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:
1331.84 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity = 0
s = Displacement = 490 km
a = Acceleration
g = Acceleration due to gravity = 1.81 m/s² = a
From equation of linear motion
The speed of the material must be 1331.84 m/s in order to reach the height of 490 km
Answer:
a rock of 50kg should be placed =drock=0.5m from the pivot point of see saw
Explanation:
τchild=τrock
Use the equation for torque in this equation.
(F)child(d)child)=(F)rock(d)rock)
The force of each object will be equal to the force of gravity.
(m)childg(d)child)=(m)rockg(d)rock)
Gravity can be canceled from each side of the equation. for simplicity.
(m)child(d)child)=(m)rock(d)rock)
Now we can use the mass of the rock and the mass of the child. The total length of the seesaw is two meters, and the child sits at one end. The child's distance from the center of the seesaw will be one meter.
(25kg)(1m)=(50kg)drock
Solve for the distance between the rock and the center of the seesaw.
drock=25kg⋅m50kg
drock=0.5m
I think it would be B because it is matter, since it has atoms, and it contains subatomic particles, which are smaller than atoms
Answer:
1) 64.2 mi/h
2) 3.31 seconds
3) 47.5 m
4) 5.26 seconds
Explanation:
t = Time taken = 2.5 s
u = Initial velocity = 0 m/s
v = Final velocity = 21.7 m/s
s = Displacement
a = Acceleration
1) Top speed = 28.7 m/s
1 mile = 1609.344 m
1 hour = 60×60 seconds
Top speed of the cheetah is 64.2 mi/h
Equation of motion
Acceleration of the cheetah is 8.68 m/s²
2)
It takes a cheetah 3.31 seconds to reach its top speed.
3)
It travels 47.5 m in that time
4) When s = 120 m
The time it takes the cheetah to reach a rabbit is 120 m is 5.26 seconds
