calculate the weight of an object of mass 100 kg on the surface of a Planet whose mass and diameter are 4.8*10^24 kg and 12000km
1 answer:
The weight of an object mass of 100 kg on the surface of a Planet willl is 981 N.
<h3>What is weight?</h3>
The weight of matter is found as the product of the mass and the gravitational acceleration;
Given data;
Weight of an object,W =?
Mass,m = 100 kg
Hence the weight of an object willl be 981 N.
To learn more about the weight refer;
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Answer:
The box displacement after 6 seconds is 66 meters.
Explanation:
Let suppose that velocity given in statement represents the initial velocity of the box and, likewise, the box accelerates at constant rate. Then, the displacement of the object (), in meters, can be determined by the following expression:
(1)
Where:
- Initial velocity, in meters per second.
- Time, in seconds.
- Acceleration, in meters per square second.
If we know that ,
and
, then the box displacement after 6 seconds is:
The box displacement after 6 seconds is 66 meters.
Answer:
Explanation:
What problem says can be written mathematically as:
Where:
The problem itself it's really simple, we only need to replace the data provided in the previous equation, but first, let's convert the units of the velocity from cm/s to m/s because we have to work with the same units and working in meters is the most apropiate action, because is the base unit of length in the International System of Units:
Now, we can replace the data in the equation and find the time it will take the bird to travel 3.7 m:
Solving for t, multiplying by t both sides, and dividing by 0.52 both sides:
Answer:
T₂ = 123.9 N, θ = 66.2º
Explanation:
To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.
The tension T1 = 100 N, we create a reference frame centered on the pole
X axis
T₁ₓ - = 0
T_{2x}= T₁ₓ
Y axis y
T_{1y} + T_{2y} - 200N = 0
T_{2y} = 200 -T_{1y}
let's use trigonometry to find the component of the stresses
sin 60 = T_{1y} / T₁
cos 60 = t₁ₓ / T₁
T_{1y} = T₁ sin 60
T1x = T₁ cos 60
T_{1y}y = 100 sin 60 = 86.6 N
T₁ₓ = 100 cos 60 = 50 N
for voltage 2 it is done in the same way
T_{2y} = T₂ sin θ
T₂ₓ = T₂ cos θ
we substitute
T₂ sin θ= 200 - 86.6 = 113.4
T₂ cos θ = 50 (1)
to solve the system we divide the two equations
tan θ = 113.4 / 50
θ = tan⁻¹ 2,268
θ = 66.2º
we caption in equation 1
T₂ cos 66.2 = 50
T₂ = 50 / cos 66.2
T₂ = 123.9 N
