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zimovet [89]
3 years ago
6

An electron is accelerated from rest through a potential difference. After acceleration the electron has a de Broglie wavelength

of 880 nm. What is the potential difference though which this electron was accelerated
Physics
1 answer:
Vladimir [108]3 years ago
4 0

Answer:

3x10⁴v

Explanation:

Using

Wavelength= h/ √(2m.Ke)

880nm = 6.6E-34/√ 2.9.1E-31 x me

Ke= 6.6E-34/880nm x 18.2E -31.

5.6E-27/18.2E-31

= 3 x 10⁴ Volts

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A pickup truck is traveling down the highway at a steady speed of 30.1 m/s. The truck has a drag coefficient of 0.45 and a cross
Sav [38]

Answer:

The energy that the truck lose to air resistance per hour is 87.47MJ

Explanation:

To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by

F_D=\frac{1}{2}\rhoC_dAV^2

Our values are:

V=30.1m/s

C_d=0.45

A=3.3m^2

\rho=1.2kg/m^3

Replacing,

F_D=\frac{1}{2}(1.2)(0.45)(3.3)(30.1)^2

F_D=807.25N

We need calculate now the energy lost through a time T, then,

W = F_D d

But we know that d is equal to

d=vt

Where

v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,

W=(807.25N)(30.1m/s)(3600s/1hr)

W=87.47*10^6J (per hour)

Therefore the energy that the truck lose to air resistance per hour is 87.47MJ

4 0
3 years ago
An object moving at 38m/s takes 4s to come to a stop. What is the object’s acceleration?
Rudik [331]
- 9.5 m/s^2

use the SUVAT method 

0 = 38 + 4a
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Explanation:

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7 0
2 years ago
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To practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.7 kg and length 3.0 m
Zanzabum

Answer:

T_{1} = 14.88 N

Explanation:

Let's begin by listing out the given variables:

M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,

g = 9.8 m/s²

At equilibrium, the sum of all external torque acting on an object equals zero

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Taking moment about T_{1} we have:

(M + m) g * 0.5L - T_{2}(L - d) = 0

⇒ T_{2} = [(M + m) g * 0.5L] ÷ (L - d)

T_{2} = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)

T_{2}= 59.535 ÷ 2.4

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Using sum of equilibrium in the vertical direction, we have:

T_{1} + T_{2} = W + w   ------- Eqn 1

Substituting T2, W & w into the Eqn 1

T_{1} + 24.81 = 26.46 + 13.23

T_{1} = <u>14.88</u> N

8 0
3 years ago
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