Answer:
83%
Explanation:
On the surface, the weight is:
W = GMm / R²
where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.
In orbit, the weight is:
w = GMm / (R+h)²
where h is the height of the shuttle above the surface of the Earth.
The ratio is:
w/W = R² / (R+h)²
w/W = (R / (R+h))²
Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:
w/W = (6.4×10⁶ / 7.03×10⁶)²
w/W = 0.83
The shuttle in orbit retains 83% of its weight on Earth.
Sample Response: Zamir and Talia’s total distances are different because they walked different paths in the maze. Zamir took a longer path. However, they had the same displacement because they both ended at the same position.
Answer:
Spring constant, k = 24.1 N/m
Explanation:
Given that,
Weight of the object, W = 2.45 N
Time period of oscillation of simple harmonic motion, T = 0.64 s
To find,
Spring constant of the spring.
Solution,
In case of simple harmonic motion, the time period of oscillation is given by :

m is the mass of object


m = 0.25 kg


k = 24.09 N/m
or
k = 24.11 N/m
So, the spring constant of the spring is 24.1 N/m.
Answer:
0.0367
Explanation:
The loss in kinetic energy results into work done by friction.
Since kinetic energy is given by
KE=0.5mv^{2}
Work done by friction is given as
W= umgd
Where m is the mass of suitacase, v is velocity of the suitcase, g is acceleration due to gravity, d is perpendicular distance where force is applied and u is coefficient of kinetic friction.
Making u the subject of the formula then we deduce that

Substituting v with 1.2 m/s, d with 2m and taking g as 9.81 m/s2 then

Therefore, the coefficient of kinetic friction is approximately 0.0367
Answer:
Explanation:
Mutual inductance is equal to magnetic flux induced in the secondary coli due to unit current in the primary coil .
magnetic field in a torroid B = μ₀ n I , n is number of turns per unit length and I is current .
B = 4π x 10⁻⁷ x (1000 / 2π x .16 )x 1 ( current = 1 A)
flux in the secondary coil
= B x area of face of coil x no of turns of secondary
= 4π x 10⁻⁷ x (1000 /2π x .16 ) .25 x 10⁻⁴ x 750
= 2 x 1000 x .25 x( 750 /.16) x 10⁻¹¹
2343.75 x 10⁻⁸
= 23.43 x 0⁻⁶ H.
.