Differential portion of the invertebrate
Answer:
W = 725 J
Explanation:
given,
mass of wagon = 20 kg
distance of pull = 40 m
angle made with the horizontal = 25°
tension force on the wagon = 20 N
Work done = ?
Horizontal component of the force will help in movement of the wagon.
Horizontal component of force=
F_x = F cos θ
Work done is equal to force into displacement
W = F.s
W = F cos θ.s
W = 20 cos 25° x 40
W = 725 J
hence, work done on pulling the wagon is equal to W = 725 J
Answer:
The length of the wire is 579 m
Explanation:
Given;
current on the wire, I = 11.3-mA
magnetic field of the wire, B = (16.2i + 2.4 ĵ) T
Magnitude of force experience by the wire, F = 15.7 N
Magnitude of force experience by current carrying wire at a given a magnetic field strength is calculated as;
F = BILsinθ
Where;
B is magnitude of magnetic field
F is the force on the wire
L is length of the wire
θ is direction of the magnetic field


Length of the wire is calculated as;

Therefore, the length of the wire is 579 m
Answer:
<em>1.49 x </em>
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Explanation:
Kepler's third law states that <em>The square of the orbital period of a planet is directly proportional to the cube of its orbit.</em>
Mathematically, this can be stated as
∝ 
<em>to remove the proportionality sign we introduce a constant</em>
= k
k = 
Where T is the orbital period,
and R is the orbit around the sun.
For mars,
T = 687 days
R = 2.279 x 
for mars, constant k will be
k =
= 3.987 x 
For Earth, orbital period T is 365 days, therefore
= 3.987 x
x 
= 3.34 x 
R =<em> 1.49 x </em>
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Answer: Answer is D
I took the test little while back.