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stiks02 [169]
4 years ago
8

Two golf balls are thrown with the same speed off the top of a large bridge at the same time. ball a is thrown straight downward

, and ball b is thrown horizontally off the bridge. which ball hits the ground first?
a. ball a
b. ball b
c. both balls hit at the same time.
d. we would need to know the speed.
Physics
1 answer:
iren2701 [21]4 years ago
8 0
I think that ball  a hit the ground because it says that it went straight down.
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The number of significant figures on the measurement 0.050010 kg id
Alex17521 [72]
Its been some time so i might be wrong but i think the answer is 3 either or 2
6 0
4 years ago
Explain the method to measure the external diameter of a sphere
Anton [14]

Answer:

The sphere is that the circular objects in the two-dimensional space (1) circle

(2) disk. Two-dimensional space is a set of points and the distance of that point, The two points of Sphere that length and center.

The sphere can be constructed as the name of surface form circle about any diameter. The circle is the special type of the revolution replacing the circle,

the sphere is the distance r is the radius of the ball and the circle is the center of the mathematical ball, as the center and the radius of the sphere is to respectively.

The ball and sphere have not to be maintained mathematical references as solid references. A sphere of any radius is centered at the number of zero.

Explanation: Hope this helps and good luck :)

3 0
3 years ago
Every Saros cycle (19 eclipse years): A. All total solar eclipses occur as total solar eclipses. B. All annular solar eclipses o
pickupchik [31]

Answer:

D. All lunar eclipses reoccur.

Explanation:

Every Saros cycle (19 eclipse years)

Saro was described by Edmond Halley arround the year 1691, which is a period of 223 synodic which can be interepreted as "repitition" it is used by the scientist for eclipse prediction either for the sun or moon eclipse. It is 18years and 11days and some hours.

It should be noted that with this All lunar eclipses reoccur.

4 0
3 years ago
A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed
Vaselesa [24]

Answer:

E = 1580594.95 N/C

Explanation:

To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:

\int EdS=\frac{Q_{in}}{\epsilon_o}   (1)

dS: differential of the Gaussian surface

Qin: charge inside the Gaussian surface

εo: dielectric permittivity of vacuum =  8.85 × 10-12 C2/N ∙ m2

The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:

\int EdS=ES=E(4\pi r^2)   (2)

Qin is calculate by using the charge density:

Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho  (3)

Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.

The charge density is given by:

\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}

Next, you use the results of (3), (2) and (1):

E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})

Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:

E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}

hence, the electric field is 1580594.95 N/C

7 0
3 years ago
Convert-73°c to kelvin scale​
pantera1 [17]

Answer:

200 K

Explanation:

0 °C = 273 K

-73°C = 273 K - 73 K = 200 K

5 0
3 years ago
Read 2 more answers
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