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shtirl [24]
2 years ago
5

1. A jogger jogs 25 m East, stops to rest, then continues for 10 more meters

Physics
1 answer:
Komok [63]2 years ago
8 0

Answer:

35m

Explanation:

25m + 10m = 35m east.

*(adding since it's in the same direction)

You might be interested in
A slice of pizza has 500 kcal. If we could burn the pizza and use all the heat to warm a 50-L container of cold water, what woul
topjm [15]

Answer:

increase in temperature of water is 10° C

Explanation:

Given data

pizza = 500 kcal = 500000 calories

cold water = 50 L

to find out

increase in temperature of water

solution

we know heat formula that is

heat = m × specific heat × Δt

here m is mass = 50000 gram and Δt is change in temperature

and specific heat = 1 cal / gram C

so put here all value and find Δt

500000 = 50000 × 1  × Δt

Δt = 10° C

so increase in temperature of water is 10° C

4 0
3 years ago
Read 2 more answers
The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting
White raven [17]

Answer:

\frac{\dot Q}{A} =20.129\ W.m^{-2}

T_1=27.58\ ^{\circ}C & T_2=2.41875\ ^{\circ}C

Explanation:

Given:

  • interior temperature of box, T_i=30^{\circ}C
  • height of the walls of box, h=3\ m
  • thickness of each layer of bi-layered plywood, x_p=1.25\ cm=0.0125\ m
  • thermal conductivity of plywood, k_p=0.104\ W.m^{-1}.K^{-1}
  • thickness of sandwiched Styrofoam, x_s=5\ cm=0.05\ m
  • thermal conductivity of Styrofoam, k_s=0.04\ W.m^{-1}.K^{-1}
  • exterior temperature, T_o=0^{\circ}C

<u>From the Fourier's law of conduction:</u>

\dot Q=\frac{dT}{(\frac{x}{kA}) }

\dot Q=\frac{dT}{R_{th} } ....................................(1)

<u>Now calculating the equivalent thermal resistance for conductivity using electrical analogy:</u>

R_{th}=R_p+R_s+R_p

R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}

R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})

R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})

R_{th}=\frac{1.4904}{A} .....................(2)

Putting the value from (2) into (1):

\dot Q=\frac{30-0}{\frac{1.4904}{A} }

\dot Q=\frac{30\ A}{1.4904}

\frac{\dot Q}{A} =20.129\ W.m^{-2} is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

<u>For plywood-Styrofoam interface from inside:</u>

\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}

20.129=0.104\times \frac{30-T_1}{0.0125}

T_1=27.58\ ^{\circ}C

&<u>For Styrofoam-plywood interface from inside:</u>

\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}

20.129=0.04\times \frac{27.58-T_2}{0.05}

T_2=2.41875\ ^{\circ}C

4 0
3 years ago
One characteristic of mass wasting processes is that they a. only operate on steep slopes. b. move materials very slowly. c. ope
lord [1]

Answer:

One characteristic of mass wasting processes is that they move materials relatively short distances compared to streams. - d.

4 0
3 years ago
What is the average power consumption in watts of an appliance that uses 5.00 kWh of energy per day? How many joules of energy d
denpristay [2]

Answer:

(A)  power  = 0.208 kW = 208 watts

(B)  energy = 6.6 x 10^{9} joules

Explanation:

energy consumed per day = 5 kWh

(a) find the power consumed in a day

         1 day = 24 hours

        power = \frac{energy}{time}

        power = \frac{5}{24}

          power  = 0.208 kW = 208 watts

         

(b) find the energy consumed in a year

    assuming it is not a leap year and number of days = 365 days

     1 year = 365 x 24 x 60 x 60 = 31,536,000 seconds

            energy = power x time

            energy = 208 x 31,536,000

            energy = 6.6 x 10^{9} joules

5 0
3 years ago
Does the KE of a car change more when it accelerates from 22 km/h to 32 km/h or when it accelerates from 32 km/h to 42 km/h
mote1985 [20]

Answer:

The change in kinetic energy (KE) of the car is more in the second case.

Explanation:

Let the mass of the car = m

initial velocity of the first case, u = 22 km/h = 6.11 m/s

final velocity of the first case, v = 32 km/h = 8.89 m/s

change in kinetic energy (K.E) = ¹/₂m(v² - u²)

                                          ΔK.E = ¹/₂m(8.89² - 6.11²)

                                                    = 20.85m J

 

initial velocity of the second case, u = 32 km/h = 8.89 m/s

final velocity of the second case, v = 42 km/h = 11.67 m/s

change in kinetic energy (K.E) = ¹/₂m(v² - u²)

                                          ΔK.E = ¹/₂m(11.67² - 8.89²)

                                                    = 28.58m J

The change in kinetic energy (KE) of the car is more in the second case.

6 0
3 years ago
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