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WINSTONCH [101]
3 years ago
9

To remove a nut from an old rusty bolt, you apply a 100-N force to the

Physics
1 answer:
Aleks04 [339]3 years ago
7 0

Answer:

yes its 25cmjdj8dnen8dnejdid

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What is the magnitude of the sum of the two vectors A = 36 units at 53 degrees, and B =47 units at 157 degrees.
Thepotemich [5.8K]

Answer:

51.82

Explanation:

First of all, let's convert both vectors to cartesian coordinates:

Va = 36 < 53° = (36*cos(53), 36*sin(53))

Va = (21.67, 28.75)

Vb = 47 < 157° = (47*cos(157), 47*sin(157))

Vb = (-43.26, 18.36)

The sum of both vectors will be:

Va+Vb = (-21.59, 47.11)   Now we will calculate the module of this vector:

|Va+Vb| = \sqrt{(-21.59)^2+(47.11)^2}=51.82

4 0
3 years ago
When forces acting on an object are __________, the object's motion is constant.
Ivanshal [37]
The correct answer Is B-balanced
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3 years ago
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what is the magnitude of the electric force between charges of 0.25 C and 0.11 C at a separation of 0.88 m? if the separation be
leonid [27]
F = k \cdot \frac{q_1 q_2}{d} = 9 \cdot 10^{9} \cdot \frac{0.25 \cdot 0.11}{0.88} =  144 \cdot 5^{9} \ N.

If the separation between the charges is increased then the magnitude of the force will increase in fact how the distance is being used in that formula.
6 0
3 years ago
The atomic number of a nucleus descended during which nuclear reactions?
kolbaska11 [484]

Answer:

(A.)Nuclear fission and beta decay (electron emission)

8 0
3 years ago
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A typical cell phone charger is rated to transfer a maximum of 1.0 Coulomb of charge per second. Calculate the maximum number of
Alexandra [31]

Answer:

The maximum no. of electrons- 2.25\times 10^{22}

Solution:

As per the question:

Maximum rate of transfer of charge, I = 1.0 C/s

Time, t = 1.0 h = 3600 s

Rate of transfer of charge is current, I

Also,

I = \frac{Q}{t}

Q = ne

where

n = no. of electrons

Q = charge in coulomb

I = current

Thus

Q = It

Thus the charge flow in 1. 0 h:

Q = 1.0\times 3600 = 3600\ C

Maximum number of electrons, n is given by:

n = \frac{Q}{e}

where

e = charge on an electron = 1.6\times 10^{- 19}\ C

Thus

n = \frac{3600}{1.6\times 10^{- 19}} = 2.25\times 10^{22}

3 0
3 years ago
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