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Artyom0805 [142]

1 year ago

13

1 Which requires more work, lifting a 10kg sack of coal or lifting a 15kg bag of feathers?

1 answer:

d1i1m1o1n [39]1 year ago

8 0

15KG bag of feathers

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A runner runs 4875 ft in 6.85 minutes. what is the runners average speed in miles per hour?

yanalaym [24]

The average speed can be easily calculated by taking the ratio of distance and time. That is:

average speed = distance / time

so calculating:

average speed = 4875 ft / 6.85 minutes

<span>average speed = 711.68 ft / min</span>

8 0

2 years ago

Read 2 more answers

A composite wall separates combustion gases at 2400°C from a liquid coolant at 100°C, with gas and liquid-side convection coeffi

evablogger [386]

Answer:

$\text{heat loss} = 24864.05 \ W/m^2$

Explanation:

If

$T_1$ , $T_2$ are temperatures of gasses and liquid in Kelvins,

$t_1$ and $t_2$ are thicknesses of gas layer and steel slab in meters,
$h_1$ , $h_2$ are convection coefficients gas and liquid in $W/m^2 \cdot K$ ,
$R_c$ is the contact resistance in $m^2 \cdot K/W$ ,

and $k_1, k_2$ are thermal conductivities of gas and steel in $W/m \cdot K$ ,

then: part(a):

$\text{heat loss } = \frac{T_1 - T_2} { \frac{1}{h_1} + \frac{t_1}{t_2} + R_c + \frac{t_2}{k_2} + \frac{1}{h_2}}$

using known values:

$\text {heat loss} = 2486.05 W/m^2$

part(b): Using the rate equation :

$\text {heat loss} = h_1 (T_1 - T_{s1})$

the surface temperature $T_{s1} = 1678.438 \ K$

and $T_{c1} = T_{s1} - \frac {t_1 (\text{heat loss})}{k_1} = 1664.560 \ K$

Similarly

$T_{c2} = T_{c1} - R_c (\text{heat loss}) = 421.357 \ K$

$T_{s2} = T_{c2} - \frac {t_2 (\text{heat loss})}{ k_2} = 397.864 \ K$

The temperature distribution is shown in the attached image

3 0

2 years ago

Liam throws a water balloon horizontally at 8.2 m/s out of a window 18 m from the ground.

Alecsey [184]

Time taken by the water balloon to reach the bottom will be given as

$h = \frac{1}{2} gt^2$

here we know that

$h = 18 m$

$g = 9.8 m/s^2$

now by the above formula

$18 = \frac{1}{2}*9.8* t^2$

$18 = 4.9 t^2$

$t = 1.92 s$

now in the same time interval we can say the distance moved by it will be

$d = v_x * t$

$d = 8.2 * 1.92 = 15.7 m$

so it will fall at a distance 15.7 m from its initial position

5 0

2 years ago

A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the drive train, and 70% reaches the wheels.

shutvik [7]

Answer:

a = 6.53 m/s^2

v = 11.5689 m/s

Explanation:

Given data:

engine power is 217 hp

70 % power reached to wheel

total mass ( car + driver) is 1530 kg

from the data given

2/3 rd of weight is over the wheel

w = 2/3rd mg

maximum force

$F = \mu W$

we know that F = ma

$ma = \mu (2/3 mg)$

$a_{max} = 2/3(1.00) (9.8) = 6.53 m/s^2$

the new power is $p = 70\% P_[max} = 0.7 P_{max}$

$P =f_{max} v$

$0.7P_{max} = ma_{max} v$

solving for speed v

$v =0.7 \times \frac{P_{max}}{ma_{max}}$

$v = 0.7 \frac{217 [\frac{746 w}{1 hp}]}{1500 \times 6.53}$

v = 11.5689 m/s

7 0

2 years ago

The volume of air is 53.28 m 3 . Convert it into scientific notation. 5.328×101 m3 532.8×101 m3 0.5328×103 m3 Both A and C

aivan3 [116]

Explanation:

It is given that, the volume of air is 53.28 m³. We need to convert it into scientific notation.

Any number in scientific notation can be given by :

$N=a\times 10^b$

a is real no and b is any integer

We have, V = 53.28 m³

To write it into scientific notation, shift decimal before 3 such that,

$V=\dfrac{5328}{100}$

No multiply and divide by 10. So,

$V=\dfrac{5328}{100}\times \dfrac{10}{10}\\\\V=\dfrac{5328}{1000}\times 10\\\\V=5.328\times 10^1\ m^3$

So, the correct option is (a).

5 0

2 years ago