6. A child tugs on a rope attached to a 0.62-kg toy with a horizontal force
1 answer:
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Answer:
Q = 1057.5 [cal]
Explanation:
In order to solve this problem, we must use the following equation of thermal energy.
where:
Q = heat energy [cal]
Cp = specific heat = 0.47 [cal/g*°C]
T_final = final temperature = 32 [°C]
T_initial = initial temperature = 27 [°C]
m = mass of the substance = 450 [g]
Now replacing:
Refer to the diagram shown below.
Define the unit vector i to point in the eastern direction, and the unit vector j to point in the northern direction.
The first distance is 184 m west. It is represented by
d₁ = -184 i
The second distance is 220 m at 30° south of east. It is
d₂ = 220(cos 30° i - sin 30° j) = 190.53 i - 110 j
The third distance is 104 m at 80 east of north. It is
d₃ = 104(sin 80° i + cos 80° j) = 102.42 i + 18.06 j
Let the fourth distance be
d₄ = a i + b j
Because the traveler ends back at the original position, the vector sum of the distances is zero. It means that each component of the vector sum is zero.
The x-component yields
-184 + 190.53 + 102.42 + a = 0
a = -108.95
The y-component yields
0 - 110 + 18.06 + b = 0
b = 91.94
The magnitude of the fourth displacement is
√[(-108.95)² + 91.94² ] = 142.56 m
The direction is at an angle θ north of west, given by
θ = tan⁻¹ (91.94/108.95) = 40.2°
Answer:
The fourth displacement has a magnitude of 142.56 m. It is about 40° north of west.
Define the unit vector i to point in the eastern direction, and the unit vector j to point in the northern direction.
The first distance is 184 m west. It is represented by
d₁ = -184 i
The second distance is 220 m at 30° south of east. It is
d₂ = 220(cos 30° i - sin 30° j) = 190.53 i - 110 j
The third distance is 104 m at 80 east of north. It is
d₃ = 104(sin 80° i + cos 80° j) = 102.42 i + 18.06 j
Let the fourth distance be
d₄ = a i + b j
Because the traveler ends back at the original position, the vector sum of the distances is zero. It means that each component of the vector sum is zero.
The x-component yields
-184 + 190.53 + 102.42 + a = 0
a = -108.95
The y-component yields
0 - 110 + 18.06 + b = 0
b = 91.94
The magnitude of the fourth displacement is
√[(-108.95)² + 91.94² ] = 142.56 m
The direction is at an angle θ north of west, given by
θ = tan⁻¹ (91.94/108.95) = 40.2°
Answer:
The fourth displacement has a magnitude of 142.56 m. It is about 40° north of west.
Answer:
Explanation:
Hi there,
To get started, recall the kinematic equations from either a textbook, equation sheet, etc. Kinematic equations are used when acceleration is <em>constant,</em> as stated in the prompt.
Best way to use kinematic equations is to see which variable you are looking for, then which variable is unknown to you and is not needed for that equation.
a) average velocity
Takes the form of:
this is the literal definition of average velocity; initial plus final divided by 2.
We know total displacement and total time elapsed, so we will use the middle form of the equation:
b) the final velocity
We can still use the average velocity formula, as the other two equations that include final velocity have acceleration variable which is unknown as of now.
Solve for final velocity:
this makes sense, since a velocity later in time is higher than a velocity earlier in time. It is increasing with increasing time because of acceleration.
c) the acceleration
There are two equations that can be used to solve this, but we will use the less time-consuming one, but both produce same answer:
Notice, change in velocity over change in time, and acceleration is constant. When acceleration is constant, it models a linear function, and acc. is just slope!
Study well and persevere. If you liked this solution, hit Thanks or give a rating!
thanks,