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3 years ago

5. A projectile is fired in Earth's gravitational field with a horizontal velocity of y = 9.00 m/s.

Physics

1 answer:

aniked [119]3 years ago

3 0

Answer:

Explanation:

Given

Horizontal velocity = 9.00m/s

time taken = 0.550 s

Required

How far does the projectile fall in the vertical direction

Using the formula for finding the maximum height of the projectile

H = U²sin²θ/2g where;

U = 9.00m/s

θ = 90° (object launched in the vertical direction)

g = 9.81m/s²

Substituting the given parameters into the formula;

H = 9²sin²90/2(9.81)

H = 81(1)/19.62

H = 81/19.62

H = 4.128 m

H ≈ 4.13m

Hence the distance that the projectile fall in the vertical direction is 4.13m

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A father racing his son has 1/3 the kinetic energy of the son, who has 1/4 the mass of the father. The father speeds up by 1.5 m

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Explanation:

Let the speeds of father and son are $v_f\ and\ v_s$ . The kinetic energies of father and son are $K_f\ and\ K_s$ . The mass of father and son are $m_f\ and\ m_s$

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Kinetic energy of son is given by :

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From equation (1), (2) we get :

$\dfrac{v_f^2}{v_s^2}=\dfrac{1}{12}$ ..............(3)

If the speed of father is speed up by 1.5 m/s, so the ratio of kinetic energies is given by :

$\dfrac{K_f}{K_s}=\dfrac{1/2m_f(v_f+1.5)^2}{1/2m_sv_s^2}$

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3 years ago

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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)

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:

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

'θ' = 90°.

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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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