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3 years ago

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A car traveling west at 70 miles per hour for 3 hours covers a distance of

1 answer:

gtnhenbr [62]3 years ago

5 0

70x3=210 mph... its that easy

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A velocity selector is built with a magnetic field of magnitude 5.2 T and an electric field of strength 4.6 × 10 ^4 N/C. The dir

MAXImum [283]

<h2>Answer</h2>

Option A that is 8.8 × 10^3 m/s

<h2>Explanation</h2>

The magnetic field B is defined from the Lorentz Force Law, and specifically from the magnetic force on a moving charge. It says

Field-strength = BVqsinΔ

<h2>v = E/B </h2>

Since field are perpendicular so sin90 = 1

v = 4.6/10^4 / 5.2

v = 8846.15 m /s

The speed at which electrons pass through the selector without deflection = 8846.15 m /s

4 0

3 years ago

Read 2 more answers

Explain the difference between Longitudinal and Transverse waves.

masya89 [10]

The answer is C sorry if it’s wrong

4 0

3 years ago

For your final exam in electronics, you’re asked to build an LC circuit that oscillates at 10 kHz. In addition, the maximum curr

Marina CMI [18]

Answer:

L= 2 mH

$C=1.26\times 10^{-7}\ F$

Explanation:

Given that

Frequency , f= 10 kHz

Maximum current ,I = 0.1 A

Maximum energy stored ,E= 1 x 10⁻⁵ J

The maximum energy stored in the inductor is given as follows

$E=\dfrac{1}{2}LI^2$

Where ,L= Inductance

I=Current

E=Energy

Now by putting the values in the above equation

$10^{-5}=\dfrac{1}{2}\times L\times 0.1^2$

$L=\dfrac{2\times 10^{-5}}{0.1^2}\ H$

L=0.002 H

L= 2 mH

We know that frequency f is given as

$2\pi f=\dfrac{1}{\sqrt{LC}}$

C=Capacitance , f=frequency ,L=Inductance

Now by putting the values

$2\pi \times 10\times 10^3=\dfrac{1}{\sqrt{0.002\times C} }$

$62831.85=\dfrac{1}{\sqrt{0.002\times C}}$

$\sqrt{0.002\times C=\dfrac{1}{62831.85}$

$0.002\times C=0.0000159^2$

$C=\dfrac{0.0000159^2}{0.002}\ F$

$C=1.26\times 10^{-7}\ F$

Therefore the inductance and capacitance will be 2 mH and 1.26 x 10⁻⁷ F respectively.

6 0

3 years ago

How does electricity affect objects that are not in contact with one another?

Elan Coil [88]

Electrostatic forces are non-contact forces; they pull or push on objects without touching them

5 0

2 years ago

Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

6 0

3 years ago