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kupik [55]
3 years ago
15

AP physics Will give brainliest to correct answer.

Physics
2 answers:
rjkz [21]3 years ago
6 0

Answer:

the 4th one

Explanation:

tangare [24]3 years ago
6 0
The answer is the 4th one
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The difference between tension and traction?
Kitty [74]
The object may not end up in motion - either if tension is balanced in opposite directions, or if traction is greater than tension.
3 0
3 years ago
The change in momentum that occurs when a 1. 0 kg ball traveling at 4. 0 m/s strikes a wall and bounces back at 2. 0 m/s is.
Doss [256]

Answer:

The change is momentum is given by ∆p=p(inital) - p(final) =4-2=2 kg.m/s

Explanation:

momentum is the product of mass and velocity (speed)

So it's initial momentum would be:

p=mv=(1)(4)=4 kg.m/s

It's final momentum is given by:

p=mv=(1)(2)=2 kg.m/s

7 0
2 years ago
Find the electric force acting on an alpha particle in a horizontal electric field of 600N/C
Sergeeva-Olga [200]
Since an alpha particle has 2 protons and no negative particles (electrons) to balance the net charge, its charge is
Q=2(1.6e-19)=3.2e-19C.
The force on a charged particle is F=QE so
(3.2e-19C)(600N/C)=1.92e-16N
6 0
3 years ago
In Physics, work depends on two factors. What are those two<br> factors?
Aloiza [94]

Answer:

Force and displacement.

Explanation:

Work done is positive when we push table and it move in the direction of applied force.

5 0
2 years ago
a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the
tatuchka [14]
The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N
the force of the wind F, acting horizontally, with intensity
F=12 N
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
T \cos \alpha -F=0
T \sin \alpha -W=
By dividing the second equation by the first one, we get
\tan \alpha =  \frac{W}{F}= \frac{24.5 N}{12 N}=2.04
From which we find
\alpha = 63.8 ^{\circ}
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N




7 0
3 years ago
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