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Troyanec [42]
3 years ago
12

When a 40N force, parallel to and directed up the incline, is applied to a block on a frictionless incline that is 30o above the

horizontal, the acceleration of the block is 2.0 m/s2, up the incline. Find the mass of the block.

Physics
2 answers:
Vlad1618 [11]3 years ago
4 0

Answer:

5.8 kg

Explanation:

We are given that

Force,F=40 N

\theta=30^{\circ}

Acceleration,a=2 m/s^2

We have to find the mass of block.

F_{net}=ma

F-mgsin\theta=ma

F=mgsin\theta+ma=m(gsin\theta+a)

40=m(9.8sin30+2)

Where

g=9.8m/s^2

40=6.9m

m=\frac{40}{6.9}=5.8 kg

Karolina [17]3 years ago
4 0

Answer:

m=5.797\ kg

Explanation:

GIVEN:

applied force up the inclined plane, F=40\ N

angle of the plank from the horizontal, \theta=30^{\circ}

acceleration of the body up the plane, a=2\ m.s^{-2}

Now from the  schematic we balance the forces:

F-mg\sin\theta=m.a

F=m(g\sin\theta +a)

40=m(9.8\times0.5+2)

m=5.797\ kg

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A 4,000 kg boat floats with one-third of its volume submerged. if two more people get into the boat, each of whom weighs 690 n,
AveGali [126]

 

The weightiness of the added water displaced is equivalent to the joined weight of the two extra people who come to be into the boat:


<span>m water g                   = 2 x 690 N</span>

<span>                                   = 1,380 N</span>

<span>
</span>

The mass of the water displace is then


<span>m water g                   = 1,380 N</span>

<span>                                   = 1,380 N / 9.8 m/s^2</span>

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Compute the calculation for density for the volume of water displace and practice this outcome for the mass of the water displace to get the answer:


<span>p water                      = mass of water / volume of water</span>

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<span>volume of water        = mass of water / p water</span>

<span>                                  = 141 kg / 1000 kg /m^3 eliminate kilogram</span>

<span>                                  = 0.14 m^3 the additional volume of water that is displaced</span>

4 0
4 years ago
Suppose you design an apparatus in which a uniformly charged disk of radius R is to produce an electric field. The field magnitu
ANEK [815]

Answer:

The electric field will be decreased by 29%

Explanation:

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\hat {K_a} = \dfrac{\sigma}{2 \varepsilon _o} \Big( 1 - \dfrac{z}{\sqrt{z^2+R_i^2}} \Big))

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Similarly;

\hat {K_b} = \hat {k_a} - \dfrac{\sigma}{2 \varepsilon_o} \Big( 1 - \dfrac{2.0 \ R}{\sqrt{(2.0 \ r)^2 + (\dfrac{R}{2}^2)}}\Big)

However; the relative difference is: \dfrac{\hat {k_a} - \hat {k_b}}{\hat {k_a} }= \dfrac{E_a -E_a + \dfrac{\sigma}{2 \varepsilon_o  \Big[1 - \dfrac{2.0 \ R}{\sqrt{(2.0 \ R)^2 + (\dfrac{R}{2})^2}} \Big] } } { \dfrac{\sigma}{2 \varepsilon_o \Big [ 1 - \dfrac{2.0 \ R}{\sqrt{ (2.0 \ R)^2 + (R)^2}} \Big] }}

\dfrac{\hat {k_a} - \hat {k_b}}{\hat {k_a} }= \dfrac{1 - \dfrac{2.0}{\sqrt{(2.0)^2 + \dfrac{1}{4}}} }{1 - \dfrac{2.0 }{\sqrt{(2.0)^2 + 1}}}

= 0.2828 \\ \\ \mathbf{\simeq  29\%}

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