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Nesterboy [21]
3 years ago
13

Which of the following is a risk associated with texting?

Physics
1 answer:
earnstyle [38]3 years ago
4 0

possible car wrecks and harsh injuries and also distractions

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How does kinetic energy connect or relate to Newtons second law of motion?
Orlov [11]
Newtons second law of motion: "T<span>he acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object."

kinetic energy is energy that an object posses while in motion and to get that it must have potential energy.</span>
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3 years ago
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Which action is mostly likely to make the team on the left win
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D. adding one student to the team on the left
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2 years ago
When a force of 20.0N is applied to a spring, it elongates 0.20m. Determine the period of oscillation of a 4.0kg object suspende
choli [55]
So what we can do is apply the<span> Hooke's law wich states that
F = -kx ( P.S the -ve sign means opposite in direction ) 
Also we will need to determine the spring's constant with the formula:
k = F / x 
Where F = the force ( = 20 N ) 
x = the displacement of the end of the spring from it's position ( = 0.20 m ) 
k = the spring's constant ( = unknown ) 
So this would be: k = 20 / 0.20 = 100 N/m 
The period of oscillation of 4 kg : T = 2 * pi * square root m / k 
T = 2 * pi * square root 4 / 100 
T = 1.256 seconds
Hope it helps</span>
7 0
3 years ago
What happens when a substance absorbs light?
Anna11 [10]
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7 0
3 years ago
in a tall building, a certain window is 500 meters above the ground.A tennis ball is catapulted horizontally out of the window w
expeople1 [14]

The tennis ball lands at a point 40.4 m from the base of the building.

The tennis ball is projected with a horizontal velocity <em>u</em> from a window, which is at a height <em>y</em> from the ground. The ball lands at a distance <em>x</em> from the base of the building. Let the ball take a time <em>t</em> to reach the ground. In the time <em>t</em> ,the ball falls a vertical distance <em>y</em> and also travel a horizontal distance <em>x</em>.

The initial vertical velocity of the ball is zero, since the ball is projected in the horizontal direction.  The ball falls down under the action of gravitational force.

Thus, use the equation of motion,

y=\frac{1}{2} gt^2

rewrite the expression for <em>t</em> and calculate the value of <em>t</em> using 9.81 m/s²for <em>g</em> and 500 m for <em>y</em>.

t=\sqrt{\frac{2y}{g} } \\ =\sqrt\frac{(2)(500m)}{9.81m/s^2}  \\ =10.096 s

The horizontal distance <em>x</em> is traveled using the constant velocity <em>u </em>since no force acts on the ball in the horizontal direction.

Therefore,

x=ut

Substitute 4 m/s for <em>u</em> and 10.096 s for <em>t</em>

x=ut\\ =(4m/s)(10.096s)\\ =40.384m=40.4 m

Thus, the ball lands at a point 40.4 m from the base of the building.



6 0
3 years ago
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