Answer:
Explanation:
Given that,
A crate of mass M = 50kg
The crate is pulled along an horizontal floor by a string at an angle
Force pulling the crate
F = 210N
Angle θ = 20° to the horizontal
Distance moved by crate d=3m
A. Work done by force
W = FdCosθ
W = 210 × 3 × Cos20
W = 592J
B. Work done by Gravitational force?
Work is define as the dot product of force and displacement in the direction of the force.
Since the gravitational force does not cause the crate to move any distance downward
Then,
W(gravity) = mg×d
distance d=0
W(gravity) = 0 J
C. Work done by normal?
Since, the normal force did not cause the crate to move upward by any distance
Then,
W(normal) = 0J
D. The total workdone?
The only workdone on the crate is by the person
Then, the total workdone is the workdone by the person
W = 592 J
<span>If an electrostatically charged object is placed near other objects, </span>it would be attracted to objects with an opposite charge.
Answer:
(a). 2
(b). 1/3
(c). 11.11
Explanation:
(a). k= (t₍s₎-t₍o₎)/t₍o₎...............(1)
where k= retention factor,
t₍o₎=solvent time, t₍s₎= solute time.
Given t₍s₎=9.0 Minutes, t₍o₎=3.0 minutes.
∴ k= (9-3)/3
k= 2.
(b). the fraction of time the solute spend in the mobile phase in the column is the ratio of the solvent time to the solute time. = t₍o₎/t₍s₎..........(2)
= 3/9
=1/3.
(c). K=k(Vm/Vs)................(3)
where K= partition coefficient, k= retention factor, Vm=volume of mobile phase, Vs= volume of stationary phase.
∴K = k(Vm/Vs)
k=2, and Vs=0.18Vm.
∴K = 2(Vm/0.18vm)
⇒K = 2/0.18
∴K=11.11
Answer:
F = 4856.32 N
Explanation:
Given,
A satellite is orbiting earth at a distance from Earth surface, h = 35000 m
The mass of the satellite, m = 500 Kg
The radius of the Earth, R = 6.371 x 10⁶ m
The mass of the Earth, M = 5.972 x 10²⁴ Kg
The gravitational constant, G = 6.67408 x 10 ⁻¹¹ m³ kg⁻¹ s⁻²
The force between the Earth and the satellite is given by the formula
F = GMm/(R+h)² N
Substituting the values in the above equation
F = (6.67408 x 10 ⁻¹¹ X 5.972 x 10²⁴ X 500) / (6.371 x 10⁶ + 35000)²
= 4856.32 N
Hence, the force between the planet and the satellite is, F = 4856.32 N
We know, F = k * q₁ * q₂ / r²
Substitute the known values,
F = 9 * 10⁹ * 5 * 7 / (1.2)²
F = 315 * 10⁹ / 1.44
F = 218.75 * 10⁹ N
F = 2.1875 * 10¹¹ N [ Final Answer ]
Hope this helps!
