"If one increases the force on an object, its acceleration increases too because the push it feels is greater"
We have the 2nd law of Newton that relates the 3 concepts; F=m*a. We have that if the mass of an object increases (put weight in luggage), the accelearation decreases; in fact it is inversely proportional to the mass. Hence if the mass is doubled, acceleration is halved. Accelerations is proportional to force; if one doubles the force, the acceleration doubles too.
Answer:
3311N
Explanation:
r = radius = 600m
V = speed = 150m/s
Mass = weight = 70kg
The weight of pilot when calculated due to circular motion
W = tv
Fv = mv²/r
Fv = 70x150²/600
Fv = 79x22500/600
= 15750000/600
= 2625N
Real Weight of the pilot = m x g
= 70 x 9.8
= 686N
The apparent Weight is calculated by
Mv²/r + mg
= 2625N + 686N
= 3311 N
Therefore the apparent Weight is 3311N
Correct option: A
An object remains at rest until a force acts on it.
As the water moves faster, it applies greater force on the sediment, which over comes the frictional forces between the bed and the sediment. So, when the river flows faster, more and larger sediment particles are carried away. When the flow slows down, the river couldn't apply enough force on the larger sediments which can overcome the frictional force between the sediment and the river bed. So, the net force on the heavier particles become zero. Hence, the heavier particles of the load will settle out.
Answer:
<em>The number of moles of palladium and tantalum are 0.00037 mole and 0.0000404 mole respectively</em>
Explanation:
Number of mole = reacting mass/molar mass
n = R.m/m.m......................... Equation 1
Where n = number of moles, R.m = reacting mass, m.m = molar mass.
For palladium,
R.m = 0.039 g and m.m = 106.42 g/mol
Substituting theses values into equation 1
n = 0.039/106.42
n = 0.00037 mole
For tantalum,
R.m = 0.0073 and m.m = 180.9 g/mol
Substituting these values into equation 1
n = 0.0073/180.9
n = 0.0000404 mole
<em>Therefore the number of moles of palladium and tantalum are 0.00037 mole and 0.0000404 mole respectively</em>
Answer:
6.32 m/s 18.43° northeast
Explanation:
We express the velocity of hawk as:
We consider positive x towards east and positive y due north. So the magnitude is simply the square root of the square components:
![|v_{hawk}|=\sqrt[]{6^2+2^2}=\sqrt{40}](https://tex.z-dn.net/?f=%7Cv_%7Bhawk%7D%7C%3D%5Csqrt%5B%5D%7B6%5E2%2B2%5E2%7D%3D%5Csqrt%7B40%7D)
≈
And the angle with respect to the east should be with:
