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faltersainse [42]
3 years ago
5

5)

Physics
1 answer:
miskamm [114]3 years ago
4 0
B . I hope this is right
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Potential diffetence​
Stells [14]

Answer:

6v

Explanation:

V=IR

V= 2* 3

V= 6 volts

7 0
3 years ago
3. To open a soda bottle, a force of 55.0 N is applied to the bottle opener. If the bottle opener
Bogdan [553]

Answer:

0.0815

Explanation:

Mechanical advantage is output force/input force. So all you have to do is 55.0N/675N ~ 0.0815

Hope this helped!

5 0
3 years ago
A football player kicks a 0.94 kg football with a force of 2.4 N. Calculate the acceleration of the football as the player kicks
Anon25 [30]
Hello, I hope this helps :)
So the equation to figure these kinds of questions is F=MA
F refers to force, which in this situation F would be 2.4
M refers to mass, mass would be 0.94
A refers to acceleration, which we are trying to figure out
So if we put in the information we know into the equation, it is now 2.4=0.94*A
So with that equation we can figure that 2.4 divided by 0.94 equals A
So 2.4/0.94= 2.553191489362 :'D
Don't worry, the rounded and correct answer is 2.6
Have a nice day =)
7 0
3 years ago
Read 2 more answers
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
For this problem, imagine that you are on a ship that is oscillating up and down on a rough sea. Assume for simplicity that this
ikadub [295]

Answer:

no idea

Explanation:

7 0
4 years ago
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