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PilotLPTM [1.2K]
3 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]3 years ago
4 0

Answer:

D

Explanation:

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Is it possible for an object to be going 100 miles per hour but have a velocity of zero?
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It is possible for on object to be going at 100 miles per hour, but still have a velocity. This is because the object going at 100 miles per hour has speed, which is a scalar quantity, which is defined by only magnitude, but the velocity of the object can be 0, since velocity is a vector quantity which is defined by both magnitude and direction.

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A force is always needed to slow an object down.<br><br> A True<br> B false
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3 years ago
What is the physical state of water at 250°C?
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3 years ago
In an experiment designed to measure the strength of a uniform magnetic field produced by a set of coils, electrons are accelera
Arlecino [84]

Answer:

the magnitude of the magnetic field is 8.704 x 10⁻⁴ T

Explanation:

Given;

potential difference, V =  278 V

radius of the circular path, r = 6.46 cm = 0.0646 m

charge of electron, q = 1.60218 × 10⁻¹⁹ C

mass of electron, m = 9.10939 × 10⁻³¹ kg

The magnitude of the magnetic field is given as;

B = \frac{M_e*v}{q*r}

where;

B is the magnitude of the magnetic field

M_e is mass of the electron

v is velocity of the electron

r is the radius of the circular path

q is charge of the electron

Determine velocity of the electron from kinetic energy equation;

K = \frac{1}{2} M_ev^2\\\\Vq = \frac{1}{2} M_ev^2\\\\v^2 = \frac{2qV}{M_e} \\\\v = \sqrt{\frac{2qV}{M_e}} = \sqrt{\frac{2*1.602*10^{-19}*278}{9.109*10^{-31}}} = 9.8886*10^{6} \ m/s

the magnitude of the magnetic field:

B = \frac{M_e*v}{q*r} \\\\B = \frac{(9.109*10^{-31})*(9.8886*10^6)}{(1.602*10^{-19})*(0.0646)} = 8.704*10^{-4} \ T

Therefore, the magnitude of the magnetic field is 8.704 x 10⁻⁴ T

5 0
3 years ago
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