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3 years ago

What is the work done if you push a tree with 50 N of force but the tree does not move?

Physics

1 answer:

Soloha48 [4]3 years ago

4 0

Answer:

The work done is 0.

Explanation:

The reason no work is done is because the equation W = Fs.

W = work

F= force

s= displacement

In this scenario F = 50 and s= 0

Therefore.

W = 50(0)

W = 0

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OLga [1]

Answer:

ggn

Explanation:

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3 years ago

A cube of wood having an edge dimension of 20.0cm and a density of 650 kg /m³ floats on water. (b) What mass of lead should be p

schepotkina [342]

The mass added is "m" so the complete cube is submerged in the water is 2.8 kg.

<h3>What mass of lead should be placed on the cube?</h3>

Given: Side of the cube (a) = 20cm

The density of the cube (ρc) = $$$650 kg/m^3$

a) Applying the force balance, the buoyant force must be equal to the weight of the cube

ρcgV = ρg × (Ax)

Substituting the values in the above equation, we get

$(650*(0.2 m)^3)=1000*(0.2 m)^2*x$

x = 0.13

where x is the height of the cube in the water

$$$A = a^2$ is the area of the cross-section

ρ is the density of the water

V is the volume of the cube

Now, the height above the surface of the water would be

h = a − x

Substituting the values, then we get

h = 0.2 − 0.13

h = 0.07 m

b) The mass added is "m" so the complete cube is submerged in the water, therefore

ρcgV + mg = ρg × (V)

$$(650*(0.2 m)^3)+m=1000*(0.2 m)^3$

m = 2.8 kg

The mass added is "m" so the complete cube is submerged in the water is 2.8 kg.

To learn more about buoyant force refer to:

brainly.com/question/11884584

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7 0

1 year ago

The loop is in a magnetic field 0.20 T whose direction is perpendicular to the plane of the loop. At t = 0, the loop has area A

love history [14]

Answer:

Part a)

$EMF = 14 \times 10^{-3} V$

Part b)

$EMF = 15.67 \times 10^{-3} V$

Explanation:

As we know that magnetic flux through the loop is given as

$\phi = B.A$

now we have

$\phi = B\pi r^2$

now rate of change in flux is given as

$\frac{d\phi}{dt} = B(2\pi r)\frac{dr}{dt}$

now we know that

$A = \pi r^2$

$0.285 = \pi r^2$

$r = 0.30 m$

Now plug in all data

$EMF = (0.20)\times 2\pi\times (0.30) \times (0.037)$

$EMF = 14 \times 10^{-3} V$

Part b)

Now the radius of the loop after t = 1 s

$r_1 = r_0 + \frac{dr}{dt}$

$r_1 = 0.30 + 0.037$

$r_1 = 0.337 m$

Now plug in data in above equation

$EMF = (0.20)\times 2\pi\times (0.337) \times (0.037)$

$EMF = 15.67 \times 10^{-3} V$

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3 years ago

An artist working on a piece of metal in his forging studio plunges the hot metal into oil in order to harden it. The metal piec

lutik1710 [3]

Answer:

The temperature of the metal is $T_m = 376.8 ^o C$

Explanation:

From the question we are told that

The mass of the metal is $M = 60 \ kg$

The specific heat of the metal is $c_p = 0.1027 kcal/(kg \cdot ^oC)$

The mass of the oil is $M_o = 810 \ kg$

The temperature of the oil is $T_o = 35^oC$

The specific heat of oil is $c_o = 0.7167 kcal/(kg \cdot ^oC )$

The equilibrium temperature is $T_e = 39 ^oC$

According to the law of energy conservation

Heat lost by metal = heat gained by the oil

The quantity of heat lost by the metal is mathematically represented as

$Q = - Mc_p \Delta T$

=> $Q = -Mc_p (T_m - T_c)$

Where $T_ m$ the temperature of metal before immersion

The negative sign show heat lost

The quantity of gained t by the metal is mathematically represented as

$Q = M_o c_o \Delta T$

=> $Q = M_o c_o (T_c - T_o)$

$Mc_p (T_m - T_c) = M_o c_o (T_c - T_o)$

substituting values

$- 60 * 0.1027 (T_m - 39) = 810 * 0.7167 * (39 - 35)$

=> $T_m = 376.8 ^o C$

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3 years ago

N discussing engines, the ratio of output work to input work expressed as a percentage is called

djyliett [7]

The ration of output work to input work expressed as a percentage is called <u>Efficiency</u>.

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3 years ago