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EastWind [94]
3 years ago
7

A car travels at 60 miles per hour. How much time does it take the car to travel 30 miles? Select one: a. 1 hour b. 1/2 hour c.

3/4 hour d. 2 hours
Physics
2 answers:
Marrrta [24]3 years ago
8 0

Answer:

b.t=\frac{1}{2} hour

Explanation:

Conceptual analysis

The average velocity (v) is defined as the ratio between the displacement (d) that a body performed and the total time (t) that the displacement lasted and is calculated using the following formula:

v=\frac{d}{t}Formula (1)

v: velocity

d: Displacement

t: Time

Known data:

v=60 \frac{miles}{hour}

d = 30 miles

Problem development

We replace data in formula (1):

60\frac{miles}{hour}=\frac{30miles}{t}

t= \frac{30 miles}{60\frac{miles}{hour} }

t=\frac{1}{2} hour

andreev551 [17]3 years ago
4 0
The car is traveling at 60 miles per hour. Theoretically, at this pace, It'll take (B.), 1/2 an hour.


Brainliest?


Hope I helped! ^-^
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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}

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