answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Elena-2011 [213]
2 years ago
5

A healthy astronaut's heart rate is 60 beats/min. Flight doctors on Earth can monitor an astronaut's vital signs remotely while

in flight.How fast would an astronaut be flying away from Earth if the doctor measured her having a heart rate of 21 beats/min?
Physics
2 answers:
Anton [14]2 years ago
5 0
GIVEN:
   60 beats per minute
   21 beats per minute
   find x= how fast would an astronaut be flying away
 1            x
-----   *  ------ = (60x = 21)  -------> 60x = 21    ------------>  x= 0.35 
60         21                                  -------   -----
                                                     60      60

The answer is 0.35 seconds which refers to how fast would an astronaut be flying away from the earth if he has a heart rate of 21 beats/min. 
amm18122 years ago
4 0

Answer:

2.81 x 10⁸ m/s

Explanation:

R₁ = rate of heartbeat on earth = 60 beats/min

T₁ = Time period for 1 beat on earth = 1/R₁ = 1/60 min = 60/60 sec = 1 sec  

R₂ = rate of heartbeat on location of astronaut = 21 beats/min

T₂ = Time period for 1 beat at the location of astronaut = 1/R₂ = 1/21 min = 60/21 sec

v = speed of astronaut

c = speed of light = 3 x 10⁸ m/s

Using the equation

T₁ = T₂ sqrt(1 - (v/c)²)

1 = (60/21) sqrt(1 - (v/(3 x 10⁸))²)

v = 2.81 x 10⁸ m/s

You might be interested in
A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diameter pipe at
GaryK [48]

Answer:

50.93 m/s

199.5 kW

Explanation:

From the question, the nozzle exit diameter = 5 cm, Radius= diameter/2= 5cm/2= 2.5cm. we can convert it to metre for unit consistency= (2.5×0.01)=

0.025m

We can calculate the The cross sectional area of the nozzle as

A= πr^2

A= π ×0.025^2

= 1.9635 ×10^- ³ m²

From the question, the water is moving through the pipe at a rate of 0.1 m /s , then for the water to move through it at a seconds, it must move at

(0.1 / 1.9635 ×10^- ³ m²)

= 50.93 m/s

During the Operation of the pump, the Dynamic energy of the water= potential energy provided there is no loss during the Operation

mgh = 1/2mv²

We can make "h" subject of the formula, which is the height of required head of water

h = (1/2mv²)/mg

h= v² / 2g

h = 50.93² / (2 ×9.81)

h = 132.21m

From the question;

The total irreversible head loss of the system = 3 m,

the given position of nozzle = 3 m

the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )

=(3 + 3 + 132.21m)

=138.21m

mass of water pumped in a seconds can be calculated since we know that mass is a product of volume and density

Volume= 0.1m³

Density of sea water=1030 kg/m

(0.1 m^3× 1030)

= 103kg

We can calculate the Potential enegry, which is = mgh

= (103 ×9.81 × 138.21)

= 139651.5 Watts

= 139.65kW

To determine required shaft power input to the pump and the water discharge velocity

Energy= efficiency × power

But we are given efficiency of 70 percent, then

139651.5 Watts = 0.7P

=199502.18 Watts

P=199.5 kW

Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW

5 0
1 year ago
Steam undergoes an adiabatic expansion in a piston–cylinder assembly from 100 bar, 360°C to 1 bar, 160°C. What is work in kJ per
vfiekz [6]

Answer:

work is 130.5 kJ/kg

entropy change is 1.655 kJ/kg-k

maximum  theoretical work is 689.4 kJ/kg

Explanation:

piston cylinder assembly

100 bar, 360°C to 1 bar, 160°C

to find out

work  and amount of entropy  and magnitude

solution

first we calculate work i.e heat transfer - work =   specific internal energy @1 bar, 160°C  - specific internal energy @ 100 bar, 360°C    .................1

so first we get some value from steam table with the help of 100 bar @360°C and  1 bar @ 160°C

specific volume = 0.0233 m³/kg

specific enthalpy = 2961 kJ/kg

specific internal energy = 2728 kJ/kg

specific entropy = 6.004 kJ/kg-k

and respectively

specific volume = 1.9838 m³/kg

specific enthalpy = 2795.8 kJ/kg

specific internal energy = 2597.5 kJ/kg

specific entropy = 7.659 kJ/kg-k

now from equation 1 we know heat transfer q = 0

so - w =   specific internal energy @1 bar, 160°C  - specific internal energy @ 100 bar, 360°C

work = 2728 - 2597.5

work is 130.5 kJ/kg

and entropy change formula is i.e.

entropy change =  specific entropy ( 100 bar @360°C)  - specific entropy ( 1 bar @160°C )

put these value we get

entropy change =  7.659 - 6.004

entropy change is 1.655 kJ/kg-k

and we know maximum  theoretical work = isentropic work

from steam table we know specific internal energy is 2038.3 kJ/kg

maximum  theoretical work = specific internal energy - 2038.3

maximum  theoretical work = 2728 - 2038.3

maximum  theoretical work is 689.4 kJ/kg

3 0
2 years ago
A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an automobile at a distance of 0.24 m behind the mir
Semmy [17]

Answer:

The distance and height of the object  is 6 m and 2 m.

The image is virtual and upright.

Explanation:

Given that,

Focal length = 0.25 m

Length of image = 0.080 m

Image distance = 0.24 m

We need to calculate the distance of the object

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{0.24}=\dfrac{1}{0.25}+\dfrac{1}{u}

\dfrac{1}{u}=\dfrac{1}{0.24}-\dfrac{1}{0.25}

\dfrac{1}{u}=\dfrac{1}{6}

u=6\ m

We need to calculate the magnification

Using formula of magnification

m=-\dfrac{v}{u}

Put the value into the formula

m=-\dfrac{0.24}{-6}

m=0.04

We need to calculate the height of the object

Using formula of magnification

m=\dfrac{h'}{h}

h=\dfrac{0.080}{0.04}

h=2\ m

A convex mirror produce a virtual and upright image behind the mirror.

Hence, The distance and height of the object  is 6 m and 2 m.

The image is virtual and upright.

6 0
2 years ago
Read 2 more answers
A cylindrical tank of methanol has a mass of 40 kgand a volume of 51 L. Determine the methanol’s weight, density,and specific gr
mezya [45]

Answer:

Weight  W = 392.4 N

Density  \rho = 784.31 \frac{kg}{m^{3} }

Specific gravity S = 0.78431

Force required F = 10 N

Explanation:

Given data

Mass (m) = 40 kg

Volume (V) = 0.051 m^{3}

Weight W = m × g

⇒ W = 40 × 9.81

⇒ W = 392.4 N

This is the weight of the methanol.

Density \rho = \frac{mass }{volume}

⇒ \rho = \frac{40}{0.051}

⇒ \rho = 784.31 \frac{kg}{m^{3} }

This is the density of the methanol.

Specific gravity (S) = \frac{\rho}{\rho_{water} }

⇒ S = \frac{784.31}{1000}

⇒ S = 0.78431

This is the specific gravity of the methanol.

Force needed to accelerate this tank F = ma

⇒ F = 40 × 0.25

⇒ F = 10 N

This is the force required to accelerate the tank.

4 0
2 years ago
We observe that a moving charged particle experiences no magnetic force. From this we can definitely conclude that:_______
Leto [7]

Answer:

b. the particle must be moving parallel to the magnetic field.

Explanation:

The magnetic force on a moving charged particle is given by;

F = qvBsinθ

where;

q is the charge of the particle

v is the velocity of the particle

B is the magnetic field

θ is the angle between the magnetic field and velocity of the moving particle.

When is the charge is stationary the magnetic force on the charge is zero.

Also when the charge is moving parallel to the magnetic field, the magnetic force is zero.

Therefore, when a moving charged particle experiences no magnetic force, we can definitely conclude that the particle must be moving parallel to the magnetic field.

b. the particle must be moving parallel to the magnetic field.

5 0
2 years ago
Other questions:
  • How much total work is done by the force in lifting the elevator from 0.0 m to 9.0 m?
    8·1 answer
  • A physical change occurs when a material changes shape or size but the composition of the material does not change. True or Fals
    10·1 answer
  • A coin completes 18 spins in 12 seconds. The centripetal acceleration of the edge of the coin is 2.2 m/s2. The radius of the coi
    8·2 answers
  • An iron ball and an aluminum ball of mass 100 g each are heated to the same temperature and then cooled to a temperature of 20°C
    8·1 answer
  • Two disks with the same rotational inertia i are spinning about the same frictionless shaft, with the same angular speed ω, but
    8·1 answer
  • A turtle takes 3.5 minutes to walk 18 m toward the south along a deserted highway. A truck driver stops and picks up the turtle.
    11·1 answer
  • A backyard swimming pool with a circular base of diameter 6.00 m is filled to depth 1.50 m. (a) Find the absolute pres- sure at
    6·1 answer
  • A truck is using a hook to tow a car whose mass is one quarter that of the truck. If the force exerted by the truck on the car i
    12·1 answer
  • a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres
    11·1 answer
  • Consider four different oscillating systems, indexed using i = 1 , 2 , 3 , 4 . Each system consists of a block of mass mi moving
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!