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mr_godi [17]
4 years ago
11

Help please, I need it

Physics
1 answer:
love history [14]4 years ago
5 0

Answer:

1.97×10⁻²¹ J

Explanation:

Use ideal gas law to find temperature.

PV = nRT

(9 atm) (9 L) = (83.3 mol) (0.0821 L·atm/mol/K) T

T = 11.9 K

The average kinetic energy per atom is:

KE = 3/2 kT

KE = 3/2 (1.38×10⁻²³ J/K) (11.9 K)

KE = 2.46×10⁻²² J

For a mass of 5.34×10⁻²⁶ kg, the kinetic energy is:

KE = (5.34×10⁻²⁶ kg) (1 mol / 0.004 kg) (6.02×10²³ atom/mol) (2.46×10⁻²² J)

KE = 1.97×10⁻²¹ J

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An 7.9 kg crate is pulled 5.5 m up a 30∘ incline by a rope angled 19 ∘ above the incline. The tension in the rope is 140 N and t
dedylja [7]

Answer:

Wt= 0.73kJ, Wg=0.21kJ Wn= 0J

(B) Wf= 0.096kJ

Explanation:

See attachment below.

5 0
4 years ago
A movie stunt performer is filming a scene where he swings across a river on a vine. The safety crew must use a vine with enough
Julli [10]

Answer:

1125.66956 N

Explanation:

m = Mass of stunt performer

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity of the swing = 7 m/s

T = Tension

r = Radius of the swing = Length of vine = 11.5 m

From the free body diagram

T-mg-m\frac{v^2}{r}=0\\\Rightarrow T=mg+m\frac{v^2}{r}\\\Rightarrow T=m(g+\frac{v^2}{r})\\\Rightarrow T=80(9.81+\frac{7^2}{11.5})\\\Rightarrow T=1125.66956\ N

The minimum tension force the vine must be able to support without breaking is 1125.66956 N

7 0
3 years ago
25% part (c) assume that d is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression
levacccp [35]

maximum speed of cheetah is

v_1 = v_{max}

speed of gazelle is given as

v_2 = v_{g}

Now the relative speed of Cheetah with respect to Gazelle

v_{12} = v_1 - v_2

v_{12} = v_{max} - v_g

now the relative distance between Cheetah and Gazelle is given initially as "d"

now the time taken by Cheetah to catch the Gazelle is given as

d = v_{12}* t

so by rearranging the terms we can say

t = \frac{d}{v_{12}}

t = \frac{d}{v_{max} - v_g}

so above is the relation between all given variable

6 0
4 years ago
The speed of sound is 346m/s if a sound wave travels at a frequency of 55hz what would it’s wavelength be
algol [13]

Answer:

6.29 m

Explanation:

The speed of a wave is equal to frequency times wavelength, so to find wavelength you'd have to divide frequency by both sides, speed(346) divided frequency(55) = 6.29 m

I'm not so sure about this let me know if I'm wrong

3 0
3 years ago
Electric field lines moves away from positive to wards negative?
Archy [21]

The vector force on the unit positive charge placed at any location in the field defines the strength of the electric field at that point. The charge used to determine field intensity (field strength) is known as the test charge. Now, a field line is defined as a line to which the previously mentioned field strength vectors are tangents at the relevant places. When we study positive charge field lines, the field strength vectors point away from the positive charge. If there is a negative charge anywhere in the vicinity, the field lines that began from the positive charge will all terminate at the negative charge if the value of the negative charge is the same as the value of the positive charge. Remember that the number of field lines originating from positive charge is proportional to the charge's value, and similarly, the number of field lines terminating at negative charge is proportionate to the charge's value. As a result, if all charges are equivalent, all lines originating from the positive charge terminate at the negative charge. If the value of the positive charge is greater than the value of the negative charge, the number of lines ending at the negative charge will be proportionally less than the number of lines beginning at the positive charge. The remaining lines that do not end at the negative charge will go to infinity. If the positive charge is less, all lines from it terminate at a negative charge, and any other reasonable number of ines terminate at a negative charge from infinity. We should also keep in mind that the number of lines that run perpendicular to the field direction across a surface of unit area is proportional to the field strength at that location. As a result, lines are dense in the strong field zone and sparse in the low intensity region.

6 0
2 years ago
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