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2 years ago

What is the time spent by the alpha particle in the magnetic field?

1 answer:

Ber [7]2 years ago

7 0

the alpha-particle goes around a quarter of the circle, so the time it takes would be t=0.25×2.61×10−6s=6.5×10−7s.

I hope this help

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Place a small object on the number line below at the position marked zero. Draw a circle around the object. Mark the center of t

Viefleur [7K]

It’s is D bc it is and idk you should do it

6 0

2 years ago

The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at posit

const2013 [10]

Answer:

fr = ½ m v₀²/x

Explanation:

This exercise the body must be on a ramp so that a component of the weight is counteracted by the friction force.

The best way to solve this exercise is to use the energy work theorem

W = ΔK

Where work is defined as the product of force by distance

W = fr x cos 180

The angle is because the friction force opposes the movement

Δk = $K_{f}$ –K₀

ΔK = 0 - ½ m v₀²

We substitute

- fr x = - ½ m v₀²

fr = ½ m v₀²/x

8 0

2 years ago

10. A girl pulls a wagon along a level path for a distance of 44 m. The handle of

Rina8888 [55]

The work done on the wagon is 3549 J

Explanation:

The work done by a force when moving an object is given by

$W=Fd cos \theta$

where :

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

In this problem we have the following data:

F = 87 N is the magnitude of the force

d = 44 m is the displacement of the wagon

$\theta=22^{\circ}$ is the angle between the direction of the force and the displacement

Substituting, we find the work done

$W=(87)(44)(cos 22^{\circ})=3549 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0

2 years ago

When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa

MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

$\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2$

In parallel

$\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}$

$\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0$

Solving the above quadratic equation

$\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}$

$\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega$

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0

2 years ago

5. Measure: With the lights on, click Pause. Turn on Show rulers. A. The wavelength of a longitudinal wave is equal to the dista

Marysya12 [62]

Explanation:

A) The distance between the two successive compressions (or rarefactions) is actually called the wavelength of the longitudinal waves.

B) Wavelengths of longitudinal and transverse waves are comparable in the fact that in a transverse wave, the particles move perpendicular to the direction the wave travels whereas in a longitudinal wave the particles are displaced along the direction to the direction the wave travels

6 0

2 years ago