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3 years ago

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Two vectors are being added, one at an angle of 20.0 , and the other at 80.0. The only thing you know about the magnitudes is th

ey are both positive. Will the equilibrant vector be in the:
a. first quadrant
b. second quadrant
c. third quadrant
d. fourth quadrant
e. you cannot tell which quadrant from the available information

1 answer:

postnew [5]3 years ago

4 0

Answer:a

Explanation: they are all positive

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What layer of the atmosphere is hot but does not have enough gas molecules to transfer heat to you (i.e., you would not feel the

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Answer:

thermosphere...........

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3 years ago

A person has legs of length L = 1.10 m. (a) If the maximum angle between the legs during walking is ϕ = 50o , what is the person

vampirchik [111]

Answer:

a). Maximum Length L=0.929m

b). T=0.83 Hz or 1.2s

c). Longer, the effortless waling T=2.1 Hz or t=0.475s

d). t=1.2s V=0.774 $\frac{m}{s}$

t=0.475s V=1.95 $\frac{m}{s}$

Explanation:

Length legs=L=1.1m

angle=50

the step that give the person forms a triangle whose two sides are known and the angle that forms between them, then using trigonometry as the image

Divide the original triangle in two and form a right triangle so the angle is 25 and the L is hypotenuse and the opposite is the step length

a).

$sin(\alpha) =\frac{op}{h}$

$op=h*sin(\frac{\alpha }{2})\\ op=1.1m*sin(\frac{50}{2})\\op=0.464m$

Length of the step

L=0.464m*2

L=0.928m

b).

period=T

$T=\frac{1}{time}=\frac{1}{t}\\ T=\frac{1}{1.2s}\\T=0.83 s^{-1}\\ T=0.83Hz$

c).

$T1=2\pi *\sqrt{\frac{L}{g}} \\T1=2\pi *\sqrt{\frac{1.1m}{9.8\frac{m}{s^{2}}}}\\ T1=2.1 Hz$

The period is the inverse of the time of the motion so, the T1 is faster that the T because

$t=\frac{1}{T1}=t= \frac{1}{2.1}=0.47s$

d).

The speed is the relation between the distance with time so:

$Vt=\frac{0.928m}{1.2s} \\Vt=0.773 \frac{m}{s} \\Vt=\frac{0.928m}{0.475s} \\Vt=1.953 \frac{m}{s}$

3 0

3 years ago

Which of the following is true?

lesantik [10]

Answer:

D

Explanation:

What it means to be an autotroph is that you make your own food and plants do that by photosynthesis to be a heterotroph is finding food from other sources like autotrophs or even other heterotrophs

3 0

3 years ago

Read 2 more answers

Tubby and his twin brother Libby have a combined mass of 200 kg and are zooming along in a 100 kg amusement park bumper car at 1

harkovskaia [24]

Answer: 14.1 m/s

Explanation:

We can solve this with the Conservation of Linear Momentum principle, which states the initial momentum $p_{i}$ (before the elastic collision) must be equal to the final momentum $p_{f}$ (after the elastic collision):

$p_{i}=p_{f}$ (1)

Being:

$p_{i}=m_{1}V_{i} + m_{2}U_{i}$

$p_{f}=m_{1}V_{f} + m_{2}U_{f}$

Where:

$m_{1}=200 kg +100 kg=300 kg$ is the combined mass of Tubby and Libby with the car

$V_{i}=10 m/s$ is the velocity of Tubby and Libby with the car before the collision

$m_{2}=25 kg + 100 kg=125 kg$ is the combined mass of Flubby with its car

$U_{i}=0 m/s$ is the velocity of Flubby with the car before the collision

$V_{f}=4.12 m/s$ is the velocity of Tubby and Libby with the car after the collision

$U_{f}$ is the velocity of Flubby with the car after the collision

So, we have the following:

$m_{1}V_{i} + m_{2}U_{i}=m_{1}V_{f} + m_{2}U_{f}$ (2)

Finding $U_{f}$ :

$U_{f}=\frac{m_{1}(V_{i}-V_{f})}{m_{2}}$ (3)

$U_{f}=\frac{300 kg(10 m/s-4.12 m/s)}{125 kg}$ (4)

Finally:

$U_{f}=14.1 m/s$

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3 years ago

A proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. Wha

Gwar [14]

A proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. The change in the electric potential energy of the proton-field system when the proton travels to x = 2.50m is -3.40 × 10⁻¹⁶ J (Option B)

<h3 /><h3>How is the change in electric potential energy of the proton-field system calculated?</h3>

Work done on the proton =Negative of the change in the electric potential energy of the proton field
In the given case, W = -qΔV
-W = qΔV
= qEcosθ
Therefore, work done on the proton = -e(8.50× $10^2$ N/C)(2.5m)(1)
= -3.40× $10^-^1^6$ J
Any change in the potential energy indicates the work done by the proton.
Therefore the positive sign shows that the potential energy increases when the proton does the work.
The negative sign shows that the potential energy decreases when the proton does the work.

To learn more about electric potential energy, refer

brainly.com/question/14306881

#SPJ4

3 0

2 years ago