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Umnica [9.8K]
3 years ago
9

The length of the track itself is 500m. The vehicles (each of mass 15kg) are dragged

Physics
1 answer:
gogolik [260]3 years ago
5 0

Answer:

W = 100000 J = 100 KJ

Explanation:

Here we will use the most basic and general formula of work, which is as follows:

W = Fd

where,

W = Work Done = ?

F = Force Required = 200 N

d = Length of Track = 500 m

Therefore,

W = (200\ N)(500\ m)\\

<u>W = 100000 J = 100 KJ</u>

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The doctor writes a prescription for na heparin 20,000 units in 500 ml n.s. infuse over 8 hours. what is the flow rate in ml/hr?
Ne4ueva [31]
I think this type of equation could be conducted in simple division equation since it does not involve drop rate.

we know that there is 500 ml of substance and should be infused within 8 hours period.

So the flow rate in ml/hr would be: 

500/8 = 62.5 ml/hr                                                                                    
8 0
3 years ago
The temperature at a point (x, y) on a flat metal plate is given by T(x, y) = 54/(7 + x2 + y2), where T is measured in °C and x,
ANTONII [103]

Answer:

dt/dx = -0.373702

dt/dy =  -1.121107

Explanation:

Given data

T(x, y) = 54/(7 + x² + y²)

to find out

rate of change of temperature with respect to distance

solution

we know function

T(x, y) = 54 /( 7 + x² + y²)

so derivative it x and y direction i.e

dt/dx = -54× 2x / (7 +x² + y²)²    .........................1

dt/dy = -54× 2y / (7 + x² + y²)²      .........................2

now put the value point (1,3) as x = 1 and y = 3 in equation 1 and 2

dt/dx = -54× 2(1) / (7 +(1)² + (3)²)²  

dt/dx = -0.373702

and

dt/dy =  -54× 2(3) / (7 + (1)² + (3)²)²

dt/dy =  -1.121107

7 0
3 years ago
At time t=0t=0 a proton is a distance of 0.360 mm from a very large insulating sheet of charge and is moving parallel to the she
insens350 [35]

Answer:

1.34 * 10^{3}m/s

Explanation:

Parameters given:

distance of the proton form the insulating sheet = 0.360mm

speed of the proton, v_{x} = 990m/s

Surface charge density, σ = 2.34 x 10^{-9} C/m^{2}

We need to calculate the speed at time, t = 7.0 * 10^{-8}s.

We know that the proton is moving parallel to the sheet, hence, we can say it is moving in the x direction, with a speed v_{x} on the axis.

The electric force acting on the proton moves in the y direction, so this means it is moving with velocity v_{y} in the y axis.

Hence, the resultant velocity of the proton is given by:

v = \sqrt{v_{x} ^{2} + v_{y} ^{2}}

v_{x} = 990m/s from the question. We need to find v_{y} and then the resultant velocity v.

Electric field is given in terms of surface charge density, σ as:

E = σ/ε0

where ε0 = permittivity of free space

=> E = \frac{2.34 * 10^{-9} } {2 * 8.85418782 * 10^{-12} }

E =  132 N/C

Electric Force, F is given in terms of Electric field:

F = eE

where e = electronic charge

=> F = ma = eE

∴ a = eE/m

where

a = acceleration of the proton

m = mass of proton

a = \frac{1.60 * 10^{-19} * 132}{1.672 * 10^{-27} }

a = 1.3 * 10^{10} m/s^{2}

Therefore, at time, t = 7.0 * 10^{-8}, we can use one of the equations of linear motion to find the velocity in the y axis:

a = \frac{v_{y} - v_{0}}{t} \\\\=> v_{y} = v_{0} + at

v_{y} = 0 + (1.3 * 10^{10} * 7.0 * 10^{-8})

v_{y} = 910 m/s

∴ v = \sqrt{v_{x} ^{2} + v_{y} ^{2}}

v = \sqrt{990^{2} + 910^{2} }

v = \sqrt{1808200}

v = 1344.69 m/s = 1.34 * 10^{3}m/s

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Answer:

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5 0
3 years ago
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