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2 years ago

What is the smallest possible number of products in a decomposition reaction?

1 answer:

3 0

The smallest number of products in a decomposition reaction is two because there is 1 reactant that breaks down and when it does break down there must be two products. It is generally more, though.

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Calculate the mechanical advantage of a lever that has an input force of 15 N and an output force of 60 N. *

andreev551 [17]

Answer:

Explanation:

Mechanical Advantage is given as a ratio of the force a machines gives out to the force that it receives.

Simply put, it is the ratio of output force to input force:

$MA = \frac{OF}{IF}$

We are given that:

OF = 60 N

IF = 15 N

Hence, the Mechanical Advantage of the lever is:

$MA = \frac{60}{15} \\\\\\MA = 4$

It is a ratio so it has no unit.

4 0

3 years ago

Read 2 more answers

You have two springs. one has a greater spring constant than the other. you also have two objects, one with a greater mass than

neonofarm [45]

The answer should be "The object with the greater mass should be attached to the spring with the smaller <span>spring constant."

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
</span>

4 0

3 years ago

A 230 kg steel crate is being pushed along a cement floor. The force of friction is 480 N to the left and the applied force is 1

tangare [24]

It is unbalanced.......

7 0

3 years ago

Read 2 more answers

A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If

vampirchik [111]

Answer:

vcyl / vsph = 1.05

Explanation:

The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
The traslational part can be written as follows:

$K_{trans} = \frac{1}{2}* M* v_{cm} ^{2} (1)$

The rotational part can be expressed as follows:

$K_{rot} = \frac{1}{2}* I* \omega ^{2} (2)$

where I = moment of Inertia regarding the axis of rotation.
ω = angular speed of the rotating object.
If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

$v = \omega * R (3)$

For a solid cylinder, I = M*R²/2 (4)
Replacing (3) and (4) in (2), we get:

$K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2} (5)$

Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

$K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2} +\frac{1}{4}* M* v_{cmc}^{2} = \frac{3}{4}* M* v_{cmc} ^{2} (6)$

Repeating the same steps for the spherical shell:

$I_{sph} = \frac{2}{3} * M* R^{2} (7)$

$K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2} (8)$

$K_{sph} = \frac{1}{2}* M* v_{cms} ^{2} +\frac{1}{3}* M* v_{cms}^{2} = \frac{5}{6}* M* v_{cms} ^{2} (9)$

Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
Rearranging, and taking square roots on both sides, we get:

$\frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)$

This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.

3 0

3 years ago

The diameter of a 12-gauge copper wire is 0.081 in. The maximum safe current it can 17) carry (in order to prevent fire danger i

Soloha48 [4]

Answer:

0.44m/s

Explanation:

drift velocity=I/nAq

diameter 12 gauge

wire=0.081inches=0.081*2.5=0.2025cm radius=0.10125cm area=pi*R^2 =20/8.5*10^22*3.14*0.10125^2*10^-4*1.6*10^-19*

V = 0.44m/s

6 0

3 years ago