Answer:
35 mph
Explanation:
The key of this problem lies in understanding the way that projectile motion works as we are told to neglect the height of the javelin thrower and wind resistance.
When the javelin is thown, its velocity will have two components: a x component and a y component. The only acceleration that will interact with the javelin after it was thown will be the gravety, which has a -y direction. This means that the x component of the velocity will remain constant, and only the y component will be affected, and can be described with the constant acceleration motion properties.
When an object that moves in constant acceleration motion, the time neccesary for it to desaccelerate from a velocity v to 0, will be the same to accelerate the object from 0 to v. And the distance that the object will travel in both desaceleration and acceleration will be exactly the same.
So, when the javelin its thrown, it willgo up until its velocity in the y component reaches 0. Then it will go down, and it will reach reach the ground in the same amount of time it took to go up and, therefore, with the same velocity.
Answer:
B
Explanation:
the 3 electrons makes it neutral
Answer:
Net force: 20 N to the right
mass of the bag: 20.489 kg
acceleration: 0.976 m/s^2
Explanation:
Since the normal force and the weight are equal in magnitude but opposite in direction, they add up to zero in the vertical direction. In the horizontal direction, the 195 N tension to the right minus the 175 force of friction to the left render a net force towards the right of magnitude:
195 N - 175 N = 20 N
So net force on the bag is 20 N to the right.
The mass of the bag can be found using the value of the weight force: 201 N:
mass = Weight/g = 201 / 9.81 = 20.489 kg
and the acceleration of the bag can be found as the net force divided by the mass we just found:
acceleration = 20 N / 20.489 kg = 0.976 m/s^2
Answer:
At the center of the object
At the end of the object farthest away from the ground
At the center of gravity of the object
At end of the object closest to the ground
Explanation:
A) To calculate the charge of each coin, we must apply the expression of the Coulomb's Law:
F=K(q1xq2)/r²
F: The magnitud of the force between the charges. (F=2.0 N).
K: Constant of proporcionality of the Coulomb's Law (K=9x10^9 Nxm²/C²).
q1 and q2: Electrical charges.
r: The distance between the charges (r=1.35 m).
We have the values of F, K and r, so we can calculate q1xq2, because both<span> coins have identical charges:
</span>
q1xq2=(r²xF)/K
q1xq2=(1.35 m)²(2.0 N)/9x10^9 Nxm²/C²
q1xq2=3x10^-10 C
q1=q2=(<span>3x10^-10 C)/2
</span>Then, the charge of each coin, is:
<span>
q1=1.5x</span><span>10^-10 C
</span>q2=1.5x10^-10 C
B) <span>Would the force be classified as a force of attraction or repulsion?
</span>
It is a force of repulsion, because both coins have identical charges and both are postive. In others words, when two bodies have identical charges (positive charges or negative charges), the force is of repulsion.
