Answer:
-209.42J
Explanation:
Here is the complete question.
A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component Fx=−[20.0N+(3.0N/m)x]. How much work does the force you apply do on the cow during this displacement?
Solution
The work done by a force W = ∫Fdx since our force is variable.
Since the cow moves from x₁ = 0 m to x₂ = 6.9 m and F = Fx =−[20.0N+(3.0N/m)x] the force applied on the cow.
So, the workdone by the force on the cow is
W = ∫₀⁶°⁹Fx dx = ∫₀⁶°⁹−[20.0N+(3.0N/m)x] dx
= ∫₀⁶°⁹−[20.0Ndx - ∫₀⁶°⁹(3.0N/m)x] dx
= −[20.0x]₀⁶°⁹ - [3.0x²/2]₀⁶°⁹
= -[20 × 6.9 - 20 × 0] - [3.0 × 6.9²/2 - 3.0 × 0²/2]
= -[138 - 0] - [71.415 - 0] J = (-138 - 71.415) J
= -209.415 J ≅ -209.42J
Answer:
Magnetic field will be ZERO at the given position
Explanation:
As we know that the magnetic field due to moving charge is given as

so here we know that for the direction of magnetic field we will use

so we have

so magnetic field must be ZERO
So whenever charge is moving along the same direction where the position vector is given then magnetic field will be zero
Answer:
Explanation:
Wheel completes four revolution.
The linear displacement is zero.
The angular displacement is 4 x 2π = 8π radian.
So, option (c) is correct.
Answer:
correct is d) a ’= g / 2
Explanation:
For this exercise let's use the kinematics equations
On earth
v = v₀ - a t
a = (v₀- v) / T
On planet X
v = v₀ - a' t’
a ’= (v₀-v) / 2T
Let's substitute the land values in plot X
a’= a / 2
Now let's use Newton's second law
W = ma
m g = m a
a = g
We substitute
a ’= g / 2
So we see that on planet X the acceleration is half the acceleration of Earth's gravity