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3 years ago

As you brake your bicycle, your velocity changes from 20 m/s east to 10 m/s east in 5 seconds. What’s your acceleration?

Physics

1 answer:

charle [14.2K]3 years ago

8 0

Hello!

$\large\boxed{-2m/s^{2} }$

Find the acceleration using the formula a = (vf - vi) / t where:

vf = final velocity

vi = initial velocity

t = time

Plug in the points above:

vf = +10 m/s

vi = +20 m/s

t = 5 sec

(10 - 20) / 5 = -10 / 5 = -2 m/s²

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An electric heater is rated 1000watts. If a current of 5A pass through the heater, find the value of its resistance.

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Answer: R=40 Ω

Explanation: $P=I^{2} R$

R=1000/ (25)=40

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3 years ago

A plane accelerates from 25 m/s to a velocity of 80 m/s in a time of 5 seconds. What

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3 years ago

What kind of thermal energy that flows between objects due to a difference in temperature?

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3 years ago

A proton moving in the positive x direction with a speed of 9.9 105 m/s experiences zero magnetic force. When it moves in the po

Alex

Answer:

The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

Explanation:

In order to calculate the magnitude and direction of the magnetic field, you take into account the following equation for the magnetic force on the proton:

$\vec{F_B}=q\vec{v}\ X\ \vec{B}$ (1)

v: speed of the proton = 9.9*10^5 m/s

q: charge of the proton = 1.6*10^-19C

B: magnetic field = ?

FB: magnetic force on the proton = 1.6*10^-13N

When the proton travels in the positive y direction (^j), you have that the proton experiences a force in the positive z direction (+^k). To obtain this direction of the magnetic force on the proton, it is necessary that the magnetic field points in the negative x direction, in fact, you have:

^j X (-^i) = -(-^k)=^k

To obtain the magnitude of the magnetic field you use:

$F_B=qvBsin90\°=qvB\\\\B=\frac{F_B}{qv}=\frac{1.6*10^{-13}N}{(1.6*10^{-19}C)(9.9*10^5m/s)}\\\\B=1.01T$

The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

8 0

3 years ago

A proton moves through a magnetic field at 26.7 % 26.7% of the speed of light. At a location where the field has a magnitude of

iogann1982 [59]

Answer:

$8.64283\times 10^{-14}\ N$

Explanation:

q = Charge of proton = $1.6\times 10^{-19}\ C$

v = Velocity of proton = $0.267\times c$

c = Speed of light = $3\times 10^8\ m/s$

B = Magnetic field = 0.00687 T

$\theta$ = Angle = $101^{\circ}$

Magnetic force is given by

$F=qvBsin\theta\\\Rightarrow F=1.6\times 10^{-19}\times (0.267\times 3\times 10^8)\times 0.00687\times sin101\\\Rightarrow F=8.64283\times 10^{-14}\ N$

The magnetic force acting on the proton is $8.64283\times 10^{-14}\ N$

4 0

3 years ago