How much time will it take for a bug to travel 2 meters across the floor if it is traveling at 0.5 m/s?

Which of the following is defined as an area of the body surface that is innervated by a single spinal nerve?

romanna [79]

Answer:

Dermatome. (Ans. C).

Explanation:

Dermatome is defined as the area of the human anatomy skin which is supplied by single spinal sensory nerve root. At the spinal cord these spinal sensory nerve enter the nerve root, and the branches of spinal sensory reach to the periphery of the body.

The sensory nerve which is present in the periphery of the body are the type of nerve which helps to transmit signals from sensation such as pain, temperature, etc. to the spinal cord from some specific area of the anatomy.

3 0

3 years ago

What is the energy of a mole of photons that have a wavelength of 413 nm? (h = 6.626 × 10⁻³⁴ J • s and c = 3.00 × 10⁸ m/s)

Triss [41]

Answer:

Energy of one mole of photon will be $2.89\times 10^5J$

Explanation:

We have given wavelength of photon $\lambda =413nm=413\times 10^{-9}m$

Velocity of light is given $c=3\times 10^8m/sec$

Plank's constant $h=6.626\times 10^{-34}Js$

Energy of the photon is given by $E=\frac{hc}{\lambda }=\frac{6.626\times 10^{-34}\times 3\times 10^8}{413\times 10^{-9}}=0.048\times 10^{-17}J$

We have to find the energy of one mole of photon

One mole of photon is equal to $6.023\times 10^{23}photon$

So energy of one mole of photon will be equal to $6.023\times 10^{23}\times 0.048\times 10^{-17}=2.89\times 10^5J$

So energy of one mole of photon will be $2.89\times 10^5J$

6 0

3 years ago

1. Differentiate between speed and velocity.

Mazyrski [523]

Answer:

1. Speed and velocity both involve a numeric rate describing the distance traveled by a body in a unit of time. However, speed describes the rate of a body traveling in any direction in a unit of time, while velocity describes the rate of a body traveling in a particular direction in a unit of time.

2. Answers may vary, but should resemble the following:

Average velocity explains the velocity the body traveled overall, not taking into consideration each spot in the trip. If a car moves at 65 km/h on average, it may have slowed down for some parts and sped up for others. Overall though, it would have made a certain distance of travel within a specified unit of time that totals the average velocity of 65 km/h.

Instantaneous velocity explains the velocity of a body at a particular instant of the trip. The instantaneous velocity of a car stopped at a stop sign would be 0 m/s even if it was moving before and will continue to move after this stop. The velocity at that particular instant is the instantaneous velocity.

Uniform velocity is when the distance being covered is changing uniformly with time. For example, if a car moves 20 km every 30 minutes and continues to do so in the same direction, it's traveling with a uniform velocity.

3. a=v2−v1t

a=20 m/s−60 m/s6 s

a=−406

a = –6.7 m/s2

4. v2 = v1 + at

v2 = 14 m/s + (3 m/s2 × 6 s)

v2 = 14 + 18

v2 = 32 m/s

5. v=st

v=375 km5 h

v = 75 km/h

6. First, convert the minutes to seconds. Since there are 60 seconds in one minute, multiply:

60 × 15 (minutes) = 900 seconds

s = v × t

s = 6 m/s × 900 s

s = 5,400 m

7. t=sv

t=80 km35 km/hr

t = 2.29 hr

8. a=v2−v1t

a=50 m/s−15 m/s4 s

a=35 m/s4 s

a = 8.75 m/s2

9. vav=v1+v22

vav=15 m/s+50 m/s2

vav=65 m/s2

vav = 32.5 m/s

10. a=v2−v1t

a=0 m/s−11.5 m/s3.5 s

a = –3.29 m/s2

Explanation:

7 0

3 years ago

A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top

Anton [14]

Answer:

$y_{max} = 0.829\,m$

Explanation:

Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

$v = \sqrt{(0.8\,\frac{m}{s})^{2} + 2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m)}$

$v = 2.913\,\frac{m}{s}$

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

$(3\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m) + \frac{1}{2}\cdot (3\,kg)\cdot (2.913\,\frac{m}{s} )^{2} = (3\,kg) \cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s) ^{2}$

After some algebraic handling, a second-order polynomial is formed:

$12.728\,J = \frac{1}{2}\cdot (2000\,\frac{N}{m} )\cdot (\Delta s)^{2} - (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \Delta s$

$1000\cdot (\Delta s)^{2}-29.421\cdot \Delta s - 12.728 = 0$

The roots of the polynomial are, respectively:

$\Delta s_{1} \approx 0.128\,m$

$\Delta s_{2} \approx -0.099\,m$

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

$\Delta s \approx 0.128\,m$

The maximum height that the block reaches after rebound is:

$(3\,kg) \cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s)^{2} = (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot y_{max}$

$y_{max} = 0.829\,m$

4 0

3 years ago

Read 2 more answers

A very myopic man has a far point of 32.3 cm. what power contact lens (when on the eye) will correct his distant vision?

Alenkasestr [34]

p=3.0D contact lens is needed to correct his distant vision.

The person is suffering from myopia and hence need a concave lens to correct the defect. The lens should be such that an object at infinity must form its image at the far point.

Hence,f= -32.3cm

= -0.323m

we can define the power of the lens as the reciprocal of its focal length in metre.

The power of lens is the ability to converge and diverge rays of light.

The power of the lens can be obtained as:

p= 1/f

p=1/(-0.323)

p=3.0D

Thus,the power is 3.0D to correct his distant vision.

learn more about power of lens from here: brainly.com/question/17166887

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5 0

9 months ago