You wish to hit a target from several meters away with a charged coin having a mass of 4.7 g and a charge of +2400 μC . The coin
1 answer:
Answer:
Direction = perpendicular to velocity and into the plane of the motion
Explanation:
As the coin is projected directly towards the target so due to electric field present in the region the coin will have tendency to deflect in the direction of electric field
Now in order to hit the target we need to counter balance the electric field force and gravitational force with the magnetic field force
So here we can say
here we know that
E = 28 N/C
m = 4.7 g
now from above equation
Direction = perpendicular to velocity and into the plane of the motion
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Answer:
ANSWER BELOW I
I
V
Remember that w=mg where w is weight in Newtons, m is mass in kilograms, and g is gravity in
m/s2
. For example, for Earth, 445 N = 45.4 × 9.8
m/s2
:Notice that the x-axis values will be gravity in
m/s2
, which is already given in the table, and the y-axis values will be the weight in Newtons. Remember to round your weights to a whole number, and to enter the points starting with the lowest gravity (moon, then Mars, then Venus, then Earth).
Answer:
46.48m/s
Explanation:
The problem is a combination of the principle of conservation of linear momentum and projectile motion.
The principle of conservation of linear momentum states that in a closed system, the total momentum of colliding bodies before impact is equal to the total momentum after impact. The masses stated in the problem experienced an inelastic collision. In an inelastic collision, the bodies involved stick together after the collision and move with a common velocity.
For two bodies of masses and
moving with velocities
and
before impact, if they experience inelastic collision, the conservation of their momenta is as stated in equation (1);
were v is their common velocity after impact. If the second mass was at rest before the impact, then its initial velocity
. therefore
. Equation (1) then becomes;
In the problem stated, the second mass taken as the mass of the wooden block was at rest before the impact and the collision was inelastic since both the wood and the dart stuck together and moved with a common velocity after the impact. Therefore we can use equation (2) for the problem.
Given;
Substituting these values into (2), we get the following;
Their common v velocity after impact now makes both the wooden block and the dart (as a single body) to fall vertically through a height h of 1.5m over a range R of 3.5m as stated by the problem; hence by the principle of projectile motion for a body projected horizontally, the following relationship holds;
were t is the time taken to fall through the height h. To obtain t we use the second equation of free fall under gravity;
were g is acceleration due to gravity taken as . Therefore;
We then substitute R and t into equation (4) to obtain v.
We now further substitute this value of v into (3) to obtain ;