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liraira [26]
4 years ago
7

How long does it take an airplane to fly 1500 miles of it maintains a speed of 600 miles per hour?

Physics
2 answers:
prisoha [69]4 years ago
7 0
2.5 hours. divide 15000 by 600




iVinArrow [24]4 years ago
5 0

Answer: 2.5 hours

Explanation: speed = distance / time.

Speed = 600 miles/hour.

Distance. = 1500 miles.

Time = unknown (t)

Speed = distance / time.

Therefore,

Time = Distance / Speed.

Time (t) = 1500 (miles) / 600 (miles/hour)

Time (t) = 2.5 hours or 2 hours, 30 mins

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What are graphs used for? Select all that apply.
lana66690 [7]

Answer:

analyzing what data means

diagram showing the correlation between two quantities

Explanation:

Graph is the plot of two physical quantities and it describes the relation between both quantities. It gives a specific relation between the two physical quantities which is used to analyze the result.

Thus, the following options are correct,

analyzing what data means

diagram showing the correlation between two quantities


4 0
3 years ago
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True or false: Ultimate tensile strength increases as the thickness of a solid material sample increases.
Varvara68 [4.7K]

Answer:

False.

Explanation:

Tensile strength should remain constant, regardless of thickness. For larger cross sections, it can slightly increase because the atoms in the center become more constricted and therefore less responsive to the applied stress.

7 0
3 years ago
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The prismatic bar has a cross-sectional area
ludmilkaskok [199]

solution:

sum fx=0\\
n=\frac{1}{2}\times\frac{w_{0}}{a}(2a-x)^2\\
n=\frac{w_{0}}{20}(2a-x)^2\\
angle is the normal shus\\
\sigma =\frac{N}{A}\\
\frac{\frac{w_{0}}{2a}(2a-x)^2}{A}\\
\sigma(x)=\frac{w_{0}}{2aA}[2a-x]^2

8 0
3 years ago
A centrifugal pump discharges 300 gpm against a head of 55 ft when the speed is 1500 rpm. The diameter of the impeller is 15.5 i
Ksivusya [100]

Answer: Question 1: Efficiency is 0.6944

Question 2: speed of similar pump is 2067rpm

Explanation:

Question 1:

Flow rate of pump 1 (Q1) = 300gpm

Flow rate of pump 2 (Q2) = 400gpm

Head of pump (H)= 55ft

Speed of pump1 (v1)= 1500rpm

Speed of pump2(v2) = ?

Diameter of impeller in pump 1= 15.5in = 0.3937m

Diameter of impeller in pump 2= 15in = 0.381

B.H.P= 6.0

Assuming cold water, S.G = 1.0

eff= (H x Q x S.G)/ 3960 x B.H.P

= (55x 300x 1)/3960x 6

= 0.6944

Question 2:

Q = A x V. (1)

A1 x v1 = A2 x V2. (2)

Since A1 = A2 = A ( since they are geometrically similar

A = Q1/V1 = Q2/V2. (3)

V1(m/s) = r x 2π x N(rpm)/60

= (0.3937x 2 x π x 1500)/2x 60

= 30.925m/s

Using equation (3)

V2 = (400 x 30.925)/300

= 41.2335m/s

To rpm:

N(rpm) = (60 x V(m/s))/2 x π x r

= (60 x 41.2335)/ 2× π × 0.1905

= 2067rpm.

6 0
3 years ago
The specific heat of a certain type of cooking oil is 1.75 J/(g⋅°C). How much heat energy is needed to raise the temperature of
Y_Kistochka [10]

Answer:

Q = 590,940 J

Explanation:

Given:

Specific heat (c) = 1.75 J/(g⋅°C)

Mass(m) = 2.01 kg = 2,010

Change in temperature (ΔT) = 191 - 23 = 168°C

Find:

Heat required (Q)

Computation:

Q = mcΔT

Q = (2,010)(1.75)(168)

Q = 590,940 J

Q = 590.94 kJ

5 0
3 years ago
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