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ikadub [295]
3 years ago
14

Internal injuries can include which of the following? HELP ASAP (EDG ANATOMY 2020)

Physics
1 answer:
shepuryov [24]3 years ago
6 0

Answer: fractured rib, neck sprain, ruptured organ, torn tendon

Explanation: I just did it

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A wheel, starting from rest, rotates with a constant angular acceleration of 2.80 rad/s2. During a certain 5.00 s interval, it t
igomit [66]

Answer:

a) time t1 = 2.14s

b) initial angular speed w1 = 6 rad/s

Explanation:

Given that;

Initial Angular velocity = w1

Angular distance = s = 65 rad

time = t = 5 s

Angular acceleration a = 2.80 rad/s^2

Using the equation of motion;

s = w1t + (at^2)/2

w1 = (s-0.5(at^2))/t

Substituting the values;

w1 = (65 - (0.5×2.8×5^2))/5

w1 = 6rad/s

Time to reach w1 from rest;

w1 = at1

t1 = w1/a = 6/2.8 = 2.14s

a) time t1 = 2.14s

b) initial angular speed w1 = 6 rad/s

8 0
3 years ago
Read 2 more answers
From what height must an oxygen molecule fall in a vacuum so that its kinetic energy at the bottom equals the average energy of
alexandr1967 [171]

Answer:

The  value is  h  = 11930 \ m

Explanation:

From the question we are told that

    The  temperature is  T  =  300 \  K

     

Generally the root mean square speed of the  oxygen molecules is mathematically represented as

        v  =  \sqrt{\frac{3 *  R  *  T }{M} }  =  \sqrt{ 2 *  g  *  h }

Here  R is the gas constant with a value  R  =  8.314 \  J\cdot K^{-1} \cdot \  mol^{-1}

    M  is the molar mass of oxygen molecule with value M  =  0.032 \  kg /mol

So  

     \frac{3 *  8.314   *  300 }{0.032}   =  2 *  9.8  *  h

=>    h  = 11930 \ m

   

4 0
3 years ago
A horse was grazing at the top of a small hill. The horse got thirsty and ran toward a
Phoenix [80]

Answer:

I think it may be decreased, but i'm in 6th grade so-

Explanation:

7 0
3 years ago
A billiard ball is dropped from a height of 64 feet. Use the position function s(t) = –16???? 2 + ????0???? + ????0 to answer th
Delicious77 [7]

Answer:

s(t) = -16*t^2 + 64

v(t) = -32*t

a(t) = -32 ft/s^2

v(t) = 64 ft/s ... At impact

Explanation:

Given:-

- The height of the billiard ball t = 0 , h = 64 ft.

- The position function of an object under gravity is given by:

                                    s(t) = -16*t^2 + v_o*t + s_o

Find:-

a. Determine the position function s(t),

b. the velocity function v(t),

c. the acceleration function a(t).

d. What is the velocity of the ball at impact?

Solution:-

- To determine the position function we must initialize our problem and use the given general equation.

- s(t) is the position of the billiard ball from the ground at time t. So when t = 0, then s(t) = h. Hence, we have:

                                  s(t) = s_o = h = 64 ft

- Similarly we know that v_o is the initial velocity of the ball. Since, the ball was dropped we say that the initial velocity v_o = 0. Hence, the position of the ball from ground is given by following expression:

                                  s(t) = -16*t^2 + 64  

- To find the velocity expression v(t) we will take the time derivative of the position expression s(t) as follows:

                                  v(t) = d s(t) / dt

                                  v(t) = -16*2*t + 0

                                  v(t) = -32*t ft/s

- Similarly, the expression for acceleration a(t) is given by the time derivative of the velocity expression v(t) as follows:

                                  a(t) = d v(t) / dt

                                  a(t) = -32*t

                                  a(t) = -32 ft/s^2

- The velocity of ball at impact can be determined by evaluating s(t) = 0 and find the value for time t. Then that time t can be substituted in the velocity expression v(t) for final velocity. Or we could use the following 3rd kinematic equation as follows:

                                 v(t)^2 - 0^2 = 2*a(t)*s_o

                                 v(t)^2 = 2*(32)*(64)

                                 v(t) = 64 ft/s

- The ball has a velocity of 64 ft/s at impact!

6 0
3 years ago
Two equal charges with magnitude Q and Q experience a force of 12.3442 when held at a distance r. What is the force between two
Pepsi [2]

Answer:197.504 N

Explanation:

Given

Two Charges with magnitude Q experience a  force of 12.344 N

at distance r

and we know Electrostatic force is given

F=\frac{kq_1q_2}{r^2}

F=\frac{kQ\cdot Q}{r^2}

F=\frac{kQ^2}{r^2}

Now the magnitude of charge is 2Q and is at a distance of \frac{r}{2}

F'=\frac{k2Q\cdot 2Q}{\frac{r^2}{2^2}}

F'=16F

F'=197.504 N

4 0
3 years ago
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