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iris [78.8K]
3 years ago
10

Please could someone help me?

Physics
1 answer:
Talja [164]3 years ago
6 0

Answer:

I) 18.75 m/s

ii) for 10 second

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(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas
aleksley [76]

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = \frac{k*q}{r}

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = \frac{k*q1}{x} +\frac{k*q2}{(1.63m-x)} = 0

⇒k*q1* (1.63m - x) = -k*q2*x:

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

3 0
3 years ago
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True or False? Sound travels faster through metal than air​
nadezda [96]

Answer:

Yes.

Explanation:

Sound waves travel through metal 17 times faster than air.

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Table sugar has the chemical formula
Gelneren [198K]

Explanation:

This information is contained in the subscripts after each element. A molecule of sucrose (C12H22O11) has 12 carbon atoms, 22 hydrogen atoms and 11 oxygen atoms. The subscripts also indicate the ratios of the elements. A dozen CO2 molecules have one dozen carbon atoms and two dozen oxygen atoms.

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3 years ago
A 5.30 m long pole is balanced vertically on its tip. What will be the speed (in meters/second) of the tip of the pole just befo
suter [353]

Answer:

v = 17.66 m/s

Explanation:

As we know that the lower end of the pole is fixed in the ground and it start rotating about that end

so here we can say that the gravitational potential energy of the pole will convert into rotational kinetic energy of the pole about its one end

so we have

mgL = \frac{1}{2}(\frac{mL^2}{3})\omega^2

so we have

\omega = \sqrt{\frac{6g}{L}}

now we have

\omega = \sqrt{\frac{6(9.81)}{5.30}}

\omega = 3.33 rad/s

now the speed of the other tip of the pole is given as

v = \omega L

v = (3.33)(5.30) = 17.66 m/s

8 0
4 years ago
A projectile is launched at an angle of 30° and lands 20 s later at the same height as it was launched. (a) What is the initial
Elina [12.6K]

Answer:

a)Initial speed of the projectile = 196.2 m/s

b)Maximum altitude = 490.5 m

c) Range of projectile = 3398.28 m

d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

Explanation:

Time of flight of a projectile is given by the expression,

               t=\frac{2usin\theta}{g}

           Here θ = 30° and t = 20 s

a) t=\frac{2usin\theta}{g}\\\\20=\frac{2\times usin30}{9.81}\\\\u=196.2m/s

  Initial speed of the projectile = 196.2 m/s

b) Maximum altitude is given by

                  H=\frac{u^2sin^2\theta}{2g}=\frac{196.2^2\times sin^230}{2\times 9.81}=490.5m

      Maximum altitude = 490.5 m

c) Range of projectile is given by

                              R=\frac{u^2sin2\theta}{g}=\frac{196.2^2\times sin(2\times 30)}{9.81}=3398.28m

    Range of projectile = 3398.28 m

d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s

   Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s

   We have equation of motion s = ut + 0.5 at²

   Horizontal motion

                         u = 169.91 m/s

                         a = 0 m/s²

                          t = 15 s

                Substituting

                          s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m

      Vertical motion

                         u = 98.1 m/s

                         a = -9.81 m/s²

                          t = 15 s

                Substituting

                          s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m

   \texttt{Total displacement =}\sqrt{2548.71^2+367.88^2}=2575.12m

   Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

7 0
4 years ago
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