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3 years ago

Unmoving objects are being acted on by balance forces. TRUE or FALSE?

2 answers:

8 0

True
hope this helps also their is a quizlet that can help you if you are still confused

https://quizlet.com/104253745/quizguide-flash-cards/

Reil [10]3 years ago

3 0

True. There are forced acting on it, but as they're balanced it is unmoving

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A horizontal force, F1 = 65 N, and a force, F2 = 12.4 N acting at an angle of θ to the horizontal, are applied to a block of mas

Nezavi [6.7K]

Answer:

(a) FN = 24.18 N

(b) a = 22.87 m/s²

Explanation:

Newton's second law of the block:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the block on the surface and the y-axis in the direction perpendicular to it.

F₁ : Horizontal force

F₂ : acting at an angle of θ to the horizontal,

W: Weight of the block : In vertical direction

FN : Normal force : perpendicular to the direction the surface

fk : Friction force: parallel to the direction to the surface

Known data

m =3.1 kg : mass of the block

F₁ = 65 N, horizontal force

F₂ = 12.4 N acting at an angle of θ to the horizontal

θ = 30° angle θ of F₂ with respect to the horizontal

μk = 0.2 : coefficient of kinetic friction between the block and the surface

g = 9.8 m/s² : acceleration due to gravity

Calculated of the weight of the block

W= m*g = (3.1 kg)*(9.8 m/s²) = 30.38 N

x-y F₂ components

F₂x = F₂cos θ= (12.4)*cos(30)° = 10.74 N

F₂y = F₂sin θ= (12.4)*sin(30)° = 6.2 N

a)Calculated of the Normal force (FN)

We apply the formula (1)

∑Fy = m*ay ay = 0

FN+6.2-30.38 = 0

FN = -6.2+30.38

FN = 24.18 N

Calculated of the Friction force:

fk=μk*N= 0.2* 24.18 N = 4.836 N

b) We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax , ax= a : acceleration of the block

F₁ + F₂x -fk = ( m)*a

65 N + 10.74 -4.836 = ( 3.1)*a

70.904 = ( 3.1)*a

a = (70.904 ) / ( 3.1)

a = 22.87 m/s²

4 0

2 years ago

A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

4 0

3 years ago

How much heat is evolved when 1234 g of water condenses to a liquid at 100.°c?

lbvjy [14]

During the phase transition vapour --> liquid water, the temperature of the water does not change; the molecules of water release heat and the amounf of heat released is equal to
$Q=mL_e$
where
m is the mass of the water
$L_e$ is the latent heat of evaporation.

For water, the latent heat of evaporation is $L=2257 kJ/kg$ , while the mass of the water is
$m=1234 g=1.234 kg$

so, the amount of heat released in the process is
$Q=mL_e = (1.234 kg)(2257 kJ/kg)=2785 kJ$

6 0

3 years ago

Suppose a firm is producing 2,475 units of output by hiring 50 workers (W = $20 per hour) and 25 units of capital (R = $10 per h

Neko [114]

Answer

given,

firm is producing = 2,475 units

output by hiring 50 workers W = $20 per hour

25 units of capital R = $10 per hour

marginal product of labor = 40

marginal product of capital = 25

$\dfrac{MP_l}{MP_c}=\dfrac{40}{25}$