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Lyrx
10 days ago
9

The equation for the change in the position of a train (measured in units of length) is given by the following expression: x = ½

at² + bt³ where a and b are constants and t is time (measured in units of time). If (mass), (length), and (time) represent units of mass, length and time respectively then b must have units of:
(A) (length)/(time)
(B) (length)/(time³)
(C) (mass)(length)
(D) (length)/(mass)
(E) (mass2)(time2)
(F) 1/(mass)(time)
(G) 1/(length)(time)
(H) (time)(mass)
(I) (mass)(time2)
(J) 1/(time³)
(K) (time)/(mass)
(L) (length)/(time²)

(mass)(length2)
Physics
2 answers:
Keith_Richards [2.2K]10 days ago
6 0
(B) (length)/(time³) Explanation: The term x = ½ at² + bt³ should meet dimensional consistency. This means that both bt³ and ½ at² must share the same units, which are length. To find the dimension of b, we rearrange the equation: [x] = [b]*[t]³ leads to length = [b]*time³, hence [b] = length/time³.
ValentinkaMS [2.4K]10 days ago
3 0
The formula representing a train's position as a function of time is given by x(t) = ½at² + bt³ where a and b are constants. We wish to identify the dimension of b. The variable x signifies length and is measured in meters, while t represents time measured in seconds. The constant b has an unknown unit, as does the constant a. With x = ½at² + bt³, we find that for dimensional correctness, units of a and b must eliminate the time units. Therefore, a should be in m/s² units. Conversely, for b to negate seconds cubed, b must be in meters/seconds³, leading to the conclusion that the unit of b is represented as meters/seconds³, which implies length/time³. Thus, the suitable answer is B.
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Calculate the minimum average power output necessary for a person to run up a 12.0 m long hillside, which is inclined at 25.0° a
Ostrovityanka [2204]

Answer:

Power output, P = 924.15 watts

Explanation:

We have the following parameters:

Length of the ramp, l = 12 m

Weight of the individual, m = 55.8 kg

Incline angle with respect to the horizontal, \theta=25^{\circ}

Elapsed time, t = 3 s

Let h represent the vertical height of the hill:

h=l\ sin\theta

h=12\times \ sin(25)

h = 5.07 m

Power P required for a person to ascend the hill can be expressed as:

P=\dfrac{E}{t}

P=\dfrac{mgh}{t}

P=\dfrac{55.8\times 9.8\times 5.07}{3}

P = 924.15 watts

This indicates that a minimum average power output of 924.15 watts is essential for an individual to ascend this elevation. Thus, this is the answer sought.

3 0
17 days ago
The gas tank of Dave’s car has a capacity of 12 gallons. The tank was 38 full before Dave filled it to capacity. It cost him $2.
Sav [2226]

Answer:

$ 18.75

Explanation:

Given:

Dave's car has a capacity of 12 gallons.

If the tank is at 3/8 full prior to filling:

The cost of gasoline per gallon is $2.50.

The volume Dave needs to fill:

=1 - \dfrac{3}{8}

=\dfrac{5}{8} of full tank

=\dfrac{5}{8}\times 12

Dave must fill 7.5 gallons.

Total amount Dave will pay:

7.5 gallons x $2.50 =

$ 18.75

Thus, Dave will spend $ 18.75 to completely fill the tank.

5 0
29 days ago
Read 2 more answers
As computer structures get smaller and smaller, quantum rules start to create difficulties. Suppose electrons move through a cha
Softa [2029]

Answer:

a) ∆x∆v = 5.78*10^-5

∆v = 1157.08 m/s

b) 4.32*10^{-11}

Explanation:

This problem can be addressed using Heisenberg's uncertainty principle, which is expressed as:

\Delta x\Delta p \geq \frac{\hbar}{2}

Where h represents Planck’s constant (6.62*10^-34 J s).

Assuming that the electron's mass remains the same, we proceed as follows:

\Delta x \Delta (m_ev)=m_e\Delta x\Delta v \geq \frac{\hbar}{2}

Utilizing the electron's mass (9.61*10^-31 kg) and the uncertainty in position (50 nm), we can compute ∆x∆v and ∆v:

\Delta x \Delta v\geq\frac{\hbar}{2m_e}=\frac{(1.055*10^{-34}Js)}{2(9.1*10^{-31}kg)}=5.78*10^{-5}\ m^2/s

\Delta v\geq\frac{5.78*10^{-5}}{50*10^{-9}m}=1157.08\frac{m}{s}

If we treat the electron like a classic particle, the time required to cross the channel is determined using the upper limit of the uncertainty in velocity:

t=\frac{x}{v}=\frac{50*10^{-9}m}{1157.08m/s}=4.32*10^{-11}s

6 0
27 days ago
A wire, of length L = 3.8 mm, on a circuit board carries a current of I = 2.54 μA in the j direction. A nearby circuit element g
Keith_Richards [2256]

It equals 5z

Explanation:

5z

5 0
3 days ago
A jogger accelerates from rest to 3.0 m/s in 2.0 s. A car accelerates from 38.0 to 41.0 m/s also in 2.0 s. (a) Find the accelera
Sav [2226]

Answer:

a) The jogger's acceleration is 1.5 m/s²

b) The car's acceleration is also 1.5 m/s²

c) Yes, the car covers a distance 76 m greater than the jogger.

Explanation:

a) Acceleration is the change in velocity over a given time interval:

a = (final velocity - initial velocity) / time

For the jogger:

a = (3.0 m/s - 0 m/s) / 2.0 s = 1.5 m/s²

b) For the car:

a = (41.0 m/s - 38.0 m/s) / 2.0 s = 1.5 m/s²

c) To find how far the car has traveled after 2 seconds, use the formula for position under acceleration along a straight path:

x = x₀ + v₀ t + ½ a t²

where

x = position at time t

x₀ = initial position

v₀ = initial velocity

t = elapsed time

a = acceleration

Assuming x₀ = 0 (origin at car's starting point):

x = 38.0 m/s × 2 s + ½ × 1.5 m/s² × (2.0 s)²

x = 79 m

Similarly, position of the jogger after 2 seconds is:

x = 0 m/s × 2 s + ½ × 1.5 m/s² × (2.0 s)² = 3 m

The difference traveled by the car compared to the jogger is 79 m - 3 m = 76 m

4 0
1 month ago
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