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2 years ago

Consider the common measurement speed = 55 mph. What is the magnitude quantity?

Physics

1 answer:

noname [10]2 years ago

3 0

Answer:

Magnitude = 55

Quantity = Speed

Explanation:

We are told that the speed is 55 mph. This implies that the magnitude is 55 while the unit is mph.

Also, physical quantities are usually things that can be measured such as distance, voltage, temperature e.t.c

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Using knowledge of states of matter,write a message about the importance of science in our society

oee [108]

Answer:

उव्ग्वुव ह्व्झ एउएइहे एइएइएइएएइ सिसुब्स्सी बीस सिस इस्ब एइब

Explanation:

?उग्व्ब्वु विब्सिए इसिग्व विद्बिअब्द सिह्व्व इस्ब्व दिव्ब्स विह्द ऐद्जिइ सुउगव्दी सिइगैगे क्ज्गैइव अजिव्व्ज्व्स कैह्द अजि ह्ज्फ्ज इअह इकुगै ईग इअबे अजिव्ब जैइअब इऐहे ऐइहे ऐइग्गे अत्व्ब ओप्झब रोज दिधिए ऊइफ्ब इसुहद ईउहे सिउउअ दिइब्द स्सिउए ऐइहे सिएय्व एउविये एइव्वे

5 0

1 year ago

Hey I need help do animals take in oxygen and give off carbon dioxide or Plants take in oxygen and give off carbon dioxide or pl

Sophie [7]

Answer:

Animals take in oxygen and give off carbon dioxide.

Explanation:

This is because plants take in our carbon dioxide and give off the oxygen that we as people and animals need to breathe.

3 0

2 years ago

Read 2 more answers

When 560,000 is written in scientific notation, the power of 10 is

attashe74 [19]

The answer would be 5.6x10^5

4 0

3 years ago

Read 2 more answers

Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc

Crazy boy [7]

Answer:

Temperature at the exit = $267.3 C$

Explanation:

For the steady energy flow through a control volume, the power output is given as

$W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

Inlet area of the turbine = $60cm^{2}= 0.006m^{2}$

To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume $v_{1}$

$v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg$

$m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec$

for Ideal gasses, the enthalpy change can be calculated using the formula

$h_{2}-h_{1}=C_{p}(T_{2}-T_{1})$

hence we have

$W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

$250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})$

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have $T_{2}=267.3C$

Hence, the temperature at the exit = $267.3 C$

5 0

3 years ago

if you run around a circle at 4.5 m/s and the circle has a radius of 7.7 m, what is your centripetal acceleration?

madreJ [45]

Answer:

Centripetal acceleration,

$a_{c} =2.63\ m/s^{2} }$

Explanation:

Centripetal acceleration:

Centripetal acceleration is the idea that any object moving in a circle, in something called circular motion, will have an acceleration vector pointed towards the center of that circle.

Centripetal means towards the center.

Examples of centripetal acceleration (acceleration pointing towards the center of rotation) include such situations as cars moving on the cicular part of the road.

An acceleration is a change in velocity.

Formula for Centripetal acceleration:

$a_{c} =\frac{(velocity)^{2} }{radius}$

Given here,

Velocity = 4.5 m/s

radius = 7.7 m

To Find :

$a_{c} = ?$

Solution:

We have,

$a_{c} =\frac{(velocity)^{2} }{radius}$

Substituting given value in it we get

$a_{c} =\frac{(4.5)^{2}}{7.7} \\\\a_{c} =\frac{20.25}{7.7}\\\\a_{c} =2.629\ m/s^{2} \\\\\therefore a_{c} =2.63\ m/s^{2$

Centripetal acceleration,

$a_{c} =2.63\ m/s^{2} }$

7 0

3 years ago