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3 years ago

13

A piece of gym equipment states that the maximum load it can hold is 300 kg. Why do you think it is important not to go over thi

s limit?

1 answer:

Crazy boy [7]3 years ago

5 0

Answer:

The weight limit of 300kg is the maximum amount the machine can handle so it can be dangerous to exceed the maximum load.

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NEED HELP ASAP!!!!!!!!!!!!!

love history [14]

Answer: Decreased risk of heart attack

Explanation:

thats the answer because it actually

increase rish of heart attack

Hope this helps :)

5 0

3 years ago

Read 2 more answers

Which of the following best defines mass?

Radda [10]

"<span>The amount of matter in a substance" is the one among the following choices given in the question that best defines mass. The correct option among all the options that are given in the question is the first option or option "A". I hope that this is the answer that has actually come to your desired help.</span>

5 0

3 years ago

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if you use the compound pulley, you will need to pull twice the distance but with less force. the force you need is equal to one

algol13

Answer:

F= 25/2 = 12.5N

Explanation:

When you use a compound pulley the force required depends on the mechanical advantage of the compound pulley. This is known as rate of loss of distance or the ratio of the force to the load.

M.A = Effort distance /Load distance. OR M.A = Load/Effort

6 0

4 years ago

If μs is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. calculate this

fiasKO [112]

weight = mg acts downwards <span>
normal force = N acts upwards.
and force F acts at an angle θ below the horizontal.
(Let us assume that the woman pushes from the left, so F is acted towards the right, which is below the horizontal)
so that, Frictional force, f=us*N acts towards the left

Now we balance the forces along x and y directions:
y direction: N = mg + F sinΘ
x direction: us * N = F cosΘ

We let the value of µs be equal to a value such that any F will not be able to move the crate. Then, if we increase F by an amount F', then the force pushing the crate towards the right also increases by F' cosΘ. Additionally, the frictional force f must raise by exactly this amount.
Since f can’t exceed us*N, so the normal force must increase by F' cosΘ/us.
Also, from the y direction equation, the normal force exceeds by F' sin Θ.

<span>These two values must be the same, therefore:
<span>us = cot θ</span></span></span>

4 0

3 years ago

Based on the following equation, answer the questions below. ρ = (2γϕ + ψ)/rg where ρ [=] moles per cubic foot [mol/ft3] γ [=] j

AlekseyPX

1) Fundamental units of $\Psi$ are $[\frac{mol}{m\cdot s^2}]$

2) Fundamental units of $\Phi$ are $[\frac{mol}{m^3}]$

Explanation:

The equation for the variable $\rho$ is

$\rho =\frac{2\gamma \Phi+\Psi}{rg}$

where we have:

$\rho$ measured in $[\frac{mol}{ft^3}]$

$\gamma$ measured in $[\frac{J}{kg}]$

$r$ measured in $[in]$

$g$ measured in $[\frac{m}{s^2}]$

We can re-write the equation as

$\rho rg = 2\gamma \Phi + \Psi$

And we notice that the units of the term on the left must be equal to the units of the term on the right.

This means that:

1) First of all, $\Psi$ must have the same units of $\rho r g$ . So,

$[\rho r g]=[\frac{mol}{ft^3}][in][\frac{m}{s^2}]$

However, both ft (feet) and in (inches) are not fundamental dimensions: this means that they can be expressed as meters. Therefore, the fundamental units of $\Psi$ are

$[\Psi]=[\frac{mol}{m^3}][m][\frac{m}{s^2}]=[\frac{mol}{m\cdot s^2}]$

2)

The term $2\gamma \Phi$ must have the same units of $\Psi$ in order to be added to it. Therefore,

$[\gamma \Phi] = [\frac{mol}{m\cdot s^2}]$

We also know that the units of $\gamma$ are $[\frac{J}{kg}]$ , therefore

$[\frac{J}{kg}][\Phi]= [\frac{mol}{m\cdot s^2}]$

And so, the fundamental units of $\Phi$ are

$[\Phi]= [\frac{mol\cdot kg}{J\cdot m\cdot s^2}]$

However, the Joules can be written as

$[J]=[kg][\frac{m^2}{s^2}]$

Therefore

$[\Phi]= [\frac{mol\cdot kg}{(kg \frac{m^2}{s^2})\cdot m\cdot s^2}]=[\Phi]= [\frac{mol}{m^3}]$

#LearnwithBrainly

7 0

3 years ago