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3 years ago

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An object is thrown vertically upward such that it has a speed of 25 m/s when it reaches two thirds of its maximum height above

the launch point. Determine this maximum height.
A) 96m
B) 32m
C) 48 m
D) 75m
E) 64 m

1 answer:

AlekseyPX3 years ago

4 0

Answer:

v0 = 25 m/s

vf = 0 m/s

a = -9.80 m/s^2

change in x = 31.89m

but that's only 1/3 of the hight, so i time it by 3 to get 96m

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If the current in a wire increases from 5 A to 10 A, what happens to its magnetic field? If the distance of a charged particle f

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1. The magnitude of the magnetic field doubles

Explanation: the intensity of the magnetic field produced by a current-carrying wire is given by:

$I=\frac{\mu_0 I}{2 \pi r}$

where $\mu_0$ is the vacuum permeability, I is the current in the wire, r is the distance from the wire.

As we see from the formula, the intensity of the magnetic field is directly proportional to the current: if the current increases from 5 A to 10 A, it means it doubles, so the magnetic field doubles as well.

2. The magnitude of the magnetic field halves

Explanation: the intensity of the magnetic field produced by a current-carrying wire is given by:

$I=\frac{\mu_0 I}{2 \pi r}$

We see that the magnitude of the magnetic field is inversely proportional to the distance from the wire (r). In this case, the distance of the particle is changed from 10 cm to 20 cm, so it is doubled: therefore, the magnitude of the field will become half of the initial value.

3. The force reverses direction

Explanation: the force exerted on a charged particle in a magnetic field is:

$F=qvB sin \theta$

where q is the charge, v is the speed of the particle, B is the magnetic field intensity and $\theta$ the angle between the direction of v and B. If the charge of the particle is switched from 2 µC to –2µC, the magnitude of the force does not change (because the absolute value of q does not change), however the charge q gets a negative sign (-), so the sign of the force changes and gets a negative sign too, so the force reverses direction.

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3 years ago

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As your skateboard coasts uphill, your speed changes from 3 m/s to 1 m/s in

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Answer:

$a=-0.33\ m/s^2$

Explanation:

<u>Accelerated Motion</u>

The acceleration of a moving body is defined as the relation of change of speed (or velocity in vector form) with the time taken. The formula is given by

$\displaystyle a=\frac{\Delta v}{t}$

Or, equivalently

$\displaystyle a=\frac{v_f-v_o}{t}$

Where vf and vo are the final and initial speeds respectively. The problem gives us these values: v0 = 3 m/s, vf = 1 m/s, t = 3 seconds. Computing a

$\displaystyle a=\frac{1-3}{3}=-0.33\ m/s^2$

The negative sing of a indicates there is deceleration or decreasing speed

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3 years ago

A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a

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Answer:

a) x = v₀² sin 2θ / g

b) t_total = 2 v₀ sin θ / g

c) x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

sin θ = $v_{oy}$ / vo

cos θ = v₀ₓ / vo

v_{oy} = v_{o} sin θ

v₀ₓ = v₀ cos θ

v_{oy} = 13.5 sin 32 = 7.15 m / s

v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

v₀ₓ = x / t

x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

v_{y} = v_{oy} - gt

0 = v₀ sin θ - gt

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we substitute

x = v₀ cos θ (2 v_{o} sin θ / g)

x = v₀² /g 2 cos θ sin θ

x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

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v_{y} = v_{oy} - gt

v_{y} = 0

t = v_{oy} / g

t = v₀ sin θ / g

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t_total = 2 t

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x = 13.5 2 sin (2 32) / 9.8

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False as oxygen is the second most abundant and nitrogen is the most abundant at 78%.

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