Answer:
94.13 ft/s
Explanation:
<u>Given:</u>
= time interval in which the rock hits the opponent = 10 s - 5 s = 5 s
= distance to be moved by the rock long the horizontal = 98 yards
= displacement to be moved by the rock during the time of flight along the vertical = 0 yard
<u>Assume:</u>
= magnitude of initial velocity of the rock
= angle of the initial velocity with the horizontal.
For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

On dividing equation (1) by (2), we have

Now, putting this value in equation (2), we have

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.
<span>A boat would
produce the highest concentration of carbon monoxide in the exhaust system.
</span>Carbon monoxide<span> (CO) is a colorless, odorless, and tasteless gas that is
slightly less dense than air. It is toxic to </span>hemoglobic<span> <span>animals (both </span></span>invertebrate<span> <span>and
vertebrate, including humans) when encountered in concentrations above about 35 </span></span>ppm<span>.</span>
Answer:
D would be it
Explanation:
cause none of the rest Makes sense to the book of the picture and I'm 100% sure =)
Answer:
Orange , yellow, green and blue
red coat absorbs all colors of visible light except red, so red light
is the only light left to bounce off of the coat toward our eyes.
Answer:
The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s
Explanation:
From Newton's second law, F = mg and also from coulomb's law F= Eq
Dividing both equations by mass;
F/m = Eq/m = mg/m, then
g = Eq/m --------equation 1
Again, in a projectile motion, the time of flight (T) is given as
T = (2usinθ/g) ---------equation 2
Substitute in the value of g into equation 2

Charge of proton = 1.6 X 10⁻¹⁹ C
Mass of proton = 1.67 X 10⁻²⁷ kg
E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°
Solving for T;

T = 7.83 X10⁻⁷ s