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e-lub [12.9K]
3 years ago
12

¿Qué trabajo hace una fuerza de 110 N cuando mueve su punto de aplicación 20 mt en su misma dirección? *

Physics
1 answer:
lys-0071 [83]3 years ago
7 0

Answer:

W = 2.200 Joules

Explanation:

Datos (data):

  • Fuerza [force] (F) = 110 N
  • Metros [meters] (m) = 20 m
  • Trabajo [work] (W) = ?

Usar la fórmula (use formula):

  • \boxed{\bold{W = F*d}}

Reemplazar (replace):

  • \boxed{\bold{W = 110\ N*20\ m}}

Resolver la multiplicación, recuerda que 1 N * 1 m = 1 J (resolve the multiplication, remember that 1 N * 1 m = 1 J:

  • \boxed{\boxed{\bold{W =2.200\ J}}}

Greetings.

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Two moles of helium are initially at a temperature of 21.0 ∘Cand occupy a volume of 3.30×10−2 m3 . The helium first expands at c
Anettt [7]

Answer:

(B) The total internal energy of the helium is 4888.6 Joules

(C) The total work done by the helium is 2959.25 Joules

(D) The final volume of the helium is 0.066 cubic meter

Explanation:

(B) ∆U = P(V2 - V1)

From ideal gas equation, PV = nRT

T1 = 21°C = 294K, V1 = 0.033m^3, n = 2moles, V2 = 2× 0.033=0.066m^3

P = nRT ÷ V = (2×8.314×294) ÷ 0.033 = 148140.4 Pascal

∆U = 148140.4(0.066 - 0.033) = 4888.6 Joules

(C) P2 = P1(V1÷V2)^1.4 =148140.4(0.033÷0.066)^1.4= 148140.4×0.379=56134.7 Pascal

Assuming a closed system

(C) Wc = (P1V1 - P2V2) ÷ 0.4 = (148140.4×0.033 - 56134.7×0.066) ÷ 0.4 = (4888.6 - 3704.9) ÷ 0.4 = 1183.7 ÷ 0.4 = 2959.25 Joules

(C) Final volume = 2×initial volume = 2×0.033= 0.066 cubic meter

6 0
3 years ago
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Answer:

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Explanation:

3 0
3 years ago
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Two protons in an atomic nucleus are typically separated by a distance of 2 ✕ 10-15 m. The electric repulsion force between the
castortr0y [4]

Answer:

The magnitude of the electric force between the to protons will be 57.536 N.

Explanation:

We can use Coulomb's law to find out the force, in scalar form, will be:

F \ = \ \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{d^2}.

Now, making the substitutions

d \ = \ 2.00 * 10 ^{-15} \ m,

q_1 = q_2 = 1.60 * 10 ^ {-19} \ C,

\frac{1}{4\pi\epsilon_0}=8.99 * 10^9 \frac{Nm^2}{C^2},

we can find:

F \ = \ 8.99 * 10^9 \frac{Nm^2}{C^2} \frac{(1.60 * 10 ^ {-19} \ C)^2}{(2.00 * 10 ^{-15} \ m)^2}.

F \ = 57.536 N.

Not so big for everyday life, but enormous for subatomic particles.

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3 years ago
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A big wheel at a theme park has a diameter of 14m and people on the ride complete one revolution in 24s. calculate the distance
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Explanation:

We'll call the radius r and the diameter d:

We also assume that the riders are at a distance r = d/2 = 7m from the center of the wheel.

The period of the wheel is 24s. The tangent velocity of the wheel (and the riders) will be: (2pi/T)*r = 0.8 m/s (circa).

It means that in 3 minutes (180 seconds) they'll run 0.8 m/s * 180s = 144m.

Hopefully I understood the question. If yes, that's the answer.

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2 years ago
While a roofer is working on a roof that slants at 39.0 degrees above the horizontal, he accidentally nudges his 88.0 N toolbox,
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Answer:

V= 6.974 m/s

Explanation:

Component( box) weight acting parallel and down roof 88(sin39.0°)=55.4 N

Force of kinetic friction acting parallel and up roof = 18.0 N

Fnet force acting on tool box acting parallel and down roof

Fnet= 55.4 - 18.0

Fnet=37.4 N

acceleration of tool box down roof

a = 37.4(9.81)/88.0

a= 4.169 m/s²

d = 4.90 m

t = √2d/a

t= √2(4.90)/4.169

t= 1.662 s

V = at

V= 4.169(1.662)

V= 6.974 m/s

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