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3 years ago

9

An electron that is moving to the right experiences a magnetic field of 2.5 T directed upward. If the force on the electron is 2

.4 x 10^-12N, what is the speed of the electron?

1 answer:

Zinaida [17]3 years ago

5 0

Answer:

5.99×10⁶ m/s

Explanation:

We look for the speed of the electron by applying,

F = Bqv.................. Equation 1

Where F = Force on the electron, B = magnetic field, q = charge of an electron, v = speed of the electron.

make v the subject of the equation

v = F/Bq.............. Equation 2

Given: F = 2.4×10⁻¹² N, B = 2.5 T.

Constant: q = 1.602×10⁻¹⁹ C

Substitute into equation 2

v = 2.4×10⁻¹² /(2.5× 1.602×10⁻¹⁹ )

v = 5.99×10⁶ m/s

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3 years ago

The concrete post (Ec = 3.6 × 106 psi and αc = 5.5 × 10-6/°F) is reinforced with six steel bars, each of 78-in. diameter (Es = 2

masha68 [24]

Answer:

the normal stress induced by the concrete post $\sigma_c = 67.26 psi$

the normal stress induced by the steel $\sigma_s = - 1795.84 psi$

Explanation:

Given that:

Modulus for elasticity for concrete post $E_c$ = $3.6 *10^6$ psi

Thermal coefficient for concrete post $\alpha _c$ = $5.5 *10^{-6}/^0F$

Modulus for elasticity of steel bar $E_s$ = $29*10^6psi$

Thermal coefficient of steel bar $\alpha _2$ = $6.5*10^{-6}/^0F$

Change in temperature ΔT = 80°F

Diameter of the steel rood = 7/8-in

Area of the steel rod $A_s$ = $6(\frac{ \pi}{4} )(d_s)^2$

= $6(\frac{ \pi}{4} )(\frac{7}{8} )^2$

= 3.61 in²

Area of concrete parts $A_c$ = (10)(10) - $A_s$

= (100 - 3.61) in²

= 96.39 in²

The total strain developed in the concrete post can be expressed as:

= $[\frac{1}{E_cA_c}+\frac{1}{E_sA_s}]P=(\alpha_s-\alpha_c)(\delta T)$

= $[\frac{1}{(3.6*10^6)(96.39)}+\frac{1}{(29*10^6)(3.61)}]P=(6.5*10^{-6}-5.5*10^{-6})(80)$

= $[(2.88*10^{-9}) +(9.55*10^{-9}]P = 8.0*10^{-5}$

= $1.234*10^{-8}P = 8.0*10^{-5}$

$P = \frac {8.0*10^{-5}}{1.234*10^{-8}}$

P = 6482.98 lb

Since, the normal stress in concrete is induced as a result of temperature rise; we have the expression :

$\sigma_c =\frac{P}{A_c}$

$\sigma_c = \frac{6482.98}{96.39}$

$\sigma_c = 67.26 psi$

Thus, the normal stress induced by the concrete post $\sigma_c = 67.26 psi$

Also; the normal stress in the steel bars induced as a result of temperature rise is as follows:

$\sigma_s = \frac{-P}{A_s}$

$\sigma_s =\frac{-6482.98}{3.61}$

$\sigma_s = - 1795.84 psi$

Thus, the normal stress induced by the steel $\sigma_s = - 1795.84 psi$

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4 years ago

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jek_recluse [69]

Coupla things wrong with this question, Sam.
Let's clean those up first, and then we'll work on the answer.

-- The car is NOT moving with uniform velocity.
'Velocity' includes both speed and direction. If either of these
changes, it's a change of velocity.
On a circular track, the car's direction is CONSTANTLY changing,
so its velocity is too.
The thing that's uniform is its speed, not its velocity.

-- A 'neutron' is a subatomic particle found in the nucleus of most
atoms. It's not a unit of force. The unit of force is the 'Newton'.
_______________________

OK. A centripetal force of 6,000 newtons keeps 1,200 kg of mass
moving in a circle at 20 m/s.

The formula:

   Centripetal force = (mass) (speed)² / (radius)

Multiply each side
by 'radius': (centripetal force) x (radius) = (mass) x (speed)²

Divide each side by
'centripetal force':   Radius = (mass) x (speed)² / (centripetal force)

Write in the numbers
that we know: Radius = (1200 kg) (20 m/s)² / (6000 Newtons)

   = (1200 kg) (400 m²/s²) / (6000 Newtons)

   = (480,000 kg-m²/s²) / (6000 kg-m/s²)

   = (480,000 / 6000) meters

   = 80 meters .

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3 years ago

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