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Alchen [17]
3 years ago
14

A family leaves home at 1:00pm and arrives at their vacation spot 200 miles away at 5:00pm on the same day. What is their averag

e speed?
Physics
1 answer:
Shkiper50 [21]3 years ago
4 0
The answer is 40 because you just have to do 200 divided by 5
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Calculate the weight of an apple of mass 100grams on Earth
11111nata11111 [884]

Answer:

In physics the standard unit of weight is Newton, and the standard unit of mass is the kilogram. On Earth, a 1 kg object weighs 9.8 N, so to find the weight of an object in N simply multiply the mass by 9.8 N. Or, to find the mass in kg, divide the weight by 9.8 N.

Explanation:

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8 0
2 years ago
A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the
Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

E=\dfrac{Q}{4\pi\epsilon_0 r^2}

Substitute numerical values:

E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

brainly.com/question/2004529

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6 0
2 years ago
Please help with this!!!!!
34kurt
(1.00 atm) (0.1156 L) = (n) (0.08206 L atm / mol K) (273 K)  I hoped that helped
4 0
3 years ago
In an application, Germanium is
Zigmanuir [339]

Answer:

produce electronics

Explanation:

The uses of Germanium are recorded beneath: Germanium's principle use is to deliver strong state hardware, semiconductors and fiber optic frameworks. As a phosphor in fluorescent lights.

7 0
3 years ago
A father fashions a swing for his children out of a long rope that he fastens to the limb of a tall tree. As one of the children
trasher [3.6K]

Answer:

The centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

                     

Explanation:

The centripetal acceleration is given by:

a_{c} = \frac{v^{2}}{r}

Where:

v^{2}: is the tangential speed = 9.50 m/s

r: is the distance = 6.00 m

Hence, the centripetal acceleration is:

a_{c} = \frac{v^{2}}{r} = \frac{(9.50 m/s)^{2}}{6.00 m} = 15.04 m/s^{2}

Therefore, the centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

I hope it helps you!

3 0
3 years ago
Read 2 more answers
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