The magnitude of the work done by force experience by the object is (2a²b + 3b²)J.
<h3>
Work done by the force experienced by the object</h3>
The magnitude of the work done by force experience by the object is calculated as follows;
W = f.d
where;
- F is the applied force (2xyi + 3yj), where x and y are in meters
- d is the displacement of the object = (a, b)
The work done by the force is determined from the dot product of the force and the displacement of the object.
F = (2xyi + 3yj).(a + b)
W = (2abi + 3bj).(ai + bj)
W = (2a²b + 3b²)J
Thus, the magnitude of the work done by force experience by the object is (2a²b + 3b²)J.
The complete question is below:
The particle moves from the origin to the point with coordinates (a, b) by moving first along the x-axis to (a, 0), then parallel to the y-axis.
How much work does the force do?
Learn more about work done here: brainly.com/question/8119756
If a car crashes into another car like this, the wreck should go nowhere. Besides this being an unrealistic question, the physics of it would look like this:
Momentum before and after the collision is conserved.
Momentum before the collision:
p = m * v = 50000kg * 24m/s + 55000kg * 0m/s = 50000kg * 24m/s
Momentum after the collision:
p = m * v = (50000kg + 55000kg) * v
Setting both momenta equal:
50000kg * 24m/s = (50000kg + 55000kg) * v
Solving for the velocity v:
v = 50000kg * 24m/s/(50000kg + 55000kg) = 11,43m/s
Answer:
Explanation:
wave length of light λ = 502 nm
screen distance D = 1.2 m
width of one fringe = 10.2 mm / 20
= .51 mm
fringe width = λ D / a , a is separation of slits
Puting the values given
.51 x 10⁻³ = 502 x 10⁻⁹ x 1.2 / a
a = 502 x 10⁻⁹ x 1.2 / .51 x 10⁻³
= 1181.17 x 10⁻⁶ m
1.18 x 10⁻³ m
= 1.18 mm .
The ball accelerates because of gravity.
Answer:
A)6.15 cm to the left of the lens
Explanation:
We can solve the problem by using the lens equation:
where
q is the distance of the image from the lens
f is the focal length
p is the distance of the object from the lens
In this problem, we have
(the focal length is negative for a diverging lens)
is the distance of the object from the lens
Solvign the equation for q, we find
And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is
A)6.15 cm to the left of the lens
