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Aloiza [94]
3 years ago
8

What type of image is formed by a lens if m= 3.2?

Physics
1 answer:
Leokris [45]3 years ago
5 0

Answer:

c

Explanation:

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Answer: its b bro

Explanation:

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2 years ago
"4.What determines how many figures are significant in reported measurement values? What
MAXImum [283]

As the number of significant figures increases, the more accurate or precise the measurement is.

<h3>What is significant figure?</h3>

The term significant figures refers to the number of important single digits in the coefficient of an expression in scientific notation.

Significant figures are the digits in a value that are known with some degree of confidence.

The effect of reporting more or fewer figures or digits than are significant;

As the number of significant figures increases, the more accurate or precise the measurement is.

As precision of a measurement increases, so does the number of significant figures.

Learn more about significant figures here: brainly.com/question/24491627

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1 year ago
What is the effect that distance has on gravity?
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The forces of gravity between two objects are smaller when the objects are farther apart.
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Protons and neutrons grouped in a specific pattern
alexgriva [62]
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3 years ago
A 49 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 3.3 m/s just before hitting the ground.
hichkok12 [17]

Answer:

a) \Delta U_g=-5.3kJ

b) K=0.27kJ

c) F_f=0.45kN

Explanation:

the gravitational potential energy is given by:

U_g=m.g.h\\

\Delta U_g=m.g.h_f-m.g.h_i\\\Delta U_g=49kg*9.8m/s^2*(0m-11m)\\\Delta U_g=-5.3kJ

The kinetic energy is given by:

K=\frac{1}{2}m.v^2\\

the initial kinetic energy is zero because the motion started from rest, so:

K=\frac{1}{2}*49kg*(3.3m/s^2)^2\\K=0.27kJ

applying the conservation of energy theorem:

U_g-W_f=K_f\\W_f=-(\Delta K+\Delta U)\\W_F=5.3kJ-0.27kJ\\W_F=-5.0kJ

The work done by the friction force is given by:

W_f=F_f.h.cos(\theta)\\

the angle of the force is 180 degrees because it's against the movement:

F_f=\frac{W_f}{h.cos(\theta)}\\\\F_f=\frac{-5.0kJ}{11m.cos(180^o)}\\\\F_f=0.45kN

8 0
3 years ago
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