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spayn [35]
3 years ago
11

I NEED THIS FILLED ASAP

Physics
1 answer:
Gnesinka [82]3 years ago
3 0
What class is that in ?????
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In the 4 X 100M relay, how long is the acceleration zone..
valkas [14]

Answer:it’s 400 meters

Explanation:

3 0
3 years ago
A train moves at constant velocity of 50km/h. how far will it move in 0.5h?
Kobotan [32]
The formula for getting the distance will be distance = speed x time
                                                                               D = S x T
   speed or velocity = 50km/h
                       time = 0.5 h
    the equation will be done directly because it's already in it's SI units
            distance = 50km/h  x  0.5h
            hour cancels hour and the equation remains = 50km x 0.5
                               Ans = 25 km
the train will move 25 km far in 0.5h
7 0
3 years ago
A(n) ____ is a place or object used for comparison to determine of something is in motion
Andre45 [30]
<span>reference point is the answer.</span>
4 0
4 years ago
Read 2 more answers
Sphere A with mass 80 kg is located at the origin of an xy coordinate system; sphere B with mass 60 kg is located at coordinates
IRINA_888 [86]

Answer:

Fc = [ - 4.45 * 10^-8 j ] N  

Explanation:

Given:-

- The masses and the position coordinates from ( 0 , 0 ) are:

       Sphere A : ma = 80 kg , ( 0 , 0 )

       Sphere B : ma = 60 kg , ( 0.25 , 0 )

       Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15

- The gravitational constant G = 6.674×10−11 m3⋅kg−1⋅s−2

Find:-

what is the gravitational force on C due to A and B?

Solution:-

- The gravitational force between spheres is given by:

                       F = G*m1*m2 / r^2

Where, r : The distance between two bodies (sphere).

- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-

 Determine the angle (α) between vectors rac and rab using cosine rule:

                   cos ( \alpha ) = \frac{rab^2 + rac^2 - rbc^2}{2*rab*rac} \\\\cos ( \alpha ) = \frac{0.25^2 + 0.2^2 - 0.15^2}{2*0.25*0.2}\\\\cos ( \alpha ) = 0.8\\\\\alpha = 36.87^{\circ \:}

 Determine the angle (β) between vectors rbc and rab using cosine rule:

                   cos ( \beta  ) = \frac{rab^2 + rbc^2 - rac^2}{2*rab*rbc} \\\\cos ( \beta  ) = \frac{0.25^2 + 0.15^2 - 0.2^2}{2*0.25*0.15}\\\\cos ( \beta  ) = 0.6\\\\\beta  = 53.13^{\circ \:}

- Now determine the scalar gravitational forces due to sphere A and B on C:

       Between sphere A and C:

                  Fac = G*ma*mc / rac^2

                  Fac = (6.674×10−11)*80*0.2 / 0.2^2  

                  Fac = 2.67*10^-8 N

                  vector Fac = Fac* [ - cos (α) i + - sin (α) j ]

                  vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]

                  vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N

       Between sphere B and C:

                  Fbc = G*mb*mc / rbc^2

                  Fbc = (6.674×10−11)*60*0.2 / 0.15^2  

                  Fbc = 3.56*10^-8 N

                  vector Fbc = Fbc* [ cos (β) i - sin (β) j ]

                  vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]

                  vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N

- The Net gravitational force can now be determined from vector additon of Fac and Fbc:

                  Fc = vector Fac + vector Fbc

                  Fc = [ - 2.136 i - 1.602 j ]*10^-8  + [ 2.136 i - 2.848 j ]*10^-8

                  Fc = [ - 4.45 * 10^-8 j ] N  

3 0
4 years ago
An object is placed in front of a convex mirror with a radius of curvature of magnitude 10 cm. The mirror produces an image that
jek_recluse [69]

Answer:

u = - 20 cm

m =\frac{1}{5}

Given:

Radius of curvature, R = 10 cm

image distance, v = 4 cm

Solution:

Focal length of the convex mirror, f:

f = \frac{R}{2} = \frac{10}{2} = 5 cm

Using Lens' maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

Substitute the given values in the above formula:

\frac{1}{5} = \frac{1}{u} + \frac{1}{4}

\frac{1}{u} = \frac{1}{5} - \frac{1}{4}

u = - 20 cm

where

u = object distance

Now, magnification is the ratio of image distance to the object distance:

magnification, m =\frac{|v|}{|u|}

magnification, m =\frac{|4|}{|-20|}

m =\frac{4}{20}

m =\frac{1}{5}

4 0
3 years ago
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