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Afina-wow [57]
3 years ago
12

Two objects of different masses are sitting on different balance scales. Object A has a greater mass than object B. How will the

weights of the objects compare to each other?
Physics
1 answer:
kodGreya [7K]3 years ago
5 0

As long as they're both on the same planet, the greater mass always has the greater weight.  In this question, Object-A has the greater mass, so it weighs more that Object-B does.

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A ball is dropped from a height of 20 meters. At what height does the ball have a velocity of 10 meters/second?
borishaifa [10]

Answer:B

Explanation:

Initial velocity, u=0m/s

Distance,s=20m

a=+g=9.8m/s*s

Using v*v=u*u+2gs

v*v=0+2*9.8*20

v*v=392

v=19.8

When s=20m, v = 19.8m/s

Therefore when v = 10m/s, s= 10*20/19.8

s =10.1m

6 0
3 years ago
17 points
const2013 [10]

The velocity of the cannonball is 150 m/s,  the right option is B. 150 m/s.

The question can be solved, using Newton's second law of motion.

Note: Momentum of the cannon = momentum of the cannonball.

<h3>Formula:</h3>
  • MV = mv................. Equation 1

<h3>Where:</h3>
  • M = mass of the cannon
  • m = mass of the cannonball
  • V = velocity of the cannon
  • v = velocity of the cannonball

Make v the subject of the equation.

  • v = MV/m................ Equation 2

From the question,

<h3>Given: </h3>
  • M = 500 kg
  • V = 3 m/s
  • m = 10 kg.

Substitute these values into equation 2.

  • v = (500×3)/10
  • v = 150 m/s.

 

Hence, The velocity of the cannonball is 150 m/s,  the right option is B. 150 m/s.

Learn more about Newton's second law here: brainly.com/question/25545050

3 0
2 years ago
The _______________ of two objects and their distance from each other determine the gravitational force between them. (Please he
Kryger [21]
" ... product of the masses ... "
8 0
3 years ago
Read 2 more answers
The viewing screen in a double-slit experiment with monochromatic light. Fringe C is the central maximum. The fringe separation
makvit [3.9K]

Answer:

<em>Part A</em><em>:</em>

a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.

<em>Part B</em><em>:</em>

b) If the spacing between the slits is decreased the fringe spacing Δy will increase.

<em>Part C</em><em>:</em>

a) If the distance to the screen is decreased the fringe spacing will decrease.

<em>Part D</em><em>:</em>

The dot in the center of fringe E is 920\ x\ 10^{-9} m farther from the left slit than from the right slit.

Explanation:

In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.

The position of bright fringes in the screen where the pattern is formed can be calculated with

                      \vartriangle y =\frac{m \lambda L}{d}

                      m=0,\pm 1,\pm 2,\pm 3,.....

  1. m is the order number.
  2. \lambda is the wavelength of the monochromatic light.
  3. L is the distance between the screen and the two slits.
  4. d is the distance between the slits.
  • Part A:  a) In the above equation for the position of bright fringes we can see that if the wavelength of the light \lambda is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.
  • Part B:  b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
  • Part C:  a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.
  • Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at  the center of the fringe E in the screen we use the condition for constructive interference. That says that the  path length difference Δr between rays coming from the left and right slit must be \vartriangle r=m \lambda

        We simply replace the values in that equation :

                      \vartriangle r= m \lambda =2.\ 460\ nm

                      \vartriangle r= 920\ x\ 10^{-9} m

         The dot in the center of fringe E is 920\ x\ 10^{-9}m farther from the left slit than from the right slit.

     

       

       

     

3 0
3 years ago
Two 25.0N weights are suspended at opposite ends of a rope that passes over a frictionless pulley. What is the tension in the ro
goldenfox [79]

Answer:

tension in rope = 25.0 N

Explanation:

  • Two forces act on the suspended weight. The force coming down is the gravitational force and the upward force by the tension in the rope.
  • Since the suspended weight is not accelerating so that the net force will be zero. Therefore the tension in the rope should be 25 N.

       ∑F = F - W = 0

       so

       F = W

       so tension in rope = F  = T  = 25 N

8 0
2 years ago
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