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Lera25 [3.4K]
3 years ago
12

A person is attracted toward the centerof the earth by a 500 n gravitational force. the force with which the earth is attracted

toward the person is
Physics
1 answer:
nignag [31]3 years ago
4 0
500 N  is the answer, you just tell that the moon is attracted towards the person because of the Earth's huge mass. 
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A ball of mass is released from rest at a height of 30 how fast is it going when it hits the ground
elena55 [62]

Answer:

If the height is in metres, the speed is 24.25m/s

7 0
3 years ago
(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
Read 2 more answers
Andy is waiting at the signal. As soon as the light turns green, he accelerates his car at a uniform rate of 8.00 meters/second2
Tatiana [17]
You can reason it out like this:

-- The car starts from rest, and goes 8 m/s faster every second.

-- After 30 seconds, it's going (30 x 8) = 240 m/s.

-- Its average speed during that 30 sec is  (1/2) (0 + 240) = 120 m/s

-- Distance covered in 30 sec at an average speed of 120 m/s

                                                                           =  3,600 meters .
___________________________________

The formula that has all of this in it is the formula for
distance covered when accelerating from rest:

       Distance = (1/2) · (acceleration) · (time)²

                       = (1/2) ·      (8 m/s²)     · (30 sec)²

                       =      (4 m/s²)          ·      (900 sec²)

                       =            3600 meters.

_________________________________

When you translate these numbers into units for which
we have an intuitive feeling, you find that this problem is
quite bogus, but entertaining nonetheless.

When the light turns green, Andy mashes the pedal to the metal
and covers almost 2.25 miles in 30 seconds.

How does he do that ?

By accelerating at 8 m/s².  That's about 0.82 G  !

He does zero to 60 mph in 3.4 seconds, and at the end
of the 30 seconds, he's moving at 534 mph ! 

He doesn't need to worry about getting a speeding ticket.
Police cars and helicopters can't go that fast, and his local
police department doesn't have a jet fighter plane to chase
cars with.
3 0
3 years ago
A single slit of width 0.3 mm is illuminated by a mercury light of wavelength 254 nm. Find the intensity at an 11° angle to the
MArishka [77]

Answer:

The  the intensity at an 11° angle to the axis in terms of the intensity of the central maximum is  

   I_c = \frac{I}{I_o}  =8.48 *10^{-8}

Explanation:

From the question we are told that

   The  width of the slit is  D  =  0.3 \ mm =  0.3 *10^{-3} \ m

    The  wavelength is  \lambda =  254 \ nm =  254 *10^{-9} \ m

     The angle is  \theta  =  11^o

The intensity of at 11^o to the axis in terms of the intensity of the central maximum. is mathematically represented as

        I_c = \frac{I}{I_o}  = [ \frac{sin \beta  }{\beta }] ^2

Where \beta is mathematically represented as

        \beta  =  \frac{D sin (\theta ) *  \pi}{\lambda }

substituting values

      \beta  =  \frac{0.3 *10^{-3} sin (11 ) * 3.142}{254 *10^{-9} }

     \beta  =  708.1 \ rad

So

  I_c = \frac{I}{I_o}  = [ \frac{sin (708.1)  }{(708.1)}] ^2

   I_c = \frac{I}{I_o}  =8.48 *10^{-8}

6 0
4 years ago
An electromagnetic wave is propagating towards the west. At a certain moment the direction of the magnetic field vector associat
-BARSIC- [3]

Answer:

either +z direction or -z direction.

Explanation:

The direction of the electric field, in an electromagnetic wave always is perpendicular to the direction of the magnetic field and the direction of propagation of the wave.

You assume a system of coordinates with the negative x axis as the west direction, and the y axis as the up direction

In this case, the wave is propagating toward the west (- x direction), and the magnetic field vector points up (+ y direction), then, it is mandatory that the electric field vector points either +z direction or -z direction.

5 0
3 years ago
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