A projectile<span> is an object upon which the only force acting is gravity. Many </span>projectiles<span> not only undergo a vertical </span>motion<span>, but also undergo a horizontal </span>motion<span>. That is, as they move upward or downward they are also moving horizontally.</span>
The combined-gas law relates which temperature, pressure and volume.
Temperature=T
Pressure=P
Volume=V
(P₁*V₁) / T₁=(P₂*V₂) / T₂
D. Temperature, pressuere and volume.
Answer:

Explanation:
R = Horizontal range of projectile = 75 m
v = Velocity of projectile = 37 m/s
g = Acceleration due to gravity = 
Horizontal range is given by

The angle at which the arrow is to be released is
.
<span>The rule of inertia states that an object in motion will stay in motion unless another force has acted upon it. Because the person doesn't have their seatbelt on, they will keep moving. But if they were wearing a seatbelt, that would work as the force that is supposed to stop the person from flying forward.
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Please give brainliest</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.