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Helga [31]
3 years ago
9

A 2 kg block slides on a rough horizontal surface with muk=0.6. It has an initial velocity of 5 m/s. Use g = 10 m/s2

Physics
1 answer:
Irina18 [472]3 years ago
4 0

Answer:

360000

Explanation:

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Can anyone please tell me the properties of image formed in a spherical mirror?
notsponge [240]

Answer:

The two rays, CY and DM are diverging rays and when extended behind the mirror, they appear to intersect each other at point M'. Therefore, the properties of the images formed here are formed behind the mirror, between the pole and principal focus (f), the images are diminished and are virtual and erect.

Explanation:

<h2>Spherical Mirrors</h2>

  • There are two kinds of spherical mirrors, concave and convex.

  • The focal point (F) of a concave mirror is the point at which a parallel beam of light is "focussed" after reflection in the mirror. ...

  • The focal length (f) and radius of curvature (R) are defined in the diagram at the right.

<h3>hope it helps and thanks for following </h3><h2>please give brainliest </h2>
6 0
3 years ago
Read 2 more answers
At one point in space, the electric potential energy of a 15 nC charge is 42 μJ . Part A) What is the electric potential at this
Anastasy [175]

Answer:

Part A:

\rm 2.8\times 10^3\ Volts.

Part B:

\rm 5.6\times 10^{-5}\ J.

Explanation:

<u> Part A:</u>

  • Potential energy of charge at the given point, \rm U=42\ \mu J=42\times 10^{-6}\ J.
  • Charge, \rm q=15\ nC = 15\times 10^{-9}\ C.

The potential energy at a point due to a charge is defined as

\rm U=qV.

<em>where</em>,

V = electric potential at that point.

Therefore,

\rm V=\dfrac{U}{q}=\dfrac{42\times 10^{-6}}{15\times 10^{-9}}=2.8\times 10^3\ Volts.

<u>Part B:</u>

Now, if the charge at that point is replaced with \rm q_1 = 20\ nC = 20\times 10^{-9}\ C., then the electric potential energy at that point is given by

\rm U=q_1V = 20\times 10^{-9}\times 2.8\times 10^3=5.6\times 10^{-5}\ J.

5 0
3 years ago
Given the quantities a= 9.7 m, b= 4.2 s, c= 69 m/s, what is the value of the quantity d = a^3/(cb^2)?
nexus9112 [7]
D= 9.7^3/(69)(4.2)^2
d=912.673/289.8^2
d=912.673/83984.04
d=0.01086721953362...
I hope this helped!
7 0
4 years ago
Read 2 more answers
The water behind Grand Coulee Dam is 1000 m wide and 200 m deep. Find the hydrostatic force on the back of the dam. (Hint: the t
Vika [28.1K]

Answer:

The hydro static force on the back of the dam is 1.96\times10^{11}\ N

Explanation:

Given that,

Width b= 1000 m

Depth d= 200 m

We need to calculate the average pressure

Using formula of  average pressure

P_{avg}=\rho\times g\times d_{avg}

Put the value into the formula

P_{avg}=1000\times9.8\times100

P_{avg}=980000\ Pa

We need to calculate  the hydro static force on the back of the dam

Using formula of force

F = P_{avg}\times A

Put the value into the formula

F = 980000\times1000\times200

F=1.96\times10^{11}\ N

Hence, The hydro static force on the back of the dam is 1.96\times10^{11}\ N

7 0
3 years ago
Who was the first man to go into the orbit?
kkurt [141]
D. Yuri Gagarin

(Typing this to circumnavigate the 20 character minimum requirement)
4 0
4 years ago
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