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stellarik [79]
3 years ago
12

A physical count of merchandise inventory on november 30

Physics
1 answer:
Zigmanuir [339]3 years ago
4 0
It depends on the indirect taxes and the share market values
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An automobile is sitting on a hill which is 20 m higher than ground level. Find the mass of the automobile if it contains 362,60
mr Goodwill [35]
M= ?
g=9.8 m/s (2)
h=20 m

Eg=362,600 J
Eg/mg

362,600 J/9.8 m/s (2) x 20 m
=1,850 m
6 0
3 years ago
Hello people ~
lozanna [386]

Answer:

It’s called a conservative field.

Explanation:

I think it’s going to be the conservative field because in the question it talks about how it is able to become possible to define potential at a point in an electric field because electric field.

3 0
2 years ago
Read 2 more answers
A long string is wrapped around a 6.6-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is
lys-0071 [83]

Answer:

\omega_f=571.42\ rpm

Explanation:

It is given that,

Diameter of cylinder, d = 6.6 cm

Radius of cylinder, r = 3.3 cm = 0.033 m

Acceleration of the string, a=1.5\ m/s^2

Displacement, d = 1.3 m

The angular acceleration is given by :

\alpha =\dfrac{a}{r}

\alpha =\dfrac{1.5}{0.033}

\alpha =45.46\ rad/s^2

The angular displacement is given by :

\theta=\dfrac{d}{r}

\theta=\dfrac{1.3}{0.033}

\theta=39.39\ rad

Using the third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta

Here, \omega_i=0

\omega_f=\sqrt{2\alpha \theta}

\omega_f=\sqrt{2\times 45.46\times 39.39}

\omega_f=59.84\ rad/s

Since, 1 rad/s = 9.54 rpm

So,

\omega_f=571.42\ rpm

So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.

5 0
3 years ago
What terms are needed to completely describe velocity?
kotegsom [21]
In order to completely describe a velocity,
you need a speed and a direction.
4 0
3 years ago
If the truck has a mass of 2,000 kilograms , what's its momentum?(v=35 m/s)
katen-ka-za [31]

Answer:

\boxed {\boxed {\sf  70,000 \ kg*m/s}}

Explanation:

Momentum is the product of mass and velocity.

p=m*v

The mass of the truck is 2,000 kilograms and the velocity is 35 meters per second.

m= 2000 \ kg \\v= 35 \ m/s

Substitute the values into the formula and multiply.

p= 2000 \ kg * 35 \ m/s \\p= 70,000 \ kg*m/s

The truck's momentum is <u>70,000 kilograms meters per second.</u>

8 0
3 years ago
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