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Maksim231197 [3]
3 years ago
9

While you are looking at the image of your feet in a plane mirror, you see a scratch in the glass. What is the approximate heigh

t of the scratch from the floor? Assume your eyes are 1.75m above your feet.​
Physics
1 answer:
Montano1993 [528]3 years ago
3 0

Answer:

1.75 / 2 = 0.875 m

Explanation:

Provided the mirror is vertical and you are standing upright.

It could be most any altitude if the mirror is allowed any orientation.

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A man in a gym is holding an 8.0-kg weight at arm's length, a distance of 0.55 m from his shoulder joint. What is the torque abo
kipiarov [429]

Answer:

  \tau =37.34\ N m

Explanation:

given,

mass of the weight = 8 Kg

distance = 0.55 m

angle below horizontal = 30°

torque about shoulder

  \tau = \vec{r} \times \vec{F}

  \tau = r \times F cos \theta

  \tau = 0.55 \times 8 \times 9.8 \times cos 30^0

  \tau =37.34\ N m

torque about his shoulder join is equal to   \tau =37.34\ N m

5 0
3 years ago
When you stand on tiptoes on a bathroom scale, there is an increase in
pychu [463]

Answer:

B) Pressure on the scale, not registered as weight.

Explanation:

This is because energy (derived from weight) becomes compiled on the tips of your toes, and therefore does not increase your weight, but simply the pressure at a smaller point

3 0
2 years ago
What is a concern about recovered memories?
Fittoniya [83]

Answer:

D because those are both concerning.

4 0
3 years ago
Read 2 more answers
A 8.4-mH inductor carries a current I = Imaxsin ωt, with Imax = 4.00 A and f = ω/2π = 60.0 Hz. What is the self-induced emf as a
Aleks04 [339]

Answer:

E= -3.166 cosωt   V

Explanation:

Given that

I = Imax sinωt

L= 8.4 m H

Imax= 4 A

f = ω/2π = 60.0 Hz

ω = 120π  rad/s

We know that self induce E given as

E=-L\dfrac{dI}{dt}

\dfrac{dI}{dt}= Imax \ \omega\ cos\omega t

E=-L\times Imax \ \omega\ cos\omega t

E=-8.4\times 120\times \pi \ cos\omega t

E= -3166.72 cosωt  m V

E= -3.166 cosωt   V

This is the induce emf.

3 0
2 years ago
In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole.
Fiesta28 [93]

Answer:

6.0 m/s

Explanation:

According to the law of conservation of energy, the total mechanical energy (potential, PE, + kinetic, KE) of the athlete must be conserved.

Therefore, we can write:

KE_i+PE_i =KE_f+PE_f

or

\frac{1}{2}mu^2+0=\frac{1}{2}mv^2+mgh

where:

m is the mass of the athlete

u is the initial speed of the athlete (at the bottom)

0 is the initial potential energy of the athlete (at the bottom)

v = 0.80 m/s is the final speed of the athlete (at the top)

g=9.8 m/s^2 is the acceleration due to gravity

h = 1.80 m is the final height of the athlete (at the top)

Solving the equation for u, we find the initial speed at which the athlete must jump:

u=\sqrt{v^2+2gh}=\sqrt{0.80^2+2(9.8)(1.80)}=6.0 m/s

4 0
3 years ago
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