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Molodets [167]
3 years ago
13

Mary's friend observes the investigation and claims that block B has more mass. If there were no errors in the investigation, wh

ich of the following must be TRUE for this claim to be correct?
Physics
2 answers:
Elena L [17]3 years ago
6 0
This is a true claim.
Keith_Richards [23]3 years ago
3 0

Answer:

Block B must have greater density than block A.

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At 15°C air is transmitted <br>at 340 m/s. Express this speed<br>in Kilometers per hour.​
EleoNora [17]

Answer:

1224km/hr

Explanation:

To convert from m/s to km/hr

1000m = 1km

Divide both sides by 1000

1m = 1/1000 km................. (1)

60×60 seconds = 1 hr

3600s = 1hr

Divide both sides by 3600

1s = 1/3600 .............(2)

Divide (2) by (1)

1m/s =  1/1000 ÷ 1/3600 km/hr

1m/s = 1/1000 × 3600/1  km/hr

1m/s = 3600/1000  km/hr

1m/s = 3.6 km/hr .............(3)

To convert 340m/s to km/hr

Multiply (3) by 340

1× 340m/s = 3.6 × 340 km/hr

340m/s = 1224km/hr

I hope this was helpful, please mark as brainliest

7 0
3 years ago
Read 2 more answers
Give an example of when a conscientious objector might exercise their right:
Mariana [72]
Am sorry what can you be more specific
7 0
3 years ago
What is angle between the vectors 2i -3j +K
Gnom [1K]

Answer:

the answer is B

Explanation:

edg 2020

3 0
3 years ago
A tennis ball is thrown from a 25 m tall building with a zero initial velocity. At the same moment, another ball is thrown from
Nataly [62]

Answer:

The two balls meet in 1.47 sec.

Explanation:

Given that,

Height = 25 m

Initial velocity of ball= 0

Initial velocity of another ball = 17 m/s

We need to calculate the ball

Using equation of motion

s=ut+\dfrac{1}{2}gt^2+h

Where, u = initial velocity

h = height

g = acceleration due to gravity

Put the value in the equation

For first ball

s_{1}=0-\dfrac{1}{2}gt^2+25....(I)

For second ball

s_{2}=17t-\dfrac{1}{2}gt^2+0....(II)

From equation (I) and (II)

-\dfrac{1}{2}gt^2+25=17t-\dfrac{1}{2}gt^2+0

t=\dfrac{25}{17}

t=1.47\ sec

Hence, The two balls meet in 1.47 sec.

6 0
3 years ago
Water enters the turbine of an ideal Rankine cycle as a superheated vapor at 18 MPa and 550°C. If the condenser pressure is 9 kP
ivolga24 [154]

Answer:

For this ideal Rankine cycle:

A) The rate of heat addition into the cycle Qh is 3214.59KJ/Kg.

B) The thermal efficiency of the system εt=0.4354.

C) The rate of heat rejection from the cycle Qc is 1833.32KJ/Kg.

D) The back work ratio is 0.013

Explanation:

To solve the ideal Rankine cycle we have to determinate the thermodynamic information of each point of the cycle. We will use a water thermodynamic properties table. In an ideal Rankine cycle, the process in the turbine and the pump must be isentropic. Therefore S₃=S₄ and S₁=S₂.

We will start with the point 3:

P₃=18000KPa  T₃=550ºc ⇒ h₃=3416.12 KJ/Kg  S₃=6.40690 KJ/Kg

Point 4:

P₄=9KPa S₄=S₃ ⇒ h₄=2016.60 (x₄=0.7645 is wet steam)

Point 1:

P₁=P₄ in the endpoint of the steam curve. ⇒ h₁= 193.28KJ/Kg  S₁=0.62235KJ/Kg  x₁=0

Point 2:

P₂=P₃ and S₂=S₁ ⇒ h₂=201.53KJ/Kg

With this information we can obtain the heat rates, the turbine, and the pump work:

Q_h=Q_{2-3}=h_3-h_2=3214.59KJ/Kg

Q_c=Q_{4-1}=h_4-h_1=1833.32KJ/Kg

W_{turb}=W_{3-4}=h_3-h_4=1399.52KJ/Kg

W_{pump}=W_{1-2}=h_2-h_1=18.25KJ/Kg

We can answer the questions with this data:

A) Q_h=Q_{2-3}=h_3-h_2=3214.59KJ/Kg

B) \epsilon_{ther}=\displaystyle\frac{W_{turb}}{Q_h}=0.4354

C)Q_c=Q_{4-1}=h_4-h_1=1833.32KJ/Kg

D) r_{bW}=\displaystyle\frac{W_{pump}}{W_{turb}}=0.013

3 0
3 years ago
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