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mote1985 [20]
2 years ago
9

A rare and valuable antique chest is being moved into a truck using a 4.00 m long ramp. the kj weight of the chest plus packing

material is 1,500 n. if the truck bed is 1.00 m above the ground, find the work done by the movers as they slide the chest up the ramp if the coefficient of friction between the chest and the ramp is 0.200.
Physics
1 answer:
Lisa [10]2 years ago
7 0

First let us calculate for the angle of inclination using the sin function,

sin θ = 1 m / 4 m

θ = 14.48°

 

Then we calculate the work done by the movers using the formula:

W = Fnet * d

 

So we must calculate for the value of Fnet first. Fnet is force due to weight minus the frictional force.

Fnet = m g sinθ – μ m g cosθ

Fnet = 1,500 sin14.48 – 0.2 * 1,500 * cos14.48

Fnet = 84.526 N

 

So the work exerted is equal to:

W = 84.526 N * 4 m

<span>W = 338.10 J</span>

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2H2S(g)⇌2H2(g)+S2(g),Kc=1.67×10−7 at 800∘C is carried out at the same temperature with the following initial concentrations: [H2
scoray [572]

Answer : The concentration of S_2 at equilibrium will be, 1.67\times 10^{-7}M

Explanation :  Given,

Equilibrium constant = 1.67\times 10^{-7}

Initial concentration of H_2S = 0.100 M

Initial concentration of H_2 = 0.100 M

Initial concentration of S_2 = 0.00 M

The balanced equilibrium reaction is,

                      2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)

Initial conc.    0.1                0.1          0

At eqm.         (0.1-2x)       (0.1+2x)      x

The expression of equilibrium constant for the reaction will be:

K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}

Now put all the values in this expression, we get :

1.67\times 10^{-7}=\frac{(0.1+2x)^2\times (x)}{(0.1-2x)^2}

By solving the term 'x' by quadratic equation, we get two value of 'x'.

x=1.67\times 10^{-7}M

Concentration of S_2 at equilibrium = x=1.67\times 10^{-7}M

Therefore, the concentration of S_2 at equilibrium will be, 1.67\times 10^{-7}M

6 0
2 years ago
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A satellite that orbits Earth with a speed of v0 must be in an orbit of radius 8RE to maintain a circular orbit, where RE is the
NISA [10]

Answer:

1.024 × 10⁸ m

Explanation:

The velocity v₀ of the orbit 8RE is v₀ = 8REω where ω = angular speed.

So, ω =  v₀/8RE

For the orbit with radius R for it to maintain a circular orbit and velocity 2v₀, we have

2v₀ = Rω

substituting ω =  v₀/8RE into the equation, we have

2v₀ = v₀R/8RE

dividing both sides by v₀, we have

2v₀/v₀ = R/8RE

2 = R/8RE

So, R = 2 × 8RE

R = 16RE

substituting RE = 6.4 × 10⁶ m

R = 16RE

= 16 × 6.4 × 10⁶ m

= 102.4 × 10⁶ m

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8 0
1 year ago
Calculate the energy released in joules when one mole of polonium-214 decays according to the following equation21484 Po --&gt;
GuDViN [60]

Answer:

ΔE = 8.77 × 10¹¹ J

Explanation:

given,

²¹⁴₈₄Po -----> ²¹⁰₈₂Pb + 42 He

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Po-214 = 213.99519 amu

He-4 = 4.00260 amu

1 kg = 6.022 × 10²⁶ amu;

NA = 6.022 × 10²³ mol⁻¹

c = 2.99792458 × 10⁸ m/s

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- 0.00975 amu = - 0.00975 x 1.66 x 10⁻²⁷ Kg

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total mass = 6.022 × 10²³ x -0.016185 x 10⁻²⁷

                 = -0.097467 x 10⁻⁴ Kg

ΔE = -(0.097467 x 10⁻⁴) (3 x 10^8)²

ΔE = - 8.77 × 10¹¹

ΔE = 8.77 × 10¹¹ J

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igor_vitrenko [27]

Answer:

Hydrogen has one electron and one proton

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All the weight of the wooden board is bear by the support located at the centre of the rod, and the other support which is located at the end, will have no reaction force, or 0 reaction force.

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