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marysya
3 days ago
15

A bathtub contains 65 gallons of water and the total weight of the tub and water is approximately 931.925 pounds. You pull the p

lug and the water begins to drain. Let v v represent the number of gallons of water that has drained from the tub since the plug was pulled. Note that water weights 8.345 pounds per gallon. Write an expression in terms of v v that represents the weight of the water that has drained from the tub (in pounds).
Physics
1 answer:
Keith_Richards [1K]3 days ago
8 0

Response:

Q = 8,345 * v

Clarification:

We need an expression that shows how much water has been drained from the tub. This is represented by v, which indicates how many gallons have flowed out since the plug was taken out. Each gallon removed equates to 8.345 pounds of water, so the weight of the drained water Q in pounds as a function of v can be expressed as:

Q = 8,345 * v

Where v signifies the number of gallons emptied from the tub.

Have a great day! Let me know if there's anything else I can assist with.

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A Chevrolet Corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. What
kicyunya [1025]

Response:

83.1946504051 m

Rationale:

u = Starting velocity = 60\ mph=\dfrac{60\times 1609.34}{3600}=26.82233\ m/s

s = Distance traveled = 123\ ft=\dfrac{123}{3.281}=37.4885705578\ m

\theta = Incline = 26^{\circ}

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-26.82233^2}{2\times 37.4885705578}\\\Rightarrow a=-9.5954230306\ m/s^2

Friction coefficient

\mu=-\dfrac{a}{g}\\\Rightarrow \mu=\dfrac{9.5954230306}{9.81}\\\Rightarrow \mu=0.978126710561

mg sin\theta - u mg cos\theta = ma\\\Rightarrow a=9.81(sin26-0.978126710561cos26)\\\Rightarrow a=-4.32382\ m/s^2

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-26.82233^2}{2\times -4.32382}\\\Rightarrow s=83.1946504051\ m

The calculated stopping distance is 83.1946504051 m

6 0
15 days ago
Help asap please!! An aluminum block of mass 12.00 kg is heated from 20 C to 118 C. If the specific heat of aluminum is 913 J-1
Sav [1105]
Q = mCΔT, in which Q = energy required, m = mass of the block, C = specific heat, ΔT = temperature change.

Utilizing the values provided;

Q = 12*913*(118-20) = 1073688 J = 1073.688 kJ.

The correct option is B.
7 0
12 days ago
Part a consider another special case in which the inclined plane is vertical (θ=π/2). in this case, for what value of m1 would t
serg [1198]

The solution leads to the conclusion that m1 = m2

For mass m1, the force balance in the y direction equals zero:
0 = T - m1*g
Rearranging gives:
m1*g = T

For mass m2, the force balance in the y direction equals zero:
0 = T - m2*g
Rearranging provides:
m2*g = T

Setting these two equal allows us to solve for m1:
m1*g = m2*g

= m1 = m2

 

Explanation:

The force acting on each individual mass pulls down while the tension created by the other mass exerts an upward force due to the operation of the pulley system, resulting in balanced forces on both masses.

6 0
7 days ago
A transformer is to be designed to increase the 30 kV-rms output of a generator to the transmission-line voltage of 345 kV-rms.
inna [987]

Answer:

n_s = 920 \turns

Explanation:

Given,

Voltage of the primary coil (V_p) = 30 kV-rms

Voltage of the secondary coils (V_s) = 345 kV-rms

number of turns in the primary coil (n_p) = 80 turns

number of turns in the secondary coil (n_s) =?

the ratio of turns between primary and secondary coils

     \dfrac{n_p}{n_s} = \dfrac{V_p}{V_s}

     \dfrac{n_s}{n_p} = \dfrac{V_s}{V_p}

     n_s = n_p \dfrac{V_s}{V_p}

     n_s = 80\times \dfrac{345}{30}

     n_s = 80\times 11.5

     n_s = 920 \turns

The number of turns in the secondary coil is equal to n_s = 920 \turns

4 0
1 day ago
2. On January 21 in 1918, Granville, North Dakota, had a surprising change in temperature. Within 12 hours, the temperature chan
Yuliya22 [1153]

Respuesta:

El cambio de temperatura en Celsius equivale a 46°C.

El cambio de temperatura en Fahrenheit equivale a 82.8°F.

Explicación:

Un grado Celsius es equivalente a un grado Kelvin; por lo tanto,

\Delta C = \Delta K = 283K-237K\\\\ \boxed{\Delta C = 46^oC}

Un grado Fahrenheit es 1.8 veces un grado Celsius; por lo tanto

\Delta F = 1.8(46^o)

\boxed{\Delta F = 82.8^oF}

Por lo tanto, el cambio en Celsius es de 46°C y en Fahrenheit es de 82.8°F.

7 0
11 days ago
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