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zlopas [31]
2 years ago
14

A 4 kg and 6 kg bowling ball are dropped from the same height at the same time. The two balls strike the ground at the same time

. Which statement is true about the force due to gravity on the balls?
Physics
2 answers:
Ray Of Light [21]2 years ago
5 0
The force of gravity on objects is proportional to the mass of each object.

(That's a big part of the reason why, when you eat more and your mass
increases, you weigh more.)

The forces of gravity between the Earth and the 6kg ball are 50% greater
than the forces of gravity between the Earth and the 4kg ball.

(The gravitational forces between the 4kg ball and the 6kg ball, or between
both bowling balls and you, are so small that they may be ignored.)

mart [117]2 years ago
3 0

Answer:

B) The 6 kg ball has a greater force of gravity exerted on it.

Explanation: got it from usa tp .

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A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. T
slamgirl [31]

Answer:

u=14.48m/s

Explanation:

From the question we are told that:

Height of window h=2m

Height of window off the ground h_g=7.5m

Time to fall and drop t=1.3s

 

Generally the Newton's equation motion  is mathematically given by

 s=ut+\frac{1}{2}at^2

Where

h=ut+\frac{1}{2}at^2

2=u1.3-\frac{1}{2}*9.8*1.3^2

2=u1.3-8.281

u=7.91m/s^2  

Generally the Newton's equation motion  is mathematically given by

2as=v^2-u^2

Where

-2gh_g=v^2-u^2

-2*9.8*7.5=(7.91)^2-u^2

-147=62.5681-u^2

u=\sqrt{209.5681}

u=14.48m/s

Therefore the  ball’s initial speed

u=14.48m/s

8 0
2 years ago
Which of the following is an example of mechanical waves?
Nastasia [14]
Sound Waves will be an example of mechanical waves.. hope this helps!
8 0
3 years ago
Plzzzzzz help!!!!!!struggling
wel
Metamorphic rock
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Changes in pressure.
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4 0
3 years ago
Read 2 more answers
A cutting tool several forces acting on it. One force is F=-axy^2 j , a force in the negative y-direction whose magnitude depend
liq [111]
The force on the tool is entirely in the negative-y direction.
So no work is done during any moves in the x-direction.

The work will be completely defined by

                     (Force) x (distance in the y-direction),

and it won't matter what route the tool follows to get anywhere. 
Only the initial and final y-coordinates matter.

We know that    F = - 2.85 y².  (I have no idea what that ' j ' is doing there.)
Remember that 'F' is pointing down.

From  y=0  to  y=2.40  is a distance of  2.40  upward.

Sadly, since the force is not linear over the distance, I don't think
we can use the usual formula for  Work = (force) x (distance).
I think instead we'll need to integrate the force over the distance,
and I can't wait to see whether I still know how to do that.

        Work  =  integral of (F·dy) evaluated from  0  to  2.40

                  =  integral of (-2.85 y² dy) evaluated from  0  to  2.40

                 =  (-2.85) · integral of  (y² dy)  evaluated from  0  to  2.40 .


Now, integral of (y² dy)  =  1/3  y³ .

Evaluated from  0  to  2.40 , it's  (1/3 · 2.40³) - (1/3 · 0³)

                                            =  1/3 · 13.824  =  4.608 .

And the work  =  (-2.85) · the integral

                     =  (-2.85) · (4.608)

                     =      - 13.133  .

-- There are no units in the question (except for that mysterious ' j ' after the 'F',
which totally doesn't make any sense at all).
If the ' F ' is newtons and the 2.40 is meters, then the  -13.133 is joules.

-- The work done by the force is negative, because the force points
DOWN but we lifted the tool UP to 2.40.  Somebody had to provide
13.133 of positive work to lift the tool up against the force, and the force
itself did 13.133 of negative work to 'allow' the tool to move up. 

-- It doesn't matter whether the tool goes there along the line  x=y , or
by some other route.  WHATEVER the route is, the work done by ' F ' 
is going to total up to be  -13.133 joules at the end of the day.

As I hinted earlier, the last time I actually studied integration was in 1972,
and I haven't really used it too much since then.  But that's my answer
and I'm stickin to it.  If I'm wrong, then I'm wrong, and I hope somebody
will show me where I'm wrong.
3 0
3 years ago
The tension in a string from which a 4.0 kg object is suspended in an elevator is equal to 44 N. What is the acceleration of the
Ronch [10]

Answer: 1

Explanation:

Given

Tension is the string T=44\ N

mass of object m=4\ kg

Tension is greater than the weight of the object i.e. elevator is moving upward

we can write

\Rightarrow T-mg=ma\\\Rightarrow T=m(g+a)\\\Rightarrow 44=4(10+a)\\\Rightarrow 11=10+a\\\Rightarrow a=1\ m/s^2

8 0
2 years ago
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