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3 years ago

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QUESTION: What is the relationship between the masses of the objects and the gravitational force between them? (Another way of a

sking: What happens to the gravitational force as the masses get larger?)
helpppp due at 9:30am... n no mfkn links -_-

1 answer:

OLga [1]3 years ago

4 0

Answer:

As the masses get larger the force gets larger

Explanation:

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The pressure in a container will stay the same if you ——?

elena55 [62]

I think it’s D...not sure tho !

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3 years ago

An Iowa class warship holds the record for the fastest warship. If the ship accelerates uniformly from rest at 0.15 m/s² for 2 m

Mumz [18]

The equation to be used is the derived formulas for rectilinear motion at a constant acceleration. The formula for acceleration is

a = (v - v₀)/t
where
v and v₀ are the initial and final velocities, respectively
t is the time
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Since it started from rest, v₀ = 0. Using the formula:

0.15 m/s² = (v - 0)/[2 minutes*(60 s/1 min)]
Solving for v,
v = 18 m/s

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3 years ago

A 100-kg tackler moving at a speed of 2.6 m/s meets head-on (and holds on to) an 92-kg halfback moving at a speed of 5.0 m/s. Pa

DIA [1.3K]

Given that,

Mass of trackler, m₁ = 100 kg

Speed of trackler, u₁ = 2.6 m/s

Mass of halfback, m₂ = 92 kg

Speed of halfback, u₂ = -5 m/s (direction is opposite)

To find,

Mutual speed immediately after the collision.

Solution,

The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{100\times 2.6+92\times (-5)}{(100+92)}\\\\V=-1.04\ m/s$

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4 years ago

Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 23.5 m above water wit

Elodia [21]

As stated in the statement, we will apply energy conservation to solve this problem.

From this concept we know that the kinetic energy gained is equivalent to the potential energy lost and vice versa. Mathematically said equilibrium can be expressed as

$\Delta KE = \Delta PE$

$\frac{1}{2}mv_f^2-\frac{1}{2} mv_0^2 = mgh_2-mgh_1$

Where,

m = mass

$v_{f,i}$ = initial and final velocity

g = Gravity

h = height

As the mass is tHe same and the final height is zero we have that the expression is now:

$\frac{1}{2}v_f^2-\frac{1}{2} v_0^2 = gh_2$

$\frac{1}{2} (v_f^2-v_0^2) = gh_2$

$(v_f^2-v_0^2) = 2gh_2$

$v_f = \sqrt{2gh_2+v_0^2}$

$v_f = \sqrt{2(9.8)(23.5)+13.6^2}$

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3 years ago

Which two points on the wave shown in the diagram below are in phase with each other?

NNADVOKAT [17]

Answer:

4. B and D

Explanation:

Two points along a transverse wave (such as the one in the figure) are said to be in phase when:

- the vertical position of the two points is the same

- The oscillation of the wave is going in the same way for both points

Basically, we say that two points are in phase when they are separated by a complete cycle (one complete oscillation) of the wave.

For this wave, we see that point B and C have same displacement, but they are not in phase since in B the oscillation is going down while in C is going up.

Instead, B and D are in phase, because they are separated by one complete cycle: both points have same displacement and the oscillation is going in the same way for both of them.

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3 years ago