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Vaselesa [24]
3 years ago
8

1. The term that describes where the supply curve intersects the demand curve is known as

Physics
1 answer:
nadezda [96]3 years ago
6 0
Equilibrium is the answer
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What is the squirrels escape velocity in mph if the squirrel accelerates at a constant 1.5 m/s squared from rest for 2.5s
Zina [86]

Answer:

3.75 m/s

Explanation:

From the question given above, the following data were obtained:

Acceleration (a) = 1.5 m/s²

Initial velocity (u) = 0 m/s

Time (t) = 2.5 s

Final velocity (v) =?

a = (v – u) / t

1.5 = (v – 0) / 2.5

1.5 = v / 2.5

Cross multiply

v = 1.5 × 2.5

v = 3.75 m/s

Hence, the escape velocity of the squirrel is 3.75 m/s

7 0
3 years ago
(15 POINTS) Victor drew a diagram to show the life cycle of a low-mass star.
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The answer is c so the area can grow
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You are a surgeon operating on a broken bone. You find a large area, swollen with blood, that surrounds the break site. What is
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In which of these situations is convection most likely the main form of heat transfer? A.Warm air from a heater on the first flo
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The answer is D.An ice cube melts when a person holds it in his hand 
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3 years ago
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A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha
USPshnik [31]

Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

\Phi_E=\frac{Q}{\epsilon_o}

ФE: electric flux trough the Gaussian surface

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C

Finally, you obtain for E:

E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

3 0
3 years ago
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