All categories

1 year ago

A stone is projected horizontally from the

Physics

1 answer:

Nadya [2.5K]1 year ago

6 0

Answer:

80 m tall

Explanation:

*** EDITED *** (I left out the '1/2' in the equation....D'Oh!)

It was descending 4 seconds ...because it landed 20 m away (at 5 m/s ...so 4 seconds)

displacement = 1/2 a t^2

= 1/2 * 10 ( 4)^2 = 80 m tall

You might be interested in

The

andrey2020 [161]

Answer:

500 in unitivector notion

6 0

2 years ago

The land between two normal faults moves upward to form a

Leya [2.2K]

The land between two normal faults moves upward to form a

Answer:D
fault-block mountain.

5 0

3 years ago

Read 2 more answers

Please , physics. A and B

stepladder [879]

Answer:

a. -5,4

b. 28

Explanation:

(see how to solve in pic)

7 0

2 years ago

After the double-blind review policy was instituted, what percentage of published papers had a male first author?

Arte-miy333 [17]

Answer:

hesadghtyou is you here boiii

Explanation:

4 0

2 years ago

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated

Lelu [443]

Incomplete question as we have not told to find what quantity.The complete question is here

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 5.00 cm.calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

Answer:

(a) $C=16.7pF$

(b) $r_{a} =3.749cm$

Explanation:

Given data

$Q=3.50nC\\V=210V\\r_{b}=5.0cm$

For part (a)

The Capacitance given by:

$C=\frac{Q}{V}\\ C=\frac{3.50*10^{-9} C}{210V}\\C=1.6666*10^{-11}F\\or\\C=16.7pF$

For part (b)

The Capacitance of coordinates is given as

$C=\frac{4\pi e}{\frac{1}{r_{a} }-\frac{1}{r_{b} } }\\ So\\{\frac{1}{r_{a} }-\frac{1}{r_{b} } }=\frac{4\pi *8.85*10^{-12} }{1.666*10^{-11}}=6.672m^{-1} \\ \frac{1}{r_{a} }=6.672+(1 /0.05)\\\frac{1}{r_{a} }=26.672\\r_{a} =1/26.672\\r_{a} =0.0375m\\r_{a} =3.749cm$

For part (c)

The electric field according to Gauss Law is given by:

$EA=\frac{Q}{e}\\ E=\frac{Q}{4\pi er_{a}^{2} }=\frac{kQ}{r_{a}^{2}}\\ E=\frac{9*10^{9}*3.50*10^{-9} }{(0.0375m)^{2} }\\ E=2.24*10^{4} N/C$

7 0

3 years ago