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2 years ago

A net force of 125 N accelerates a 25.0 kg mass. What is the resulting acceleration?

1 answer:

8 0

The correct answer is 5.00 m/s^2

Explanation:
According to Newton's second law:
F = ma ----- (1)

Where F = net force acting on the body = 125N
m = mass of the body = 25N
a = Acceleration of the body due to the applied force = ?

Plug-in the values in equation (1):
(1) => a = F/m
a = 125N/25kg
Hence a = 5.00 m/s^2

-i

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Answer:

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2 years ago

Perform the calculation and report your answer using sig figs. 657.70 - 26.543

Anton [14]

Answer:

The answer is 631.157

Explanation:

The question requested that the answer to the subtraction of 26.543 from 657.70 must be written using significant figures.

Here are a few tips about how to Identify significant figures.

1) It should be noted that the number "0" is what is usually (but not always) affected while trying to identify significant figures. Hence, all other numbers/digits are always significant. For example, 26.543 has five significant figures.

2) The zeros found between these "other numbers/digits" are also significant. For example, 2202 has four significant figures.

3) In the case of a decimal, the tailing zeros or the final zero is also significant. 657.70 and 657.07 have five significant figures.

Now, back to the question

657.70 - 26.543 = 631.157.

Our final answer does not have a zero, hence all the digits (six) are significant.

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2 years ago

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castortr0y [4]

Only answering to ask a question

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2 years ago

A long, straight metal rod has a radius of 5.75 cm and a charge per unit length of 33.3 nC/m. Find the electric field at the fol

PIT_PIT [208]

Answer:

Explanation:

From the question;

We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.

We are to calculate the following task, i.e. to determine the electric field at the distances:

a) at 4.75 cm

b) at 20.5 cm

c) at 125.0 cm

Given that:

the charge (q) = 33.3 nC/m

= 33.3 × 10⁻⁹ c/m

radius of rod = 5.75 cm

a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.

Then, the electric field will be zero.

b) The electric field formula $E = \dfrac{kq }{d}$

$E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}$

E = 1461.95 N/C

c) The electric field E is calculated as:

$E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}$

E = 239.76 N/C

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2 years ago

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kobusy [5.1K]

On the y-axis (the bottom of the table) hope this helps

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2 years ago

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