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PilotLPTM [1.2K]
4 years ago
13

A woman wearing athletic clothing outside. The woman is stretching her legs in a lunge position. Based on the nonverbal messages

you see in this photograph, which of the following most likely describes this person’s situation? a. This woman is a successful and confident business owner. b. This woman is an overworked and overly stressed medical doctor. c. This woman is a stay-at-home mom who is very conscious of her health. d. There are not enough nonverbal messages to say for sure. Please select the best answer from the choices provided A B C D
Physics
1 answer:
Inessa05 [86]4 years ago
4 0

Answer:

D

Explanation:

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A spaceship approaches the earth with a speed 0.50c. A passenger in the spaceship measures his heartbeat as 9- beats per minute.
JulijaS [17]

Answer:

D) 61 beats per minute

Explanation:

5 0
3 years ago
Helpp .....meeee pleaseee
Lady bird [3.3K]

Answer:

hydrogen

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Explanation:

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5 0
3 years ago
A 100-kg astronaut is floating in outer space. If the astronaut throws a 2-kilogram wrench at a speed of 10 meters per second, w
sergey [27]

Since Astronaut and wrench system is isolated in the space and there is no external force on it

So here momentum of the system will remain conserved

so here we can say

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

initially both are at rest

so here plug in all values

0 = 100 v_{1f} + 2\times 10

v_{1f} = -0.20 m/s

so here the astronaut will move in opposite direction and its speed will be equal to 0.20 m/s

3 0
3 years ago
By experiment, determine what makes a force attractive or repulsive. Describe your experiments and observations with some exampl
drek231 [11]

The charge present determines a force to be attractive or repulsive.

The charges acquired by two bodies determines the Force as Attractive Or Repulsive.

Electric Force applied due to Electrical charges is same in magnitude but opposite in direction. This corresponds this phenomenon equivalent to the Newton's Third Law.

Examples of the experiments and observations:

  • On combing hair through a comb and then keeping it close to small pieces of paper shows attraction of paper pieces towards the comb.

This occurs due to the Electric charges present in the comb that induces charge in paper pieces leading to their attraction.

  • In both Gravitational Force and Coulomb force, the force remains inversely proportional to the square of the distance following the Inverse Square Law being the Central Force system. This only differs by the fact that in Gravitational Force, masses are used and in Coulomb force, charges are used.

The more the distance between the charges, the less is the Electric Force.

The lesser the distance between the charges, the more is the Electric Force.

If both the objects are charged the same i.e. either positive or negative then the Force is Repulsive and if the charges are Oppositely charged then the force is attractive.

Hence, the charge present determines a force to be attractive or repulsive.

Learn more about Coulomb Force here, brainly.com/question/15451944

#SPJ4

6 0
2 years ago
A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 152 om and makes an angl
aliya0001 [1]

Answer:

D₂= 167,21 cm : Magnitude  of the second displacement

β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement

Explanation:

We find the x-y components for the given vectors:

i:  unit vector in x direction

j:unit vector in y direction

D₁: Displacement Vector 1

D₂: Displacement Vector 2

R= resulta displacement vector

D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j

D₂= -D₂(i)-D₂(j)

R=  131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j

We propose the vector equation for sum of vectors:

D₁+ D₂= R

-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j

-51.99i+D₂x(i)=103.23i

D₂x=103.23+51.99=155.22 cm

+142.83j-D₂y(j) =+80.65j

D₂y=142.83-80.65=62.18 cm

Magnitude and direction of the second displacement

D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2}  }

D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2}  }

D₂= 167.21 cm

Direction of the second displacement

\beta = tan^{-1} \frac{D_{y}}{D_{x} }

\beta = tan^{-1} \frac{62.18}{155.22 }

β= 21.8°

D₂= 167,21 cm : Magnitude  of the second displacement

β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

6 0
4 years ago
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