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seropon [69]
1 year ago
14

(a) What is the sum of the following four vectors in unit-vector notation? For that sum, what are (b) the magnitude, (c) the ang

le in degrees, and (d) the angle in radians? Positive angles are counterclockwise from the positive direction of the x axis; negative angles are clockwise.
: 6.00 m at + 0.900 rad
: 5.00 m at - 75.0°
: 4.00 m at + 1.20 rad
: 6.00 m at - 210°
Physics
1 answer:
avanturin [10]1 year ago
7 0

6m at 0.9 radian means 6m at 51.57^0

Since positive angles are counter clockwise

6m at 51.57^0 can be written as 6*cos51.57^0i+6*sin51.57^0j = 3.73i+4.70j

5m at -75^0 can be written as 5*cos(-75)^0i+5*sin(-75)^0j = 1.294i-4.83j

4m at 1.2 radian means 4m at 68.75^0

4m at 68.75^0 can be written as 4*cos68.75^0i+4*sin68.75^0j = 1.45i+3.73j

6m at -210^0 can be written as 6*cos(-210)^0i+6*sin(-210)^0j = -5.20i-3j

a) So sum of the vector = 1.274i+0.6j

b) Magnitude = \sqrt{1.274^2+0.6^2} = 1.98 m

c) Angle,  tan\theta = \frac{0.6}{1.274} \\ \\ \theta = 25.22^0

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Ronch [10]

Answer

given,

change in enthalpy = 51 kJ/mole

change in activation energy = 109 kJ/mole

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where as activation energy of the product and the reactant decreases.

example:

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E_a= 83 kJ/mole

here activation energy decrease whereas change in enthalpy remains same.

5 0
2 years ago
The young tree was bent and has been brought into a vertical position by the three guy cables. If tension at AB = 0, AC = 10 lb,
KATRIN_1 [288]

Answer:

The young tree, originally bent, has been brought into the vertical position by adjusting the three guy-wire tensions to AB = 7 lb, AC = 8 lb, and AD = 10 lb. Determine the force and moment reactions at the trunk base point O. Neglect the weight of the tree.

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Explanation:

See attached picture.

3 0
2 years ago
You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves along a frictionless track. however, af
djyliett [7]
I attached the missing picture.
We can figure this one out using the law of conservation of energy.
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A: mgh_1+\frac{mv_1^2}{2}
Then, while the car is traveling down the track it loses some of its initial energy due to friction:
W_f=F_f\cdot L
So, we know that the car is approaching the point B with the following amount of energy:
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The law of conservation of energy tells us that this energy must the same as the energy at point B. 
The energy at point B is the sum of car's kinetic and potential energy:
B: mgh_2+\frac{mv_2}{2}
As said before this energy must be the same as the energy of a car approaching the loop:
mgh_2+\frac{mv_2}{2}=mgh_1+\frac{mv_1^2}{2}- F_fL
Now we solve the equation for v_1:
v_1^2=2g(h_2-h_1)+v_2^2+\frac{2F_fL}{m}\\
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4 0
1 year ago
Read 2 more answers
Lamar has been running sprints to prepare for his next football game.He has found that he can maintain his maximum speed for 45
Svet_ta [14]

Answer:

Kindly check explanation

Explanation:

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x = 45 yards

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x = 45 / 1093.613

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Therefore, with the available information, it can be concluded that Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.

5 0
2 years ago
Consider four different oscillating systems, indexed using i = 1 , 2 , 3 , 4 . Each system consists of a block of mass mi moving
Rzqust [24]

Answer:

The order is 2>4>3>1 (TE)

Explanation:

Look up attached file

4 0
2 years ago
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