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laila [671]
3 years ago
13

Arrange the skin layers according to the order in which a physician would cut them off during surgery, starting with the first a

t the top.
Physics
1 answer:
Mamont248 [21]3 years ago
5 0
Epidermis is the top layer of the skin, the part of the skin you see.

Dermis is the second layer of skin. It's much thicker and does a lot for your body.

Subcutaneous fat is the bottom layer.

Hope I helped.
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A vehicle travelling at an initial velocity of 20km/hr,accelerates at 4m/s².calculate its final velocity after 10 seconds.​
Nikitich [7]

acceleration = Velocity changes ÷ time of the velocity changes

4 m/s^2 =

4 × 10^(-3) × 3600 km / h =

4 × 3.6 =

14.4 km / h

Thus :

14.4 = V(2) - V(1) / t(2) - t(1)

14.4 = V(2) - 20 / 10

Multiply both sides by 10

10 × 14.4 = 10 × ( V(2) - 20 ) / 10

144 = V(2) - 20

Add both sides 20

144 + 20 = V(2) - 20 + 20

V(2) = 164 Km/h

Thus the final velocity after 10 seconds is 164 Km/h .

4 0
3 years ago
In humans, which part of the brain contains 80 percent of its weight? cerebrum cerebellum spiral cord brain stem
scZoUnD [109]
I believe it's the Cerebrum
6 0
3 years ago
Explain why secondary succession usually occurs more rapidly than primary succession.
kiruha [24]

Answer:

Explanation:

As it mostly occurs on a surface where an ecosystem has previously existed. Primary succession only occurs on a surface where no ecosystem existed ever before!

6 0
3 years ago
A motorcycle starts to move from rest. If the velocity of the motorcycle becomes 90 km/hr
bagirrra123 [75]

Explanation:

90 kmhr—1 x 1000/3600 = 25ms—1

U = 0 ms—1

V = 25ms—1

t = 10 s

a = ?

a = V - U/t

a = 25 - 0/10

a = 25/10

a = 2.5 ms—1

6 0
3 years ago
A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05
7nadin3 [17]

Answer:

W_{drag} = 4.223\,J

Explanation:

The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:

U_{g,A}+K_{A} = U_{g,B} + K_{B} + W_{drag}

The work done on the ball due to drag is:

W_{drag} = (U_{g,A}-U_{g,B})+(K_{A}-K_{B})

W_{drag} = m\cdot g\cdot (h_{A}-h_{B})+ \frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2})

W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]

W_{drag} = 4.223\,J

7 0
3 years ago
Read 2 more answers
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