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max2010maxim [7]
3 years ago
5

What is the velocity (m/s) after 5 seconds of fall

Physics
1 answer:
gregori [183]3 years ago
3 0
5 second fall starting at 0 m/s
ball strikes ground at a speed = 49 meters per second.
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5. A current of 3.00 A flows through a resistor when it is connected
viktelen [127]
I think it is D
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7 0
2 years ago
I need help, please!
marta [7]

NO. it is not balanced

7 0
3 years ago
The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, in
bezimeni [28]

1) Potential difference: 1 V

2) V_b-V_a = -1 V

Explanation:

1)

When a charge moves in an electric field, its electric potential energy is entirely converted into kinetic energy; this change in electric potential energy is given by

\Delta U=q\Delta V

where

q is the charge's magnitude

\Delta V is the potential difference between the initial and final position

In this problem, we have:

q=4.80\cdot 10^{-19}Cis the magnitude of the charge

\Delta U = 4.80\cdot 10^{-19}J is the change in kinetic energy of the particle

Therefore, the potential difference (in magnitude) is

\Delta V=\frac{\Delta U}{q}=\frac{4.80\cdot 10^{-19}}{4.80\cdot 10^{-19}}=1 V

2)

Here we have to evaluate the direction of motion of the particle.

We have the following informations:

- The electric potential increases in the +x direction

- The particle is positively charged and moves from point a to b

Since the particle is positively charged, it means that it is moving from higher potential to lower potential (because a positive charge follows the direction of the electric field, so it moves away from the source of the field)

This means that the final position b of the charge is at lower potential than the initial position a; therefore, the potential difference must be negative:

V_b-V_a = - 1V

8 0
3 years ago
What percentage of the original kinetic energy is convertible to internal energy?
schepotkina [342]
There would be very less percentage loss<span> of the kinetic energy during </span>the conversion<span> to internal energy, assuming that there is less air in the </span>surroundings<span>. Also, the friction will contribute to the conversion where if it is, the percentage loses is negligible.</span>
7 0
3 years ago
An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of th
jeyben [28]

Answer:

the shooting angle ia 18.4º

Explanation:

For resolution of this exercise we use projectile launch expressions, let's see the scope

      R = Vo² sin (2θ) / g

      sin 2θ = g R / Vo²

      sin 2θ = 9.8 75/35²

      2θ = sin⁻¹ (0.6)

      θ = 18.4º

To know how for the arrow the tree branch we calculate the height of the arrow at this point

       X2 = 75/2 = 37.5 m

We calculate the time to reach this point since the speed is constant on the X axis

       X = Vox t

       t2 = X2 / Vox = X2 / (Vo cosθ)

        t2 = 37.5 / (35 cos 18.4)

        t2 = 1.13 s

With this time we calculate the height at this point

        Y = Voy t - ½ g t²

        Y = 35 sin 18.4   1.13 - ½ 9.8 1,13²

        Y = 6.23 m

With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch

8 0
3 years ago
Read 2 more answers
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