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3 years ago

Help?? On A, B, C and D

Physics

1 answer:

erastova [34]3 years ago

3 0

A.There are 2 good point that land on an accurate calculation 1. (6,16) and 2.(12,28)

B.The slop of the graph is (3,2)

C. The y intercept is (0.4)

D. Not sure

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Please help on this one?

koban [17]

Using the given equation:

di = 20.0 * 10.0 / 20.0 - 10.0

di = 200/10

di = 20.0 cm

The answer is A.

6 0

4 years ago

Read 2 more answers

Which shows the correct lens equation? The inverse of f equals the inverse of d Subscript o Baseline times the inverse of d Subs

musickatia [10]

Answer:

The inverse of f equals the inverse of d Subscript o Baseline plus the inverse of d Subscript I Baseline.

Explanation:

The lens equation shows the relation among focal length of the lens, image distance and object distance. It can be expressed as:

$\frac{1}{f}$ = $\frac{1}{d_{o} }$ + $\frac{1}{d_{I} }$

where: f is the focal length of the lens, $d_{o}$ is the object distance to the lens and $d_{I}$ is the image distance to the lens.

The lens equation can be used to determine the unknown value among the variables f , $d_{o}$ and $d_{o}$ .

6 0

3 years ago

Read 2 more answers

falling objects drop with an average acceleration of 9.8m/sec/sec. if an object falls from a tall building how long will it take

sveta [45]

Answer:

5 seconds

Explanation:

Acceleration = (final velocity - initial velocity) ÷ time

$a = \frac{v - u}{t}$

$9.8 = \frac{49 - 0}{t}$

$9.8 = \frac{49}{t}$

$9.8t = 49$

$t = \frac{49}{9.8}$

$t = 5$

8 0

3 years ago

20 pts.

RUDIKE [14]

The cup is acted upon by an unbalanced force which is the cars acceleration, but before it was an object at rest that stayed at rest. This jet propels their body forward.

6 0

3 years ago

Read 2 more answers

The potential energy for a certain mass moving in one dimension is given by U(x)=(2.0J/m3)x3−(15J/m2)x2+(36J/m)x−23JU(x)=(2.0J/m

Angelina_Jolie [31]

Answer:x=2 and x=3

Explanation:

Given

Potential Energy for a certain mass is

$U(x)=2x^3-15x^2+36x-23$

and we know force is given by

$F=-\frac{\mathrm{d} U}{\mathrm{d} x}$

$F=-(2\times 3x^2-15\times 2x+36)$

For Force to be zero F=0

$\Rightarrow 6x^2-30x+36=0$

$\Rightarrow x^2-5x+6=0$

$\Rightarrow x^2-2x-3x+6=0$

$\Rightarrow (x-2)(x-3)=0$

Therefore at x=2 and x=3 Force on particle is zero.

8 0

3 years ago