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Pavel [41]
3 years ago
6

What school did Ronald McNair go to and what kind of science did he work in

Physics
1 answer:
alina1380 [7]3 years ago
5 0

Answer:

McNair graduated as valedictorian of Carver High School in 1967. In 1971, he received a Bachelor of Science degree in engineering physics, magna cu.m laude, from the North Carolina Agricultural and Technical State University in Greensboro, North Carolina.

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Please, can somebody help me with this project? I'll give brainlest for the best answer! (Do not answer if you don't know or onl
Sauron [17]

The density of the nickel was greater than that of the quarter and penny, thus, the results supports the hypothesis.

<h3>What is density of substance?</h3>

The density of a substance is a measure of how tightly-packed the particles of the substance are.

Density is calculated as the ratio of the mass of the substance and the volume of the substance.

  • Density = mass/volume

The hypothesis of the lab to compare the densities of a penny, a nickel, and a quarter is:

  • If the nickel has a greater density than the quarter and penny, then it will have a greater mass to volume ratio. If the nickel has a lower density than the quarter and penny, then it will have a lower mass-to-volume ratio.

The average mass and the average volume of a penny, a nickel, and a quarter are then used to determine the density of each coin.

Based on obtained results, it would be found that the density of the nickel was greater than that of the quarter and penny. Therefore, the results supports the hypothesis.

In conclusion, the density of a substance depends on the mass and the volume.

Learn more about density at: brainly.com/question/1354972

#SPJ1

7 0
2 years ago
A 1500 kg car traveling due east at 20<br> m/s slows to a stop in 5.0 seconds. What is the impulse?
lbvjy [14]

Answer:

Hans-Georg Gadamer (1900-2002) was an influential German philosopher of the twentieth century, inspiring a variety of scholastic disciplines from aesthetics to theology. In suggesting understanding was interpretation and vice versa, Gadamer identifies language acting as the medium for understanding. Gadamer’s philosophy of hermeneutics has major implications for education and formal schooling because Hermeneutics help to know the knowledge a student has prior to the lesson. This helps in the dialogue about a subject matter and therefore, the philosophy of Hermeneutics when applied in classroom helps the teachers pass information easily and effectively, hence, the learners capture the whole content of a topic.

Explanation:

5 0
2 years ago
Read 2 more answers
A charged particle is located in an electric field where the magnitude of the electric field strength is 2.0x10^3 newtons per co
Alexxandr [17]

Answer:

20.7

Explanation:

:0 because basis  of the daily occured

6 0
3 years ago
The average distance an electron travels between collisions is 2.0 μmμm . What acceleration must an electron have to gain 2.0×10
Ilya [14]

The solution is in the attachment

4 0
3 years ago
Read 2 more answers
A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
C=4 \pi \epsilon _0  \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=
=3.67\cdot 10^{-11}F

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
E\cdot (4 \pi r^2) =  \frac{Q}{\epsilon _0}
from which
E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
E(0.111m)=2680 N/C

And so, the energy density at r=0.111 m is
U= \frac{1}{2} \epsilon _0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C

And therefore, the energy density at this distance from the center is
U= \frac{1}{2}\epsilon_0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3
8 0
3 years ago
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