heat has been released by the gas.

Learn more about thermodynamics:

brainly.com/question/4759369

brainly.com/question/3063912

brainly.com/question/3564634

#LearnwithBrainly

5 0
3 years ago
These two pls :)))) ill mark brainliest :)
Anna71 [15]

Answer:

Bowling Ball: weight on Earth = 49 N

Textbook: Mass = 2 kg; weight on the moon = 3.2 N

Large dog: weight on Earth = 490 N; weight on the moon = 80 N

Law of Universal Gravitation: F_{G}=\frac{Gm_{1}m_{2}}{r^{2}}

F_{G} = gravitational force (Newtons/N)

<em>G</em> = gravitational constant, 6.67430 × 10¹¹ \frac{N*m^{2}}{kg^{2}}

<em>m</em>₁ and <em>m</em>₂ = masses of two objects (kilograms/kg)

<em>r</em>² = square of distance between centers of the two objects (meters/m)

Have a fantastic day!

4 0
3 years ago
An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.
Olegator [25]

Hi there!

We can use Biot-Savart's Law for a moving particle:
B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }

B = \boxed{7.07 *10^{-10} T}

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

c)

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}

d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.

Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.

\boxed{B = 0 T}

3 0
2 years ago
Most cars have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made of coppe
alexandr402 [8]

Answer:

0.699 L of the fluid will overflow

Explanation:

We know that the change in volume ΔV = V₀β(T₂ - T₁) where V₀ = volume of radiator = 21.1 L, β = coefficient of volume expansion of fluid = 400 × 10⁻⁶/°C

and T₁ = initial temperature of radiator = 12.2°C and T₂ = final temperature of radiator = 95.0°C

Substituting these values into the equation, we have

ΔV = V₀β(T₂ - T₁)

= 21.1 L × 400 × 10⁻⁶/°C × (95.0°C - 12.2°C)

= 21.1 L × 400 × 10⁻⁶/°C × 82.8°C = 698832 × 10⁻⁶ L

= 0.698832 L

≅ 0.699 L = 0.7 L to the nearest tenth litre

So, 0.699 L of the fluid will overflow

6 0
3 years ago
Which situation is the best analogy for the doppler effect?
Rudiy27
The best scenario to describe the doppler effect would be listening to the siren of a passing ambulance or fire truck

then it is coming towards you, the pitch is higher, it gets higher as it approaches and peaks as it gets right in front of you. then it drop at once when it passes you and continues to drop till it fades away. this is a classic descrption of the doppler effect
8 0
3 years ago
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