Answer:
C. 441 N
Explanation:
Gravitational force between two objects can by calculated by the formula
= G m₁m₂ / r² , m₁ and m₂ are masses at distance r
= ( 6.67 x 10⁻¹¹ x 45 x 5.98 x 10²⁴) / ( 6.38 x 10⁶ )²
= 44.09 x 10
= 440.9 N
= 441 N .
The vector perpendicular to the plane of A = 3i+ 6j - 2k and B = 4i-j +3k is 16 i - 17 j - 27 k
Let r be the vector perpendicular to A and B,
r = A * B
A = 3i + 6j - 2k
B = 4i - j + 3k
a1 = 3
a2 = 6
a3 = - 2
b1 = 4
b2 = - 1
b3 = 3
a * b = ( a2 b3 - b2 a3 ) i + ( a3 b1 - b3 a1 ) j + ( a1 b2 - b1 a2 ) k
a * b = [ ( 6 * 3 ) - ( - 1 * - 2 ) ] i + [ ( - 2 * 4 ) - ( 3 * 3 ) ] j + [ ( 3 * - 1 ) - ( 4 * 6 ) ] k
a * b = 16 i - 17 j - 27 k
The perpendicular vector, r = 16 i - 17 j - 27 k
Therefore, the vector perpendicular to the plane of A = 3i + 6j - 2k and B = 4i - j + 3k is 16 i - 17 j - 27 k
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The relationship between the masses of the object and the gravitational force between them is a direct relationship
Explanation:
The gravitational force between two objects is given by the equation:
where
is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between them
We observe that:
- The gravitational force is proportional to the masses of the two objects, m1 and m2, so if the masses increase, the force will increase as well (so, this is a direct relationship)
- The gravitational force is inversely proportional to the square of the separation between the objects, so if the distance is increased, the force will decrease (so, this is an inverse relationship)
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Answer:
1.01 × 10⁵ Pa
Explanation:
At the surface, atmospheric pressure is 1.013 × 10⁵ Pa.
We need to find the total pressure on the air in the lungs of a person to a depth of 1 meter.
Pressure at a depth is given by :

Where
is the density of air, 
So,

Total pressure, P = Atmospheric pressure + 12 Pa
= 1.013 × 10⁵ Pa + 12 Pa
= 1.01 × 10⁵ Pa
Hence, the total pressure is 1.01 × 10⁵ Pa.
Answer:

Explanation:
Newton's 2nd law is given as
.
To find the acceleration in the horizontal direction, you need the horizontal component of the force being applied.
Using trigonometry to find the horizontal component of the force:

Use this horizontal component of the force to solve for for the acceleration of the object:
