What type of system is earth?<br> A.Closed<br> B.Open<br> C.Connected<br> D.Isolated

During each heartbeat, about 80 g of blood is pumped into the aorta inapproximately 0.2 s. During this time, the blood is accele

taurus [48]

Answer:

Work is done by the heart on the blood during this time is 0.04 J

Explanation:

Given :

Mass of blood pumped, m = 80 g = 0.08 kg

Initial speed of the blood, u = 0 m/s

Final speed of the blood, v = 1 m/s

Initial kinetic energy of blood is determine by the relation:

$E_{1}=\frac{1}{2} m u^{2}$

Final kinetic energy of blood is determine by the relation:

$E_{2}=\frac{1}{2} m v^{2}$

Applying work-energy theorem,

Work done = Change in kinetic energy

W = E₂ - E₁

$W=\frac{1}{2} m (v^{2}-u^{2})$

Substitute the suitable values in the above equation.

$W=\frac{1}{2}\times0.08\times (1^{2}-0^{2})$

W = 0.04 J

6 0

3 years ago

Near the surface of Earth an electric field points radially downward and has a magnitude of approximately 100 N/C. What charge (

pentagon [3]

Answer:

$q=2.997\times 10^{-4}$ C

Sign-Negative

Explanation:

We are given that

Electric field = $E=100NC^{-1}$ (Radially downward)

Acceleration= $0.19 ms^{-2}$ (Upward)

Mass of charge=3 g= $3\times 10^{-3}$ kg

1kg=1000g

We have to find the magnitude and sign of charge would have to be placed on a penny .

By newton's second law

$\sum F_y=ma$

$\sum F_y=qE-mg$

Substitute the values then we get

$qE-mg=ma$

Substitute the values then we get

$q(100)-3\times 10^{-3}(9.8)=3\times 10^{-3}(0.19)$

$100q-29.4\times 10^{-3}=0.57\times 10^{-3}$

$100q=0.57\times 10^{-3}+29.4\times 10^{-3}=29.97\times 10^{-3}$

$q=\frac{29.97\times 10^{-3}}{100}$

$q=2.997\times 10^{-4}$ C

Sign of charge =Negative

Because electric force acting in opposite direction of electric field therefore,charge on penny will be negative.

4 0

3 years ago

Two astronauts push on a satellite. One pushes in the positive x direction with a force of 42 N, and the other pushes in the pos

inessss [21]

The vectors adition we can find the magnitude of the force applied by the other astronaut is 11.25 N in the y direction

Parameters given

Force of an astronaut Fₓ = 42 N

Angle θ = 15º

To find

Force another astronaut

The force is a vector magnitude for which the addition of vectors must be used, a very efficient method to perform this sum is to add the components of each vector and devise constructing the resulting vector using trigonometry and the Pythagorean theorem.

Let's use trigonometry to find the other force

tan θ = $\frac{F_y}{F_x}$

F_ y = Fₓ tan θ

let's calculate

F_y = 42 tan 15

F_y = 11.25 N

Using the summation of vectors we can find the magnitude of the force applied by the other astronaut is 11.25 N in the y direction

Learn more about vector addition here:

brainly.com/question/15074838

4 0

3 years ago

HELP PLEASE!! An illustration of a pendulum at 3 positions of a pendulum. The equilibrium position is labeled B. The other two p

Sedbober [7]

Answer:

Both A and C

Explanation:

I just got it correct on Edg

7 0

3 years ago

Read 2 more answers

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

3 years ago

What type of system is earth? A.Closed B.Open C.Connected D.Isolated