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Rzqust [24]
3 years ago
12

A driver is moving at 17.0 m/s when she sees a child on a bike 65m ahead of her. If the car can brake at 4.9 m/s^2 and the drive

rs reaction time is 1.6 s, can she avoid hitting the child
Physics
1 answer:
Lera25 [3.4K]3 years ago
7 0

Answer:

The driver can  avoid the child because at the time she brake 1.6s the car just move 20.928 meters so is far to the 65 meters where the kind is

Distance= 20.928 m

Explanation:

V_{o} = 17 \frac{m}{s} \\a= -4.9 \frac{m}{s^{2} }\\ t= 1.6 s

Distance the kid is 65m so the car have to be less in the time she is braking

So to calculated the distance while she is braking

X_{f} = X_{o} +V_{o}*t + \frac{1}{2} * a *t^{2}\\ X_{f} = 0m +17 \frac{m}{s} *1.6 s + \frac{1}{2} * 4.9 \frac{m}{s^{2} } *(1.6s)^{2}\\X_{f} = 27.2 m+ (-6.272 m)\\X_{f} = 20.928 m

The distance of the car is less than the distance of the kid in his bike

So she didn't hit him

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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.
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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)

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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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