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vampirchik [111]
3 years ago
11

A proton has been accelerated from rest through a potential difference of -1350 V. What is the proton's kinetic energy, in elect

ron volts? What is the proton's kinetic energy, in joules? What is the proton's speed?
Physics
1 answer:
Luba_88 [7]3 years ago
8 0

Answer:

1 eV = 1.60 * 10^-19 J      work done in accelerating electron thru 1 V

KE (total energy) = 1350 ^ 1 eV     (note proton goes from +  to -)

KE = 1.60 * 10^-19 * 1350 = 2.16 * 10^-16 Joules

1/2 m v^2 = KE = 2.16 * 10^-16 J

v^2 = 4.32 * 10E-16 / 1.67 * 10-27 = 2.59 * 10^11

v = 5.09 * 10^5 m/s

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A dog runs with an initial velocity of 7.5 m/s on a waxed floor. It slides to a stop in 15 seconds. What is the acceleration?
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-0.5 m/s^2

Explanation:

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3 years ago
I need help please I will give points
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Hope it helps!

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3 years ago
The emissivity of an ideal reflector has which of the following values?
Wittaler [7]

Emissivityis a measure of how much thermal radiation a body emits to its environment. On the other hand we have that reflectivity is a measure of how much is reflected, and transmissivity  is a measure of how much passes through the object. If a body is required to be ideally reflective to its maximum efficiency, the body should NOT have the property of transmissivity or emissivity. Therefore it should be 0 its emittivity.

Correct answer would be A : ZERO.

4 0
3 years ago
Steam enters a counterflow heat exchanger operating at steady state at 0.07 MPa with a specific enthalpy of 2431.6 kJ/kg and exi
PIT_PIT [208]

Answer:

Explanation:

Given:

Steam Mass rate, ms = 1.5 kg/min

= 1.5 kg/min × 1 min/60 sec

= 0.025 kg/s

Air Mass rate, ma = 100 kg/min

= 100 kg/min × 1 min/60 sec

= 1.67 kg/s

A.

Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.

xf, quality = 0.9.

Tsat = 89.9°C

hf = 376.57 kJ/kg

hfg = 2283.38 kJ/kg

Using the equation for specific enthalpy,

hi = hf + (hfg × xf)

= 376.57 + (2283.38 × 0.9)

= 2431.552 kJ/kg

The specific enthalpy of the outlet, h2 = hf

= 376.57 kJ/kg

B.

Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy

= ms × (hi - h2)

= 0.025 × (2431.552 - 376.57)

= 0.025 × 2055.042

= 51.37455 kW

= 51.38 kW.

5 0
3 years ago
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