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2 years ago

Identify the scale of measurement used.

Physics

1 answer:

Mkey [24]2 years ago

6 0

I believe it’s b. :D

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5. A 10 kg ball is traveling at the same speed as a 1 kg ball. Compared to the 10 kg ball, the 1 kg ball has (2 points)

suter [353]

What’s the rest of the question or is that it?

4 0

3 years ago

3. The soccer kicker kicks a 2 kg football with a force of 68 N. How fast will the ball accelerate down the field?

Katen [24]

Answer:

$\boxed {\boxed {\sf m= 2 \ kg , \\a= 34 \ m/s^2, \\\F= 68 \ N}}$

Explanation:

The formula for force is:

$F=m*a$

If we rearrange the formula to solve for a (acceleration), the formula becomes

$\frac{F}{m} =a$

The force is 68 Newtons. Let's convert the units to make the problem easier later on. 1 N is equal to 1 kg*m/s², so the force of 68 N is equal to 68 kg*m/s².

The mass is 2 kilograms.

$F=68 \ kg*m/s^2 \\m= 2\ kg$

Substitute the values into the formula.

$\frac{68 \ kg*m/s^2}{2 \ kg} =a$

Divide. Note that the kilograms will cancel each other out (hence why we changed the units).

$\frac{68 \ m/s^2}{2}=a$

$34 \ m/s^2=a$

The acceleration is 34 meters per second squared.

6 0

3 years ago

WORD BANK:

Mkey [24]

Answer:

1. Vector, base

2. Vector, derived

3. Vector, ?

4. Scalar, derived

5. scalar, base

4 0

2 years ago

Use the dimensional analysis and check the correctness of given equation:- PV= nRT

Harman [31]

PV=nRT

Here

P=Pressure

V=Volume

n=Molarity

R=universal gas constant

T=Temperature.

LHS

$\\ \tt\bull\leadsto PV$

$\\ \tt\bull\leadsto [ML^2T^{-2}][M^0L^3T^0]$

$\\ \tt\bull\leadsto [ML^5T^{-2}]$

RHS

$\\ \tt\bull\leadsto nRT$

$\\ \tt\bull\leadsto [M^0L^{3}T^0][M^1 L^2 T^{-2}K^{-1}][M^0L^0T^0K^1]$

$\\ \tt\bull\leadsto [ML^5T^{-2}]$

LHS=RHS

hence verified

6 0

3 years ago

In an electric field, 0.90 joule of work is required to bring 0.45 coulomb of charge from point a to point

jarptica [38.1K]

The difference in electric potential energy between the two points is
$\Delta U = q \Delta V$
where q is the magnitude of the charge and $\Delta V$ is the electric potential difference.

But for energy conservation, the difference in electric potential energy $\Delta U$ between the two points is equal to the work done to move the charge between A and B:
$W=\Delta U$
so we have
$W=q \Delta V$

and by substituting the numbers of the problem, we find the value of $\Delta V$ :
$\Delta V = \frac{W}{q}= \frac{0.90 J}{0.45 C}=2 V$

3 0

3 years ago