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3 years ago

21. An object is moving at 5 m/s and a height of 4 m. If the object has a mass of 5 kg what is

Physics

1 answer:

Stels [109]3 years ago

6 0

Answer: 258.7 J

Explanation:

Velocity of object V=5m/s

Mass ; 5kg

Height; h=4m

Mechanical energy is equal to sum of kinetic and potential energy

E=K+Ep

Kinetic energy is calculeted using; K=m*V²/2

Potential energy is calculated using : Ep=m*G*h

-------------------------------------------------------------------------

V=5m/s

m=5kg

h=4m

E=K+Ep

E=m*V²/2+m*G*h

E=5kg*(5m/s)²/2+5kg*9.81m/s²*4m

E=62.5J+196.2J

E=258.7 J

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3 years ago

what velocity must a 1340kg car have in order to havw the same momentum as a 2680 kg truck traveling at a velocity of 15m/s to t

kykrilka [37]

Car with a mass of 1210 kg moving at a velocity of 51 m/s.
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3 years ago

What is the longest wavelength of light that will emit electrons from a metal whose work function is 3.70 eV

SSSSS [86.1K]

Answer:

h f = Wf + K

where the total energy available is h f, Wf is the work function or the work needed to remove the electron and K is the kinetic energy of the removed electron

If K = zero then hf = Wf

Wf = h f = h c / λ or

λ = h c / Wf = 6.63E-34 * 3.0E8 / (3.7 * 1.6E-19)

λ = 6.63 * 3 / (3.7 * 1.6) E-7 = 3.36E-7

This would be 3360 angstroms or 336 millimicrons

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1 year ago

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

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3 years ago

Calculate the change in total internal energy for a system that releases 2.59 × 104 kJ of heat and does 6.46 × 104 kJ of work on

mrs_skeptik [129]

Answer:

See explanation below

Explanation:

The equation to use for this is the following:

dU = q + w

As the heat is being release, this value is negative, and same here happens with the work done, because it's in the surroundings.

Therefore the change in the energy would be:

dU = -2.59x10^4 - 6.46^4

dU = -9.05x10^4 kJ

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3 years ago