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3 years ago

7

10. A 25 kg apple cart is being pushed with an applied force of 115 N. The coefficient of friction between the ground and the ca

rt is 0.35. What is the acceleration of the cart?

1 answer:

ziro4ka [17]3 years ago

8 0

Answer:

1.1 m/s²

Explanation:

From the question,

F -mgμ = ma.................... Equation 1

Where F = applied force, m = mass of the apple cart, g = acceleration due to gravity, μ = coefficient of friction., a = acceleration of the apple cart.

Given: F = 115 N, m = 25 kg, μ = 0.35

Constant: g = 10 m/s²

Substitute these values into equation 2

115-(25×10×0.35) = 25×a

115-87.5 = 25a

25a = 27.5

a = 27.5/25

a = 1.1 m/s²

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In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between st

bekas [8.4K]

Answer:

a) t = 746 s

b) t = 666 s

Explanation:

a)

Total time will be the sum of the partial times between stations plus the time stopped at the stations.
Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.
At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.
In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:

$v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s (1)$

Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:

$t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)$

Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:

$x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m (3)$

In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:

$t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)$

We can find the distance traveled while the train was decelerating as follows:

$x_{3} = (v_{1} * t_{3}) + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m (5)$

Finally, we need to know the time traveled at constant speed.
So, we need to find first the distance traveled at the constant speed of 26.4m/s.
This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:
x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)

The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

$t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s (7)$

Total time between two stations is simply the sum of the three times we have just found:
t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)
Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:
tm = t*5 = 131.6 * 5 = 658.2 s (9)
Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:
Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)

b)

Using all the same premises that for a) we know that the only difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.
Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.
We can find the distance traveled at constant speed, rewriting (6) as follows:

x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)

The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

$t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s (12)$

Total time between two stations is simply the sum of the three times we have just found:
t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)
Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:
tm = t*3 = 207.4 * 3 = 622.2 s (14)
Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:
Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)

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2 years ago

What if m1 is initially moving at 3.4 m/s while m2 is initially at rest? (a) find the maximum spring compression in this case?

Lisa [10]

<span>Ans : Initial E = KE = Â˝mvÂ˛ = Â˝ * 1.2kg * (2.2m/s)Â˛ = 2.9 J max spring compression where both velocities are the same: conserve momentum: 1.2kg * 2.2m/s = (1.2 + 3.2)kg * v â†’ v = 0.6 m/s which means the combined KE = Â˝ * (1.2 + 3.2)kg * (0.6m/s)Â˛ = 0.79 J The remaining energy went into the spring: U = (2.9 - 0.79) J = 2.1 J = Â˝kxÂ˛ = Â˝ * 554N/m * xÂ˛ x = 0.0076 m â† (a)</span>

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2 years ago

38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari

kolbaska11 [484]

Answer:

Final volumen first process $V_{2} = 98,44 cm^{3}$

Final Pressure second process $P_{3} = 1,317 * 10^{10} Pa$

Explanation:

Using the Ideal Gases Law yoy have for pressure:

$P_{1} = \frac{n_{1} R T_{1} }{V_{1} }$

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

$n_{1} = n_{2} = n$

In an isobaric process the pressure is constant so:

$P_{1} = P_{2}$

$\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }$

$\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }$

$V_{2} = \frac{T_{2} V_{1} }{T_{1} }$

Replacing : $T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}$

$V_{2} = 98,44 cm^{3}$

Replacing on the ideal gases formula the pressure at this piont is:

$P_{2} = 3,92 * 10^{9} Pa$

For Temperature the ideal gases formula is:

$T = \frac{P V }{n R }$

For the second process you have that $T_{2} = T_{3}$ So:

$\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }$

$P_{2} V_{2} = P_{3} V_{3}$

$P_{3} = \frac{P_{2} V_{2}}{V_{3}}$

$P_{3} = 1,317 * 10^{10} Pa$

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2 years ago

A family car has a mass of 1400 kg. In an accident it hits a wall and goes from a speed of 27 m/s to a standstill in 1.5 seconds

horrorfan [7]

Answer:

The force has been reduced by 8018 N

Explanation:

The impulse exerted on the car during the crash is equal to the product of the force exerted and the duration of the collision, and it is also equal to the change in momentum of the car. So we can write:

$F\Delta t = m\Delta v$

where:

F is the force exerted on the car

$\Delta t$ is the duration of the collision

m = 1400 kg is the mass of the car

$\Delta v=-27 m/s$ is the change in velocity of the car

We can re-write the equation as

$F=\frac{m\Delta v}{\Delta t}$

In the 1st collision, the time is 1.5 seconds, so the force is

$F_1=\frac{(1400)(-27)}{1.5}=-25,200 N$

In the 2nd collision, the time is increased to 2.2 seconds, so the force is

$F_2=\frac{(1400)(-27)}{2.2}=-17,182 N$

Therefore, the force has been reduced by:

$F_2-F_1=-17,182-(-25,200)=8018 N$

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Explanation:

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