The maximum pressure achieved during ventricular contraction is called

How is a wave of red light different from a wave of purple?

adoni [48]

Our eyes perceive different wavelengths of light as the rainbow hues of colors. Red light has relatively long waves, around 700 nm long. Blue and purple light have short waves, around 400 nm. Shorter waves vibrate at higher frequencies and have higher energies.

Hope it helps.

4 0

2 years ago

When a gear rotates 20 revolutions, it achieves an angular velocity of 30 rad/s, starting from rest. determine its constant angu

Temka [501]

Initial angular velocity = 0 (starting from rest)
Final angular velocity = 30 rad/s
Distance traveled = (20 rev)*(2π rad/rev) = 40π rad

Let the angular acceleration be α rad/s².
Then
(30 rad/s)² = (0 rad/s)² + 2*(α rad/s²)*(40π rad)
α = 3.58 rad/s²

Answer: 3.58 rad/s² (nearest hundredth)

4 0

3 years ago

A body is traveling at 5.0 m/s along the positive direction of an x axis; no net force acts on the body. An internal explosion s

loris [4]

To solve this problem it is necessary to apply the concepts related to the conservation of momentum and conservation of kinetic energy.

By definition kinetic energy is defined as

$KE = \frac{1}{2} mv^2$

Where,

m = Mass

v = Velocity

On the other hand we have the conservation of the moment, which for this case would be defined as

$m*V_i = m_1V_1+m_2V_2$

Here,

m = Total mass (8Kg at this case)

$m_1=m_2 =$ Mass each part

$V_i =$ Initial velocity

$V_2 =$ Final velocity particle 2

$V_1 =$ Final velocity particle 1

The initial kinetic energy would be given by,

$KE_i=\frac{1}{2}mv^2$

$KE_i = \frac{1}{2}8*5^2$

$KE_i = 100J$

In the end the energy increased 100J, that is,

$KE_f = KE_i KE_{increased}$

$KE_f = 100+100 = 200J$

By conservation of the moment then,

$m*V_i = m_1V_1+m_2V_2$

Replacing we have,

$(8)*5 = 4*V_1+4*V_2$

$40 = 4(V_1+V_2)$

$V_1+V_2 = 10$

$V_2 = 10-V_1$ (1)

In the final point the cinematic energy of EACH particle would be given by

$KE_f = \frac{1}{2}mv^2$

$KE_f = \frac{1}{2}4*(V_1^2+V_2^2)$

$200J=\frac{1}{2}4*(V_1^2+V_2^2)$ (2)

So we have a system of 2x2 equations

$V_2 = 10-V_1$

$200J=\frac{1}{2}4*(V_1^2+V_2^2)$

Replacing (1) in (2) and solving we have to,

200J=\frac{1}{2}4*(V_1^2+(10-V_1)^2)

PART A: $V_1 = 10m/s$

Then replacing in (1) we have that

PART B: $V_2 = 0m/s$

8 0

3 years ago

A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria

aleksandrvk [35]

Answer:

a) in order to be maximum the length of the square wire should be 10m and no cut for making the triangle

b) in order to be minimum the length of the square wire should be 3.84 m and the triangle wire should be 6.16 m

Explanation:

for a square with a side length of a and a equilateral triangle of length b

area square = a²

area triangle = base* height/2

for an equilateral triangle height = base * sin 60 = (√2 / 2 )* base

therefore

area triangle = base* height/2 = base² (√2 /4) =(√2 /4) b²

the total length of the wire is = square length + triangle length = 4*a + 3*b

therefore

A=a²+(√2 /4) b²

4*a + 3*b=L→ b = (L-4*a)/3

A= a²+(√2 /4) (L-4*a)²/ 9

the maximum and minimum amount can be found taking the derivative of the area respect with a:

dA/da= 2*a + (√2 /36) 2*(L-4*a)*(-4) = 0

a -(√2 /9) (L-4*a) = 0

a - √2 /9 * L + √2 /9*4*a =0

(√2 /9*4+1)*a = √2 /9 * L

a= √2 /9 * L / (√2 /9*4+1) = √2 /9 * L / (√2 /9*4+1) = √2 /9 * 10 m / (√2 /9*4+1)

a= 0.96 m

therefore

b = (L-4*a)/3 = (10 m - 4*0.96m)/3= 2.053 m

A=a²+(√2 /4) b² = (0.96 m)²+ (√2 /4) (2.053m)² = 2.411 m²

in the extreme cases

a= 10 m/4=2.5 m and b=0

thus A= (2.5 m)² = 6.25 m²

b=10 m/3= 3.33 m and a=0

thus A= (√2 /4) (3.33 m)² = 3.92 m²

therefore minimum area A=3.92 m² with Length 1=4*0.96 m=3.84 m , Length 2 =2.053 m*3 = 6.16 m

the maximum area is A=6.25 m² with Length 1=10 m and Length 2=0 m

7 0

3 years ago

(a) When t = 2.00 s, calculate the magnitude of the force exerted on an electron located at point P1, which is at a distance r1

Bezzdna [24]

The complete question is;

Within the green dashed circle shown in the figure below, the magnetic field changes with time according to the expression B = 6.00t³ − 1.00t² + 0.800, where B is in teslas, t is in seconds, and R = 2.20 cm.

(a) When t = 2.00 s, calculate the magnitude of the force exerted on an electron located at point P1, which is at a distance r1 = 5.7 cm from the center of the circular field region.

I have attached the figure the question talks about.

Answer:

Force = 4.62 x 10^(-20) N

Explanation:

First of all,

∮E.ds = (d/dt)∮B.dA

E∮ds = (d/dt)B∮dA

dA is the area where B is not equal to zero.

Thus,

E•2πr = (d/dt)B•πR²

|E| = [(d/dt)B•πR²]/2πr

F = |qE| = q[(d/dt)B•πR²]/2πr

Where q is charge on electron = 1.6 x 10^(-19) C

F is magnitude of force exerted on electron

F = 1.6 x 10^(-19)[(d/dt)B•R²]/2r

F = 0.8 x 10^(-19)[(d/dt)B•R²]/r

Now, dB/dt = 18t² - 2t

Thus,

F = 0.8 x 10^(-19)[(18t² - 2t)•R²]/r

Thus, at t=2 and R=2.2cm = 0.022m and r = 5.7cm = 0.057m

Thus,

F = 0.8 x 10^(-19)[(18x2²) - (2x2)•(0.022²)]/(0.057)

F = (0.8 x 10^(-19) x 0.032912)/0.057 = 4.62 x 10^(-20) N

3 0

3 years ago