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Serga [27]
3 years ago
7

Jax fell while running and cut his knee. He noticed at the end of the day that his body had produced a hard film over his scrape

. After two weeks, his scrape was completely gone and new skin had formed where his scrape had been. Which bodily system was repaired in order to protect the soft body tissues below Jax's skin?
Physics
1 answer:
Kruka [31]3 years ago
7 0

Answer:

INTEGUMENTARY

Explanation:

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Rocket-powered sleds are been used to test the responses of humans to acceleration. Starting from rest, one sled can reach a spe
Greeley [361]

Answer:cho  v₀ =0s  

α=Δv/Δt

Explanation:

\frac{0-495}{2,16-1,78}

=-1302,631579

chuyển động chậm dầnđều

3 0
3 years ago
When a falling meteoroid is at a distance above the Earth's surface of 3.40 times the Earth's radius, what is its acceleration d
Alchen [17]

Answer:

g = 0.85 ms^{-2}

Explanation:

g = \frac{GM}{h^{2} }

were; g is the acceleration due to Earth's gravity, G is Newton's gravitation constant (6.674 x 10^{-11} Nm^{2}kg^{-2}), M is the mass of the earth (5.972 x 10^{24} kg), and h is the distance of meteoroid to the earth.

h = 3.40 x R

  = 3.40 x 6371 km

h = 21661.4 km

  = 21661400 m

Thus,

g = \frac{6.674*10^{-11}*5.972*10^{24}  }{(21661400)^{2} }

  = \frac{3.9857 *10^{14} }{4.6922*10^{14} }

  = 0.84944

g = 0.85 ms^{-2}

The acceleration due to the Earth's gravitation is 0.85 ms^{-2}.

6 0
3 years ago
Name the 2 types of tissue that form your skin?
QveST [7]

Answer:

Epithelial tissue and Muscle tissue

Explanation:

7 0
3 years ago
Consider a Young's two-slit experiment in which the wavelength of light is 37% smaller than the distance between the slits. How
Diano4ka-milaya [45]

Answer: The answer is 3

Explanation:

4 0
3 years ago
At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There
katen-ka-za [31]

a) Constant

b) Constant

Explanation:

a)

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

\tau_{net} = I \alpha (1)

where

\tau_{net} is the net torque acting on the object in rotation

I is the moment of inertia of the object

\alpha is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

\alpha = \frac{\Delta \omega}{\Delta t}

where

\Delta \omega is the change in angular velocity

\Delta t is the time interval

So we can rewrite eq.(1) as

\tau_{net}=I\frac{\Delta \omega}{\Delta t}

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

\tau_{net}=0

From the previous equation, this implies that

\Delta \omega =0

Which means that the angular velocity at that instant does not change anymore.

b)

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

\tau_{net}=0

Therefore, this means that

\Delta \omega =0

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

So cases (a) and (b) are basically the same situation, as the net torque is zero in both cases, so the object acts in the same way.

8 0
3 years ago
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