The moons of Mars, Phobos (Fear) and Deimos (Terror), are very close to the planet compared to

A cubic sample of a new kind of artificial tissue is subject to an increase in pressure of 160 kPa which results in a reduction

grin007 [14]

Answer:

0.82 MPa

Explanation:

the change in pressure 'σ'=160kPa

K= σ/∈ $Ф_v$ => σ/3∈ $Ф_L$

K= 160/(3 x 0.065)

K=820 kPA=0.82 MPa

Thus,the bulk modulus of the tissue 'K' is 0.82 MPa

3 0

3 years ago

In which of the following elevator situations would the acceleration be positive. Select TWO answers

kap26 [50]

Both options 5 and 6

Explanation:

Let us consider option 5,

In option 5 body is moving up with initial velocity lower than that of final velocity which gets accelerated. Therefore the acceleration is positive in this case.

Let us consider option 6,

In option 6 body is moving down with initial velocity lower than that of final velocity which gets accelerated. Therefore the acceleration is positive in this case.

4 0

3 years ago

A baseball is thrown straight up from a building that is 25 meters tall with an initial velocity v = 10 m/s. How fast is it goin

Yanka [14]

Answer:-24,5m/s

Explanation: what we have here is a UALM with these gravity as acceleration (-9.8 m/s^2). The initial position is 25 m and initial speed is 10m/s.

Speed and gravity are increasing in the opposite direction, speed upwards and gravity downwards, while the position is also upwards, depending on your reference system.

The first thing I need to know is the maximum high it will reach.

Hmax=- S(0)^2/2g=

S= speed.

0= initial

G= gravity

Hm= 100/19,6= 5.1 m

So, the ball will go 5,1 m higher than the initial position, and from there it will fall free.

Then, I need to know how long it takes to fall. For that we use UALM equation:

X(t)= X(0) + S(0)*t + (A*t^2)/2.

X: position

S: speed

A: acceleration

T:time

0: initial

0 = 25m +10*t -(9.8 * t^2)/2

Solving the quadratic equation we get

T= 3,5 sec. ( Negative value for time is impossible)

So now we know that the ball to go up and then fall needs 3,5 sec.

Let's see how long it takes to go up:

30,1=25+10*t-4,9*t^2

0=-5,1+10*t-4,9*t^2

T= 1 sec. So it will take 1 sec to the ball to reach the maximum high and 0=speed and then it'll fall during the resting 2,5 sec

Finally, to know the speed just before it touches the ground, we use the following formula:

A= (St-S0)/t

-9.8m/s^2 = (St- 0m/s)/ 2,5s

-24,5 m/s= St

-24,5 m/s is the speed at 3,5 sec, which is the time just before falling

3 0

3 years ago

Which tuning forks will give rise to a beat frequency of 20 Hz when sounded together with a 240 Hz tuning fork?

Lubov Fominskaja [6]

Use the concept of beat frequency to find the applicable final freqeuncy for 20Hz beat frequency.

Beat can be defined as 'the interference pattern between two sounds of slightly different frequencies0

The expression for beat frequency is given as

$f_{beat} = |f_1-f_2|$

Where,

$f_2 =$ Final frequency

$f_1 =$ Initial frequency

The beat frequency for us is 25Hz and the initial frequency is 240Hz, then

$20= |f_2-240|$

Being an absolute value, two values are possible, both in addition and subtraction:

$f_2 = 240 \pm 20$

The two possible values are

$f_2 = 220Hz$

$f_2 = 260Hz$

3 0

3 years ago

Consider three force vectors Fi with magni- tude 53 N and direction 116º, F2 with mag- nitude 57 N and direction 217°, and F3 wi

Flauer [41]

Answer:

a. Fnet =37.67N

b. The direction = 133.4 from the x axis counter clockwise.

c. Option 2

Explanation:

Given that F1 is 53N at 116°, then it will be at a direction of 116-90=26° in the second quadrant.

Given that F2 is 57N at 116°, then it will be at a direction of 217-180=37° in the third quadrant..

Given that F1 is 71N at 20°, then it is in the first quadrant.

a. Fnet= F1+F2+F3

Fnet= -F1sin26i+F2cos26j-F2cos37i-F2sin37j+F3cos20i+F3sin20j

Fnet= 53sin26i+53cos26j-57cos37i-57sin37j+71cos20i+71sin20j

Resolving the vectors into x and y components.

Fnet= -2.04i+37.62j

Magnitude of the vector

Fnet= √((-2.04)^2+(37.62)^2)

Fnet= 37.67N

Fnet is approximately 38N.

b. Direction of the Fnet.

Angle=arctan(y/x)

Angle=arctan(-37.61/2.04)

Angle= -43.37°

The angle is in the negative x axis and positive y axis.

Then the direction becomes 180-43.37

Therefore, the direction of the net force is 133.37°.

c. The instantaneous velocity of a body is always in the direction of the net force at that instant. Option 2 is correct.

Fnet=ma

Fnet= mv/t

So the velocity is in the direction of the Fnet.

3 0

3 years ago