The best answer is A) <span>keep moving at a constant velocity until some forces act on them
As the man you're probably tired of hearing about said:
"Every object persists in its state of rest or in uniform motion in a straight line unless a new force acts upon it"
This is Isaac Newton's 1st law of motion, or the law of inertia.
Put more simply, objects in motion tend to stay in motion, and tend the maintain the same velocity (direction and speed) and objects at rest tend to stay at rest. </span>
Answer:
C) 7.35*10⁶ N/C radially outward
Explanation:
- If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
- So, we can write the following equation:

- As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
- So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
- So, on the outer surface of the shell there must be a charge that be the difference between them:

- Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

- As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
Answer:
1.73 m/s²
Explanation:
Given:
Δx = 250 m
v₀ = 0 m/s
t = 17 s
Find: a
Δx = v₀ t + ½ at²
250 m = (0 m/s) (17 s) + ½ a (17 s)²
a = 1.73 m/s²
We can't look at Figure 4. We don't know where a and b are. We can't see the blue shaded area. We don't know what our method above was. We can't see the green skinny rectangle.
Answer:
The speed of the ball is 42.5 m/s
Explanation:
The initial kinetic energy of the ball is:
= 85.75 J
The speed of the ball after leaving the bat is:

V=47.92 m/s
Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:




