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yan [13]
3 years ago
15

in a 4 kilometer race, a runner completes the first kilometer in 5.9 minutes, the second kilometer in 6.2 minutes, the third kil

ometer in 6.3 minutes, and the final kilometer in 6 minutes. the average speed of the runner for the race is approximately
Physics
2 answers:
ohaa [14]3 years ago
5 0
To find the average of data collected, add all of the measurements: 5.9+6.2+6.3+6= 12.2

Then, divide the total amount by the number of data collected which is 4: 12.2/4= 3.05

The average speed of the runner of the race is approximately 3.05 km/min

Feel free to ask me any other questions you might have :)
Kruka [31]3 years ago
5 0

Answer:

9.83 km/h

Explanation:

Given: in a 4 kilometer race, a runner completes the first kilometer in 5.9 minutes, the second kilometer in 6.2 minutes, the third kilometer in 6.3 minutes, and the final kilometer in 6 minutes.

To Find:  the average speed of the runner for the race.

Solution:

In a 4 kilometer race,

time required to travel first kilometer = 5.9 \text{minutes}

time required to travel seoond kilometer = 6.2 \text{minutes}

time required to travel third kilometer = 6.3 \text{minutes}

time required to travel final kilometer = 6 \text{minutes}

we know that,

\text{Average speed}=\frac{\text{Total distance traveled}}{\text{Total time taken}}

Total distance traveled = 4  \text{kilometer}

Total time taken = (5.9+6.2+6.3+6)

                           =  24.4 \text{minutes}

                           = \frac{24.4}{60}  \text{hours}

putting values,

\text{Average speed}=\frac{4}{\frac{24.4}{60}}

 \text{Average speed}=\frac{4\times60}{24.4}

                                            =9.83 \text{km}/ \text{h}

the average speed of the runner for the race is approximately 9.83 \text{km}/ \text{h}

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Answer:

Explanation:

When the apple is held submerged in water , it experiences a buoyant force due to which it floats in water . One has to apply downward force to keep it submerged. The lower the buoyant force , lower the force needed to submerge it in water.

When apple is held at much deeper point , it experience greater pressure due to column of water around it . So its size or its volume decreases . But its weight remains the same . Due to less volume , buoyant force also decreases ( buoyant force is equal to weight of displaced volume of water. )

Due to buoyant force becoming less , force needed on apple  in downward direction will also be less.

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4 years ago
A physicist is investigating a beam of laser light of wavelength 550 nm. The light strikes a target that is 189 meters away from
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Answer: here you go I was looking for this answer everywhere,I have it now so it’s 6.30 x 10^-7 s

Explanation:

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8 0
3 years ago
Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudina
Alecsey [184]

(a) 23.4

The fiber-to-matrix load ratio is given by

\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}

where

E_f = 131 GPa is the fiber elasticity module

E_m = 2.4 GPa is the matrix elasticity module

V_f=0.3 is the fraction of volume of the fiber

V_m=0.7 is the fraction of volume of the matrix

Substituting,

\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4 (1)

(b) 44,594 N

The longitudinal load is

F = 46500 N

And it is sum of the loads carried by the fiber phase and the matrix phase:

F=F_f + F_m (2)

We can rewrite (1) as

F_m = \frac{F_f}{23.4}

And inserting this into (2):

F=F_f + \frac{F_f}{23.4}

Solving the equation, we find the actual load carried by the fiber phase:

F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N

(c) 1,906 N

Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:

F=F_f + F_m (2)

Using

F = 46500 N

F_f = 44594 N

We can immediately find the actual load carried by the matrix phase:

F_m = F-F_f = 46,500 N - 44,594 N=1,906 N

(d) 437 MPa

The cross-sectional area of the fiber phase is

A_f = A V_f

where

A=340 mm^2=340\cdot 10^{-6}m^2 is the total cross-sectional area

Substituting V_f=0.3, we have

A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2

And the magnitude of the stress on the fiber phase is

\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa

(e) 8.0 MPa

The cross-sectional area of the matrix phase is

A_m = A V_m

where

A=340 mm^2=340\cdot 10^{-6}m^2 is the total cross-sectional area

Substituting V_m=0.7, we have

A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2

And the magnitude of the stress on the matrix phase is

\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa

(f) 3.34\cdot 10^{-3}

The longitudinal modulus of elasticity is

E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa

While the total stress experienced by the composite is

\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa

So, the strain experienced by the composite is

\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}

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timurjin [86]
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6 0
3 years ago
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Water enters a baseboard radiator at 180 °F and at a flow rate of 2.0 gpm. Assuming the radiator releases heat into the room at
beks73 [17]

Answer:

Temperature of water leaving the radiator = 160°F

Explanation:

Heat released = (ṁcΔT)

Heat released = 20000 btu/hr = 5861.42 W

ṁ = mass flowrate = density × volumetric flow rate

Volumetric flowrate = 2 gallons/min = 0.000126 m³/s; density of water = 1000 kg/m³

ṁ = 1000 × 0.000126 = 0.126 kg/s

c = specific heat capacity for water = 4200 J/kg.K

H = ṁcΔT = 5861.42

ΔT = 5861.42/(0.126 × 4200) = 11.08 K = 11.08°C

And in change in temperature terms,

10°C= 18°F

11.08°C = 11.08 × 18/10 = 20°F

ΔT = T₁ - T₂

20 = 180 - T₂

T₂ = 160°F

8 0
3 years ago
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