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Y_Kistochka [10]
4 years ago
13

How much work (in j) must be done to lift a stack of bricks, which has a mass of 400 kg, from the ground to a height of 3 m? (g

= 10 m/s2)?
Physics
1 answer:
svetoff [14.1K]4 years ago
6 0
Given: Mass m = 400 Kg;   Height h = 3 m;     g = 10 m/s²

Required:  Work = ?

Formula: Work = Force x distance  F = ma    a = g    F = mg

W = fd

W = mgh

W = (400 Kg)(10 m/s²)( 3 m)

W = 12,000 Kg.m²/s²

W = 12,000 J

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Given that the velocity of blood pumping through the aorta is about 30 cm/s, what is the total current of the blood passing thro
TiliK225 [7]

Answer:

94.248 g/sec

Explanation:

For solving the total current of the blood passing first we have to solve the cross sectional area which is given below:

A_1 = \pi R^2\\\\A_1 = \pi (1)^2\\\\A_1 = 3.1416 cm^2

And, the velocity of blood pumping is 30 cm^2

Now apply the following formula to solve the total current

Q = \rho A_1V_1\\\\Q = (1)(3.1416)(30)\\\\

Q =  94.248 g/sec

Basically we applied the above formula So, that the total current could come

3 0
4 years ago
A book is sitting on a desk. What best describes the normal force acting on the book?
Sladkaya [172]
My best guess would be:

"A force equal in magnitude but opposite in direction"

However I assume that this question is multiple choice, by the way it is introduced. Therefore it would be helpful if these options were also displayed - hence take this as my best guess only.
5 0
3 years ago
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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
4 years ago
The nuclear fuel distribution in a nuclear reactor is chosen so that when in operation the wall temperature of the reactor is a
cestrela7 [59]

Answer:

Explanation:

We Often solve the the integral neutron transport equation using the collision probability (CP) method which usually requires flat flux (FF) approach. In this research, it has been carried out in the cylindrical nuclear fuel cell with the spatial of mesh with quadratic flux approach. This simply means that the neutron flux at any region of the nuclear fuel cell is forced to follow the pattern of a quadratic function.

Furthermore The mechanism may be referred to as the process of non-flat flux (NFF) approach. The parameters that calculated in this study are the k-eff and the distribution of neutron flux. The result shows that all parameters are in accordance with the result of SRAC.

8 0
4 years ago
A particle executes simple harmonic motion with an amplitude of 2.00 cm. At what positions does its speed equal one fourth of it
White raven [17]

Answer:

The positions are  0.0194 m  and - 0.0194 m.

Explanation:

Given;

amplitude of the simple harmonic motion, A = 2.0 cm = 0.02 m

speed of simple harmonic motion is given as;

v = \omega \sqrt{A^2-x^2}

the maximum speed of the simple harmonic motion is given as;

v_{max} = \omega A

when the speed equal one fourth of its maximum speed

v =\frac{v_{max}}{4}

\omega\sqrt{A^2-x^2} = \frac{\omega A}{4} \\\\\sqrt{A^2-x^2}= \frac{A}{4}\\\\A^2-x^2 = \frac{A^2}{16} \\\\x^2 = A^2 - \frac{A^2}{16} \\\\x^2 = \frac{16A^2 - A^2}{16} \\\\x^2 = \frac{15A^2}{16} \\\\x= \sqrt{\frac{15A^2}{16} } \\\\x = \sqrt{\frac{15(0.02)^2}{16} }\\\\x = 0.0194 \ m  \ \ or\  - 0.0194  \ m

Thus, the positions are  0.0194 m and - 0.0194 m.

8 0
3 years ago
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