Change in velocity of larger moose: (1/3)v - v = -(2/3)v
<span>change in velocity of small moose: (1/3)v - (-v) = (4/3)v </span>
<span>- (change in velocity of larger moose)/(change in velocity of smaller moose) = 2
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Answer:
the correct answer is c) 23 g
Explanation:
The heat lost by the runner has two parts: the heat absorbed by sweat in evaporation and the heat given off by the body
Q_lost = - Q_absorbed
The latent heat is
Q_absorbed = m L
The heat given by the body
Q_lost = M
ΔT
where m is the mass of sweat and M is the mass of the body
m L = M c_{e} ΔT
m = M c_{e} ΔT / L
let's replace
m = 90 3.500 1.8 / 2.42 10⁶
m = 0.2343 kg
reduced to grams
m = 0.2342 kg (1000g / 1kg)
m = 23.42 g
the correct answer is c) 23 g
Answer:
0.82 m
Explanation:
The ball is in free fall - uniform accelerated motion with constant acceleration downward,
(acceleration of gravity). So we can use the following suvat equation to solve the problem:

where
v is the final velocity
u = 4 m/s is the initial velocity
a is the acceleration
s is the displacement
At the maximum displacement, v = 0 (the velocity becomes zero). Substituting and solving for s, we find:

Answer:
The x component of the resultant force is -7.27N.
Explanation:
To obtain the x component of the resultant force, first we have to know the x components of the other forces. To do this, we just have to do some trigonometry:

Since both vectors are in the left side of the y-axis, they have a negative x component. So:

Finally, we sum both components to obtain the component of the resultant force:

In words, the x component of the resultant force is -7.27N.
Answer:Velocity = 6.325m/s
Directional angle= 18.43°
Explanation:
Using Right angle triangle
Let Velocity of ballon&hawk be VHB represent the height of the triangle.
Let Velocity of balloon angle ground be VBG represent adjacent of the triangle.
Let Velocity of hawk and ground BE VHG represent the hypothesis.
Theta = opp/Adj= VHB/VBG
using pythagorean
VHG= SQRT(VHB^2+VBG^2)
VHG= sqrt(2^2+6^2)
VHG= sqrt(4+36)
VHG= 6.325m/s
Tan theta= 2/6
Tan theta =0.3333
Tan^-1 0.3333=18.43°