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mr_godi [17]
3 years ago
11

Help please, I need it

Physics
1 answer:
love history [14]3 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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The right answer for the question that is being asked and shown above is that: "B) allowing Carolina planters to expand rice cultivation into Georgia's lowlands."
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8 0
3 years ago
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At what speed does a 1,248 kg compact car have the same kinetic energy as a 18,777 kg truck going 26 km/h?
notka56 [123]

The speed of car is 100.8km/h

KE =  \frac{1}{2} m(v \: truck ) {}^{2}

= 0.5 \times 18777 \times  \frac{26}{3.6} \times  \frac{26}{3.6}

= 489708.79j

=  \frac{1}{2}  \times 1248 \times( v \: car) {}^{2}

= 624v {}^{2}

so \: v {}^{2}  =  \frac{489708.7}{624}

= 784

v =  \sqrt{784}

v = 28m/s

v car= 28×3.6

=100.8km/h

Hence, the speed of the car is 100.8km/h

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3 0
1 year ago
A ball with a mass of 5 kg is accelerating at 5 m/s/s. What is the force acting on the ball?
Genrish500 [490]
Force is 25 N
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6 0
3 years ago
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If a t-shirt gun can fire t-shirts with an initial speed of 15 m/s, what is the maximum distance (along horizontal, flat ground)
kotegsom [21]

Answer:

h = 11.47 m

Explanation:

Initial speed pf the t-shirt gun is 15 m/s

We need to find the maximum distance covered by the t-shirt. It is based on the conservation of energy. The maximum distance covered is given by :

h=\dfrac{u^2}{2g}\\\\h=\dfrac{(15)^2}{2\times 9.8}\\\\h=11.47\ m

So, it will cover a distance of 11.47 m.

7 0
3 years ago
A constant-volume perfect gas thermometer indicates a pressure of 6.69 kPa at the triple point temperature of water (273.16 K).
mina [271]

Answer:

(a) ΔP=0.0245 kPa

(b) P=9.14 kPa

(c)ΔP=0.0245 kPa

Explanation:

(a) As it is perfect gas we can use

(P₁V₁)/T₁=(P₂V₂)/T₂

Since this constant volume so

P₁/T₁=P₂/T₂

T₂ is change in temperature

T₂=1.00+273.16

T₂=274.16 K

P_{2}=(\frac{6.69}{273.16} )*274.16\\P_{2}=6.71449 kPa

ΔP=6.71449-6.69

ΔP=0.0245 kPa

(b) As

P_{2}=(\frac{6.69}{273.16} )*373.16\\P_{2}=9.14 kPa

(c) Same steps as in part (a)

P_{2}=(\frac{9.14}{373.16} )*374.16\\P_{2}=9.164kPa

ΔP=9.164-9.14

ΔP=0.0245kPa

8 0
3 years ago
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