Answer:
a) correct answer is C
, b) 14º from the west to the north, c) v_{1g} = 300.79 km / h
Explanation:
This is a relative speed exercise using the addition of speeds.
1) when it is not specified regarding what is being measured, the medicine is carried out with respect to the Z Earth, therefore the correct answer is C
2 and 3) In this case we must compose the speed using the Pythagorean Theorem.
² = 
² + 
²
where v_{1a} is the speed of the airplane with respect to the air, v_{1g} airplane speed with respect to the Earth, v_{ag} air speed with respect to the Earth
in this case let's clear the speed of the airplane with respect to the Earth
v_{1g} = √(v_{1a}² - v_{ag}²)
v_{1g} = √ (310² - 75²)
v_{1g} = 300.79 km / h
we find the direction of the airplane using trigonometry
sin θ = v_{ag} / v_{1a}
θ = sin⁻¹ (v_{ag} /v_{1a})
θ = sin⁻¹ (75/310)
θ= 14º
the pilot must direct the aircraft at an angle of 14º from the west to the north
Answer:
The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.
Explanation:
Given that,
Mass = 2.15 kg
Distance = 0.0895 m
Amplitude = 0.0235 m
We need to calculate the spring constant
Using newton's second law
Where, f = restoring force
Put the value into the formula
We need to calculate the kinetic energy of the mass
Using formula of kinetic energy
Here, 
Here, 
Put the value into the formula
Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.
The answer is the FIRST OPTION
Work occurs when a force is applied to an object and the object moves in the direction of the force applied <span />
Answer:
10
Explanation:
This is tough. The last number 0.2 has only one significant figure. So while the sum of all the numbers is 12.3, you must only leave one sig figure. Rounding to the tenths gives 10.
Answer:
<h2>3.36J</h2>
Explanation:
Step one:
given data
mass m= 1.3kg
distance moved s= 2.8m
opposing frictional force= 0.34N
assume g= 9.81m/s^2
we know that work done= force *distance moved
1. work done to push the book= 1.55*2.8=4.34J
2. Work against friction = force of friction x distance
= 0.34*2.8=0.952J
Step two:
the work done on the book is the net work, which is
Network done= work done to push the book- Work against friction
Network done= 4.32-0.952=3.36J
<u>Therefore the work of the 1.55N 3.36J</u>
