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DochEvi [55]
2 years ago
7

If a set of Bluetooth headphones has a resistance of 4.5 ohms and has a current of 1.5 amps, what is the voltage of the headphon

es?
Physics
1 answer:
FinnZ [79.3K]2 years ago
7 0

Answer:

6.75 v

Explanation:

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We run a distance of 1000 m at a speed of 4.3 m/s. Calculate the time elapsed to cover this distance
pashok25 [27]

Answer:

So if we need to cover 1000 meters. And we go at a speed of 4.3 m/s. That means that every 4.3 meters we cover is 1 second. So we divide both amd get

1000/4.3 = 232.56 is approx the answer. Also the meters cancel out because

m/(m/s) = m*s/m, cancels out giving s as a unit.

<h2><u>Therefore the answer is 232.56 seconds</u></h2>

6 0
3 years ago
Two golf balls are thrown with the same speed off the top of a large bridge at the same time. ball a is thrown straight downward
iren2701 [21]
I think that ball  a hit the ground because it says that it went straight down.
8 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
A projectile of mass 0.2 kg and an initial velocity of 50 m/s collides with the end of a blade attached to a turbine. The rotati
fgiga [73]

Answer:

5.5 rad/sec

Explanation:

8 0
2 years ago
Give two examples where friction is a nuisance
dlinn [17]

Answer:

it can cause wear and tear and reduces efficiency as energy is lost.

Explanation:

8 0
3 years ago
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