Answer:
23 m/s downward
__________________________________________________________
<em>Taking the downward direction as positive</em>
<u>We are given:</u>
Initial velocity of the marble (u) = 0 m/s
Time interval (t) = 2.3 seconds
Final velocity (v) = x m/s
<u>Solving for the Final velocity:</u>
<u>Acceleration of the Marble:</u>
We know that gravity will make the marble accelerate at a constant acceleration of 10 m/s
<u>Final velocity:</u>
v = u + at [First equation of motion]
x = 0 + (10)(2.3) [replacing the given values]
x = 23 m/s
Hence, after 2.3 seconds, the marble will move at a velocity of 23 m/s in the downward direction
Answer:
there is friction between the two things
Explanation:
Answer:
a) 4.49Hz
b) 0.536kg
c) 2.57s
Explanation:
This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

for some time t you have:
x=0.134m
v=-12.1m/s
a=-107m/s^2
If you divide the first equation and the third equation, you can calculate w:

with this value you can compute the frequency:
a)

b)
the mass of the block is given by the formula:

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

Finally, the amplitude is:

Answer:
<h3>473.8 m/s; 473.8 m/s</h3>
Explanation:
Given the initial velocity U = 670m/s
Horizontal velocity Ux = Ucos theta
Vertical component of the cannon velocity Uy = Usin theta
Given
U = 670m/s
theta = 45°
horizontal component of the cannonball’s velocity = 670 cos 45
horizontal component of the cannonball’s velocity = 670(0.7071)
horizontal component of the cannonball’s velocity = 473.757m/s
Vertical component of the cannonball’s velocity = 670 sin 45
Vertical component of the cannonball’s velocity = 670 (0.7071)
Vertical component of the cannonball’s velocity = 473.757m/s
Hence pair of answer is 473.8 m/s; 473.8 m/s
Answer:
7.401 * 10^(-15) N
Explanation:
30 electrons will have a charge:
30 * -1.6022 * 10^(-19) C
= - 4.806 * 10^(-18) C
The relationship between electric field and electric force is:
E = F/q
This means that force, F, is
|F| = |E|*|q|
|F| = |1540| * |-4.806 * 10^(-18)|
|F| = |-7401.24 * 10^(-18)|
|F| = 7.401 * 10^(-15) N