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3 years ago

If you place one staple on an electronic balance, the balance still reads 0.0 grams. However, if you place

Physics

2 answers:

larisa86 [58]3 years ago

5 0

Answer:

210/6.8 = 1/x

210x = 6.8

x = 0.03238 grams = 32.4 milligrams

Tasya [4]3 years ago

4 0

Mass of 1 staple = 6.8 g/210 staples
mass of 1 staple = 0.032380952 g
Hope that helps!!

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Jill is pushing a box across the floor. Which represents the upward force perpendicular to the floor?

Georgia [21]

Answer:

When Jill is pushing a box across a floor, it represents the upward motion and it is natural force is applied.

So it is represented as FN and normal force takes place in considering the force perpendicular to the floor.

It seems to support that forced applied on an object when the object is in contact with other.

Explanation:

5 0

3 years ago

Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.5000 kgkg when weighed in air. The density o

spin [16.1K]

Answer:

The true weight of the aluminium is $m_{alu} =$ 4.5021 kg

Explanation:

Given data

$m_{app}$ = 4.5 kg

$\rho_{air}$ = 1.29 $\frac{kg}{m^{3} }$

$\rho_{al}$ = 2.7× $10^{3} \frac{kg}{m^{3} }$

The true mass of the aluminium is given by

$m_{alu} = \frac{\rho_{alu}m_{app}}{\rho_{alu} -\rho_{air} }$

Put all the values in above equation we get

$m_{alu} = \frac{(2700)(4.5)}{2700-1.29}$

$m_{alu} =$ 4.5021 kg

Therefore the true weight of the aluminium is $m_{alu} =$ 4.5021 kg

6 0

3 years ago

A black hole can be considered a star that has...

joja [24]

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3 years ago

A close coiled helical spring of round steel wire 10 mm diameter having 10 complete turns with a mean radius of 60 mm is subject

kow [346]

Answer:

The deflection of the spring is 34.56 mm.

Explanation:

Given that,

Diameter = 10 mm

Number of turns = 10

$Radius_{mean} = 60\ mm$

$Diameter_{mean} = 120\ mm$

Load = 200 N

We need to calculate the deflection

Using formula of deflection

$\delta=\dfrac{8pD^3n}{Cd^4}$

Put the value into the formula

$\delta=\dfrac{8\times200\times(120)^3\times10}{80\times10^{3}\times10^4}$

$\delta =34.56\ mm$

Hence, The deflection of the spring is 34.56 mm.

4 0

3 years ago

You can enter compound units that are combinations of other units that are multiplied together. to enter the ⋅ explicitly, type

alexdok [17]

Answer:

$30 N \cdot m$

Explanation:

The torque applied by a force can be calculated as

$\tau = F d sin \theta$

where

F is the magnitude of the force

d is the length of the arm

$\theta$ is the angle between the direction of the force and the arm

In this problem, we have

F = 15 N

d = 2.0 m

$\theta=90^{\circ}$

Substituting into the equation, we find

$\tau = (15)(2.0) sin 90^{\circ}=30 N \cdot m$

7 0

3 years ago