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igomit [66]
3 years ago
12

How much work does this force do as the particle moves along the x-axis from x = 0 to x = l? express your answer in terms of the

variables f0 and l?
Physics
1 answer:
nydimaria [60]3 years ago
3 0
<h3><u>Answer</u>;</h3>

= F0 L ( 1 - 1/e )

<h3><u>Explanation;</u></h3>

Work done is given as the product of force and distance.

In this case;

Work done  = ∫︎ F(x) dx  

                    = F0 ∫︎ e^(-x/L) dx  

                    = F0 [ -L e^(-x/L) ] between 0 and L  

                    = F0 L ( 1 - 1/e )

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A student sees a newspaper ad for an apartment that has 2330 square feet (ft2) of floor space. How many square meters of area ar
Evgen [1.6K]
216.4641 m^2........................
3 0
3 years ago
The New York Wheel is the world's largest Ferris wheel. It's 183 meters in diameter and rotates once every 37.3 min.
bagirrra123 [75]
The radius of the Ferris wheel is r = 83/2 = 91.5 m.

The angular velocity is
ω = (2π rad)/(37.3*60 s) = 2.8075 rad/s
The tangential velocity is
v = rω = 0.2569 m/s

The arc length traveled in 8.6 min (= 8.6*60 s = 516 s) is
s = (91.5 m)*(2.8075 rad/s)*(516 s) = 132.553 m
The central angle swept is
θ = 132.553/91.5 = 1.4487 rad

From the vector diagram, the change in velocity is (from the Law of Cosines)
Δv² = v²(1 - 2 cosФ)
where Ф = Π - 1.4487 = 1.6929 rad
Δv² = 0.2569²[1 - 2*(-0.1218)] = 0.0821
Δv = 0.2865 m/s

The acceleration is
a₁ = (0.2865 m/s)/(516 s) = 5.6 x 10⁻⁴ m/s²

The actual centripetal acceleration is directed toward the center of the wheel, and its value is
a = v²/r = 0.2569²/91.5 = 7.2 x 10⁻⁴ m/s²

Answer: 
a = 7.2 x 10⁻⁴ m/s², the centripetal acceleration acting toward the center of the wheel.
The magnitude of a₁ is 5.6 x 10⁻⁴ m/s², but it is not directed toward the center of the wheel.

7 0
3 years ago
What type of industry might you expect to find on land near volcanoes
Bad White [126]
I believe it is steam.
3 0
3 years ago
A 970-kg sports car collides into the rear end of a 2300-kg SUV stopped at a red light. The bumpers lock, the brakes are locked,
vlabodo [156]

Answer:

22.73 m/s or 81.72 kph

Explanation

We can find the combined mass of both cars as

970 kg + 2300 kg = 3270 kg.

Then the normal force of the cars can be calculated as

F(n)= mg

Where g is acceleration due to gravity 9.8m/s^2

3270 kg ×9.8 = 32046 kg*m/s^2.

coefficient of kinetic friction between tires and road to be 0.80 × F(n)

Then the frictional force can be calculated as

= (32046kg*m/s^2 × 0.80 )

= 25636.8 kg*m/s^2

We can now calculate the work done that was used stopping the cars as

Frictional force × distance

(25636.8 kg*m/s^2 ) × 2.9m= 74346.72kg*m^2/s^2

From kinetic energy formula, the combined velocity of the car can be determined

E=0.5 M V²

√(2E/M) = V

√(2*74346.72kg*m^2/s^2 / 3270 kg) = V

V= √ (45.472)

V=6.743293m/s

the momentum of both cars can be determined as

6.743293m/s * 3270 kg

= 22050.57kg*m/s

Now the final momentum of both cars must be equal to the the momentum of

the sports car just prior to the collision. Therefore, the speed of the sports car at impact.

=(22050.57 kg*m/s) / 970 kg = 22.73 m/s

We can convert that to km/h.

22.73 m/s * 3600 s/h / 1000 m/km = 81.72 kph

7 0
3 years ago
Two Carnot air conditioners, A and B, are removing heat from different rooms. The outside temperature is the same for both rooms
amid [387]

Answer:

a) Work required for air conditioner A = 354.7 J

b) Work required for air conditioner B = 310.3 J

c) The magnitude of the heat deposited outside for conditioner A = 4684.7 J

d) The magnitude of the heat deposited outside for conditioner B = 4640.3 J

Explanation:

In a carnot air conditioner, it operates like a reverse carnot engine; i.e. it removes heat from the cold reservoir (making it colder) and dumps the heat in the hot reservoir (making it hotter)

For a Carnot air conditioner,

Q꜀ is the heat removed from the colder reservoir = 4330 J for both cases

T꜀ is the temperature of the colder reservoir (temperature of the rooms) = 293 K and 296 K for both cases to be considered.

Qₕ is the heat deposited in the warmer reservoir = ? for both cases

Tₕ is the temperature of the hot reservoir (temperature of outside) = 317 K for both cases.

For Carnot air conditioners,

Qₕ = W + Q꜀ (eqn 1)

And

(Qₕ/Tₕ) - (Q꜀/T꜀) = 0 (eqn 2)

Making Qₕ the subject of formula in eqn 2

Qₕ = Tₕ (Q꜀/T꜀)

Substituting this into eqn 1

Tₕ (Q꜀/T꜀) = W + Q꜀

Q꜀ (Tₕ/T꜀) - Q꜀ = W

Q꜀ [(Tₕ - T꜀)/T꜀ ] = W

W = Q꜀ [ (Tₕ - T꜀)/T꜀ ]

For the air conditioner A,

T꜀ = 293 K, Tₕ = 317 K, Q꜀ = 4330 J, W = ?

W = Q꜀ [ (Tₕ - T꜀)/T꜀ ] = 4330 [ (317 - 293)/293] = 354.7 J

For the air conditioner B,

T꜀ = 296 K, Tₕ = 317 K, Q꜀ = 4330 J, W = ?

W = Q꜀ [ (Tₕ - T꜀)/T꜀ ] = 4330 [ (317 - 296)/296] = 310.3 J

c) Qₕ = W + Q꜀

For conditioner A,

Qₕ = 354.7 + 4330 = 4684.7 J

For conditioner B,

Qₕ = 310.3 + 4330 = 4640.3 J

8 0
3 years ago
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