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astra-53 [7]
3 years ago
9

A velocity selector in a mass spectrometer uses a 0.150 T magnetic field. (a) What electric field strength (in volts per meter)

is needed to select a speed of 4.40 ✕ 106 m/s? V/m
Physics
1 answer:
Alekssandra [29.7K]3 years ago
4 0

Answer:

The electric field strength is 6.6\times10^{5}\ V/m

Explanation:

Given that,

Magnetic field = 0.150 T

Speed v= 4.40\times10^{6}\ m/s

We need to calculate the electric field strength

Using formula of velocity

v=\dfrac{E}{B}

E=v\times B

Where, v = speed

B = magnetic field

Put the value into the formula

E=4.40\times10^{6}\times0.150

E=660000\ V/m

E=6.6\times10^{5}\ V/m

Hence, The electric field strength is 6.6\times10^{5}\ V/m

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a ball is thrown straight up into the air with a speed of 13 m/s. if the ball has a mass of 0.25 kg, how high does the ball go?
evablogger [386]
<h2>Hello!</h2>

The answer is: 8.62m

<h2>Why?</h2>

There are involved two types of mechanical energy: kinetic energy and potential energy, in two different moments.

<h2>First moment:</h2>

Before the ball is thrown, where the potential energy is 0.

<h2>Second moment: </h2>

After the ball is thrown, at its maximum height, the Kinetic Energy turns to 0 (since at maximum height,the speed is equal to 0) and the PE turns to its max value.

Therefore,

E=PE+KE

Where:

PE=m.g.h

KE=\frac{1*m*v^{2}}{2}

<em>E</em> is the total energy

<em>PE</em> is the potential energy

<em>KE</em> is the kinetic energy

<em>m</em> is the mass of the object

<em>g</em> is the gravitational acceleration

<em>h </em>is the reached height of the object

<em>v</em> is the velocity of the object

Since the total energy is always constant, according to the Law of Conservation of Energy, we can write the following equation:

KE_{1}+PE_{1}=KE_{2}+PE_{2}

Remember, at the first moment the PE is equal to 0 since there is not height, and at the second moment, the KE is equal to 0 since the velocity at maximum height is 0.

\frac{1*m*v^{2}}{2}+m.g.(0)=\frac{1*m*0^{2}}{2}+m.g.h\\\frac{1*m*v_{1} ^{2}}{2}=m*g*h_{2}

So,

h_{2}=\frac{1*m*v_{1} ^{2}}{2*m*g}\\h_{2}=\frac{1*v_{1} ^{2}}{2g}=\frac{(\frac{13m}{s})^{2} }{2*\frac{9.8m}{s^{2}}}\\h_{2}=8.62m}

Hence,

The height at the second moment (maximum height) is 8.62m

Have a nice day!

5 0
4 years ago
A uniform solid disk and a uniform ring are place side by side at the top of a rough incline of height h.
nadezda [96]

Explanation:

velocity of disc =\sqrt((gh)/0.75)

lets call (h) 1 m to make it simple.

= 3.614 m/s

\sqrt((4/3) x 1 x 9.8) = 3.614 m/s pointing towards this:

4×V_d=\sqrt(4/3hg)

V_h=\sqrt(hg)

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lets call (h) 1m to make it simple again.

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The disc is the fastest.

While i'm on this subject i'll show you this:

Solid ball =0.7v^2= gh

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The above is simplified from linear KE + rotational KE, the radius or mass makes no difference to the above formula.

The solid ball will be the faster of the 3, like above i'll show you.

solid ball: velocity =\sqrt((gh)/0.7)

let (h) be 1m again to compare.

\sqrt((9.8 x 1)/0.7) = 3.741 m/s

solid disk speed =\sqrt((gh)/0.75)

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solid sphere speed =\sqrt((gh)/0.7)

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