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3 years ago

How to find a scale for a graph

1 answer:

8 0

Step 1: Identify the variables. ...Step 2: Determine the variable range. ...Step 3: Determine the scale of the graph. ...Step 4: Number and label each axis and title the graph.Step 5: Determine the data points and plot on the graph. ...Step 6: Draw the graph.

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A man does 4,780 J of work in the process of pushing his 2.70 103 kg truck from rest to a speed of v, over a distance of 25.5 m.

wolverine [178]

Answer:

(A) Velocity will be 1.88 m/sec

(b) Force will be 187.45 N

Explanation:

We have given work done = 4780 j

Distance d = 25.5 m

(A) Mass of the truck m = $m=2.70\times 10^3kg$

We know that kinetic energy is given by

$KE=\frac{1}{2}mv^2$

So $v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2\times 4780}{2.7\times 10^3}}=1.88m/sec$

(B) We know that work done is given by

W = Fd

So $F=\frac{W}{d}=\frac{4780}{25.5}=187.45N$

4 0

2 years ago

Little Billie has a mass of 25 kg. The acceleration of gravity on the moon is one-sixth of the value on Earth. What is Little Bi

sineoko [7]

Answer:

Mass is constant everywhere,

But weight is different,

If earth g = 10 then moon's is 1.6666667

Now billie's weight in moon is 41.6667

6 0

2 years ago

A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect

Stels [109]

Answer:

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

Explanation:

Given that

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field at point P

$E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta$

$sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}$

$r^2=d^2+x^2$

$E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}$

Now by integrating above equation

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

4 0

3 years ago

What is special about nitrogen, and what is its main function in the atmosphere

Vlad [161]

Nitrogen is the most abundant of the gases present in the atmosphere. 78 percent of the atmospheric air comprises of nitrogen, oxygen makes up for 21 percent and all other gases make up for the remaining one percent. Oxygen is a highly flammable gas and in the absence of nitrogen it would not have been possible to utilize this atmospheric oxygen, hence the presence of nitrogen reduces its flammability and also neutralizes the toxicity of other gases.

5 0

3 years ago

Write an expression for the potential energy UD of a particle when it is at a distance D from the force center, in terms of B an

nadezda [96]

Answer:

E+mc+UD-3

Explanation:

7 0

3 years ago