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skelet666 [1.2K]

3 years ago

14

What type of energy can take up space

1 answer:

Arada [10]3 years ago

8 0

Answer:

Yes energy does take up space.

Explanation:

Every form of energy has a defining characteristic; sound is the vibration of molecules, electricity is the movement of electrons, and mass is the thing that take up space.

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A block is at rest on the incline shown in the figure. The coefficients of static and ki- netic friction are μs = 0.62 and μk =

GREYUIT [131]

The frictional force is 218.6 N

Explanation:

The block in the problem is at rest along the inclined surface: this means that the net force acting along the direction parallel to the incline must be zero.

There are two forces acting along this direction:

- The component of the weight parallel to the incline, downward along the plane, of magnitude

$mg sin \theta$

where

m = 46 kg is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=29^{\circ}$ is the angle of the incline

- The (static) frictional force, acting upward, of magnitude $F_f$

Since the block is in equilibrium, we can write

$mg sin \theta - F_f = 0$

And substituting, we find the force of friction:

$F_f = mg sin \theta = (46)(9.8)(sin 29^{\circ})=218.6 N$

Learn more about frictional force along an inclined plane:

brainly.com/question/5884009

#LearnwithBrainly

7 0

3 years ago

You’re still in a redecorating mood and want to move a 35kg bookcase to a different place in the living room. You exert 58 N on

CaHeK987 [17]

Answer: 0.16

Explanation:

Newton's second law states that the resultant of the forces exerted on the bookcase is equal to the product between the mass (m) and the acceleration (a):

$F-F_f = ma$

where

F = 58 N is the force applied

Ff is the frictional force

Substituting, we find the frictional force

$F_f = F-ma=58 N-(35 kg)(0.12 m/s^2)=53.8 N$

The frictional force has the form:

$F_f = \mu mg$

where $\mu$ is the coefficient of kinetic friction. Re-arranging the formula, we can find the coefficient:

$\mu=\frac{F_f}{mg}=\frac{53.8 N}{(35 kg)(9.8 m/s^2)}=0.16$

4 0

4 years ago

Describe what would happen if anything to the mass and weight when you go to jupiter

Art [367]

If you were somehow magically transported to Jupiter, your mass
would not change, but your weight would become roughly 2.5 times
your Earth weight.

6 0

3 years ago

Read 2 more answers

6.P.2.1 Matter: Properties and Changes Notes

Zinaida [17]

Answer:

The __law__ conservation

of state that either energy may produced nor destroyed

8 0

4 years ago

Two particles are fixed to an x axis: particle 1 of charge −1.50 ✕ 10−7 c at x = 6.00 cm, and particle 2 of charge +1.50 ✕ 10−7

sleet_krkn [62]

Answer : $\underset{E_{R}}{\rightarrow} =-2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Explanation :

Given that,

Charge of particle 1 = $-1.50\times10^{-7} c$

Distance x = 6 cm

Charge of particle 2 = $1.50\times10^{-7} c$

Distance x = 27 cm

Total distance = $\dfrac{6+27}{2}$

$r = 16.5\ cm$

Particle 1 is at (6,0) and particle 2 is at (27,0) .

Therefore, midway (16.5, 0)

Now, $r = \dfrac{|6-16.5|}{2} = \dfrac{|27-16.5|}{2} = 10.5\ cm$

Formula of electric field

$E = \dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q}{r^{2}}$

Now, the the electric field due to particle 1

$\underset{E}{\rightarrow}\ = -\dfrac{9\times10^{9}\times1.50\times10^{-7 }}{10.5}\ \widehat{i} \dfrac{N}{C}$

$\underset{E}{\rightarrow} = \dfrac{13.5\times10^{2}}{(10.5\times10^{-2})^{2}}\widehat{i} \dfrac{N}{C}$

$\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Similarly, the electric field due to particle 2

$\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Resultant Electric field

$\underset{E_{R}}{\rightarrow} = \underset{E_{1}}{\rightarrow} + \underset{E_{2}}{\rightarrow}$

$\underset{E_{R}}{\rightarrow} = -2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Hence, this is the required answer.

3 0

3 years ago