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3 years ago

An element which has strong intermolecular forces is most likely to have

2 answers:

5 0

An element like that would have have a molecular structure that has a high tolerance to heat, so the correct answer is D

Reika [66]3 years ago

4 0

The correct answer is:
d. a very high melting point

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A horizontal force of 40N is needed to pull a 60kg box across the horizontal floor at which coefficient of friction between floo

Mazyrski [523]

Answer:

Coefficient of friction is $0.068$ .

Work done is $320~J$ .

Explanation:

Given:

Mass of the box ( $m$ ): $60$ kg

Force needed ( $F$ ): $40$ N

The formula to calculate the coefficient of friction between the floor and the box is given by

$F=\mu mg...................(1)$

Here, $\mu$ is the coefficient of friction and $g$ is the acceleration due to gravity.

Substitute $40$ N for $F$ , $60$ kg for $m$ and $9.80$ m/s² for $g$ into equation (1) and solve to calculate the value of the coefficient of friction.

$40 N=\mu\times60 kg\times9.80 m/s^{2} \\~~~~~\mu=\frac{40 N}{60 kg\times9.80 m/s^{2}}\\~~~~~~~=0.068$

The formula to calculate the work done in overcoming the friction is given by

$W=Fd..........................(2)$

Here, $W$ is the work done and $d$ is the distance travelled.

Substitute $40$ N for $F$ and $8 m$ for $d$ into equation (2) to calculate the work done.

$W=40~N\times8~m\\~~~~= 320~J$

8 0

3 years ago

Dose any one have a old nitro type account they don’t want<br>Ill power it up for you<br>Please

erica [24]

Answer:

no I dont sorry lol. ............

4 0

3 years ago

Read 2 more answers

What effort force will be required to lift the 20 N object using the pulley above?

Umnica [9.8K]

Answer:

I think it is 5N.

Explanation:

4 0

3 years ago

What factor contributes the most to the warm climate of Kaula Lumpur?

inna [77]

Answer:

Latitude

Explanation:

8 0

3 years ago

If you increase the charge on a parallel-plate capacitor from 3 mu or micro CC to 9 mu or micro CC and increase the plate separa

aliya0001 [1]

Explanation:

The energy stored in a capacitor is given by

$U = \dfrac{1}{2}QV = \dfrac{Q^2}{2C}$

In the case of a parallel plate capacitor, the capacitance C is given by

$C = \dfrac{\epsilon_0A}{d}$

so we can rewrite the expression for the energy as

$U = \dfrac{Q^2d}{2\epsilon_0 A}$

Increasing the charge from $3\:\mu\text{C}\:\text{to}\:9\:\mu\text{C}$ means that you're tripling the charge. The same thing is true when you increase the distance from 1.8 mm to 5.4 mm, i.e., you triple the separation distance. So the new energy U' is given by

$U' = \dfrac{Q'^2d'}{2\epsilon_0 A}$

$\:\:\:\:\:\:= \dfrac{(3Q)^2(3d)}{2\epsilon_0 A}$

$\:\:\:\:\:\:= 27\left(\dfrac{Q^2d}{2\epsilon_0 A}\right)$

$\:\:\:\:\:\:= 27U$

As we can see, tripling both the charge and the separation distance result in the 27-fold increase in its stored energy U.

7 0

3 years ago