<span>We'll use the momentum-impulse theorem. The x-component of the total momentum in that direction is given by p_(f) = p_(1) + p_(2) + p_(3) = 0.
So p_(1x) = m1v1 = 0.2 * 2 = 0.4 Also p_(2x) = m2v2 = 0 and p_(3x) = m3v3 = 0.1 *v3 where v3 is unknown speed and m3 is the mass of the third particle with the unknown speed
Similarly, the 235g particle, y-component of the total momentum in that direction is given by p_(fy) = p_(1y) + p_(2y) + p_(3y) = 0.
So p_(1y) = 0, p_(2y) = m2v2 = 0.235 * 1.5 = 0.3525 and p_(3y) = m3v3 = 0.1 * v3 where m3 is third particle mass.
So p_(fx) = p_(1x) + p_(2x) + p_(3x) = 0.4 + 0.1v3; v3 = 0.4/-0.1 = - 4
Also p_(fy) = 0.3525 + 0.1v3; v3 = - 0.3525/0.1 = -3.525
So v_3x = -4 and v_3y = 3.525.
The speed is their resultant = âš (-4)^2 + (-3.525)^2 = 5.335</span>
Answer:
mass of the planet: 
Explanation:
When a moon keeps a circular orbit around a planet, it is the force of gravity the one that provides the centripetal force to keep it in its circular trajectory of radius R. So if we can write that in such cases (being the mass of the planet "M" and the mass of the moon "m"), we can form an equation by making the centripetal force on the moon equal the force of gravity (using the Newton's Universal Law of Gravity):
where we used here the tangential velocity (v) of the moon around the planet. This equation can be further simplified by dividing both sides by "m" and multiplying both sides by the orbital radius R:
Notice that the mass of the moon has actually disappeared from the equation, which tells us that the orbiting velocity and period do not depend on the mass of the moon, but on the mass of the actual planet.
We know the orbital radius R (
, the value of the Universal Gravitational constant G, and we can estimate the value of the tangential velocity of the moon since we know it period: 36.3 hrs = 388800 seconds.
We know that the moon makes a full circumference (
) in 388800 seconds, therefore its tangential velocity is:
where we rounded the velocity to one decimal.
Notice that we have converted all units to the SI system, so when using the formula to solve for the mass of the planet, the answer comes directly in kg.
Now we use this value for the tangential velocity to estimate the mass of the planet in the first equation we made and simplified:
As per the question Bob drops the bag full with feathers from the top of the building.
The mass of the bag(m)= 1.0 lb
Let the air resistance is neglected.As the bag is under free fall ,hence the only force that acts on the bag is the force of gravity which is in vertical downward direction.
Here the acceleration produced on bag due to the free fall will be nothing else except the acceleration due to gravity i.e g =9.8 m/s^2
Here we are asked to calculate the distance travelled by the bag at the instant 1.5 s
Hence time t= 1.5 s
From equation of kinematics we know that -
S=ut + 0.5at^2 [ here S is the distance travelled]
For motion under free fall initial velocity (u)=0.
Hence S= 0×1.5+{0.5×(-9.8)×(1.5)^2}
⇒ -S =0-11.025 m
⇒ S= 11.025 m
=11 m
Here the negative sign is taken only due to the vertical downward motion of the body .we may take is positive depending on our frame of reference .
Hence the correct option is B.
Answer:
Potential difference though which the electron was accelerated is 
Explanation:
Given :
De Broglie wavelength , 
Plank's constant , 
Charge of electron , 
Mass of electron , m=9.11\times 10^{-31}\ kg.
We know , according to de broglie equation :
Now , we know potential energy applied on electron will be equal to its kinetic energy .
Therefore ,
Putting all values in above equation we get ,
Hence , this is the required solution.
Answer:
Bank angle = 35.34o
Explanation:
Since the road is frictionless,
Tan (bank angle) = V^2/r*g
Where V = speed of the racing car in m/s, r = radius of the arc in metres and g = acceleration due to gravity in m/s^2
Tan ( bank angle) = 40^2/(230*9.81)
Tan (bank angle) = 0.7091
Bank angle = tan inverse (0.7091)
Bank angle = 35.34o
