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2 years ago

A car travels 30 miles in 1 hour on a winding mountain road. Which of the following is a true statement?

Physics

2 answers:

JulsSmile [24]2 years ago

5 0

Answer: The only true option is C

Explanation:

A) because the car travels only in one direction, the total displacement (the difference of the final position and the initial position) is equal to the distance traveled. This option is false

B) The car moves at an average speed of exactly 30 miles per hour, this statement is false.

C) For the thing said before, this statement is correct, the average speed is equal to the total distance moved divided by the lapse of time, here the car moves 30 miles in one hour, so the average speed is 30 miles per hour

D) This can be true but not necessarily, the car may have varied the velocity, we know that the car moved 30 miles in the lapse of one hour, but maybe in some parts the velocity was 20mph and in other parts was 40mph, we do not really know, so this is false.

siniylev [52]2 years ago

4 0

Answer:

The true statement is:

"(C) The magnitude of the average velocity is equal to 30 m.p.h."

Explanation:

Given that a car travels 30 miles in 1 hour on a winding mountain road.

Let' check all the statements one by one:

(A) The magnitude of the total displacement is larger than the distance traveled.

Since the entire motion of the car is not exactly given in the question, so it is not possible to tell whether the magnitude of the total displacement is larger than the distance traveled or not.

Thus, this statement is not true.

(B) The magnitude of the average velocity is greater than 30 m.p.h.

The average velocity of an object is defined as the total displacement covered by the particle divided by the total time taken in covering that displacement.

Total distance covered by the car = 30 miles.

Total time taken by the car to cover this distance = 1 hour.

Therefore, the average velocity of the car for this time interval = $\rm \dfrac{30\ miles}{1\ hour }= 30\ m.p.h.$

Thus, this statement is also not true.

(C) The magnitude of the average velocity is equal to 30 m.p.h.

As is cleared in part (B) section above, the average velocity of the car in the given time interval is 30 m.p.h.

Thus, this statement is true.

(D)The magnitude of the average velocity is less than to 30 m.p.h.

Since. the average velocity of the car is 30 m.p.h.

Thus, this statement is not true.

(E)The car traveled with a constant speed of 30 m.p.h.

The motion of the car on the mountain road is not thoroughly given in the question, so again it is not possible to tell whether the car traveled with a constant speed of 30 m.p.h. or not.

Thus, this statement is also not true.

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When a mass of 25 g is attached to a certain spring, it makes 20 complete vibrations in 4.0 s. what is the spring constant of th

earnstyle [38]

Answer: The spring of the spring is 25 N/m.

Explanation:

Mass of the body = 25 g= 0.025 kg (1 kg = 1000 g)

Oscillation is 4 sec = 20

Oscillation in 1 sec = $\frac{20}{4}=5$

Frequency of the vibration of the spring = $5 s^{-1}=5 Hz$

Force constant can be calculated bu using the relation between the frequency and, mass and spring constant 'k'

$Frequency=\frac{1}{2\pi}\times \sqrt{\frac{k}{m}}$

$5 s^{-1}=\frac{1}{2\times 3.14}\times \sqrt{\frac{k}{0.025 kg}}$

$k=24.649 N/m\approx 25 N/m$

The spring of the spring is 25 N/m.

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2 years ago

Read 2 more answers

A 5.00-pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to

Ne4ueva [31]

Answer:

a) r=4.24cm d=1 cm

b) $Q=5x10^{-10} C$

Explanation:

The capacitance depends only of the geometry of the capacitor so to design in this case knowing the Voltage and the electric field

$V=1.00x10^{2}v\\E=1.00x10^{4} \frac{N}{C}$

$V=E*d\\d=\frac{V}{E}\\d=\frac{1.0x10^{2}}{1.0x10^{4}}\\d=0.01m$

The distance must be the separation the r distance can be find also using

$C=\frac{Q}{V_{ab}}$

But now don't know the charge these plates can hold yet so

a).

d=0.01m

$C=E_{o}*\frac{A}{d}\\A=\frac{C*d}{E_{o}}$

$A=\frac{5pF*0.01m}{8.85x10^{-12}\frac{F}{m}}\\A=5.69x10^{-3}m^{2}$

$A=\pi *r^{2}\\r=\sqrt{\frac{A}{r}}\\r=\sqrt{\frac{5.64x10^{-3}m^{2} }{\pi } } \\r=42.55x^{-3}m$

b).

$C=\frac{Q}{V_{ab}}$

$Q=C*V\\Q=5x10^{-12} F*1x10^{2}\\Q=5x10^{-10}C$

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2 years ago

A permeability test was run on a compacted sample of dirty sandy gravel. The sample was 175 mm long and the diameter of the mold

LUCKY_DIMON [66]

Answer:

(a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

We need to calculate the discharge

Using formula of discharge

$Q=\dfrac{V}{t}$

Put the value into the formula

$Q=\dfrac{405}{90}$

$Q=4.5\ cm^3/s$

(a). We need to calculate the coefficient of permeability

Using formula of coefficient of permeability

$Q=kiA$

$k=\dfrac{Q}{iA}$

$k=\dfrac{Ql}{Ah}$

Where, Q=discharge

l = length

A = cross section area

h=constant head causing flow

Put the value into the formula

$k=\dfrac{4.5\times175\times10^{-1}}{\dfrac{\pi(175\times10^{-1})^2}{4}\times38}$

$k=8.6\times10^{-3}\ cm/s$

The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(c). We need to calculate the discharge velocity during the test

Using formula of discharge velocity

$v=ki$

$v=\dfrac{kh}{l}$

Put the value into the formula

$v=\dfrac{8.6\times10^{-3}\times38}{17.5}$

$v=0.0187\ cm/s$

The discharge velocity during the test is 0.0187 cm/s.

(b). We need to calculate the volume of solid in the ample

Using formula of volume

$V_{s}=\dfrac{M_{s}}{V_{s}}$

Put the value into the formula

$V_{s}=\dfrac{4950\times10^{-3}}{2710}$

$V_{s}=1826.56\ cm^3$

We need to calculate the volume of the soil specimen

Using formula of volume

$V=A\times L$

Put the value into the formula

$V=\dfrac{\pi(17.5)^2}{4}\times17.5$

$V=4209.24\ cm^3$

We need to calculate the volume of the voids

$V_{v}=V-V_{s}$

Put the value into the formula

$V_{v}=4209.24-1826.56$

$V_{v}=2382.68\ cm^3$

We need to calculate the seepage velocity

Using formula of velocity

$Av=A_{v}v_{s}$

$v_{s}=\dfrac{Av}{A_{v}}$

$v_{s}=\dfrac{V}{V_{v}}\times v$

Put the value into the formula

$v_{s}=\dfrac{4209.24}{2382.68}\times0.0187$

$v_{s}=0.0330\ cm/s$

The seepage velocity is 0.0330 cm/s.

Hence, (a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

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2 years ago

An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m

r-ruslan [8.4K]

PART A)

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here we have

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$q_2 = -1.6 \times 10^{-19} C$

$r_1 = r_2 = 1 m$

now we have

$V = \frac{Ke}{r} - \frac{Ke}{r}$

$V = 0$

Now work done to move another charge from infinite to origin is given by

$W = q(V_f - V_i)$

here we will have

$W = e(0 - 0) = 0$

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

$U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}$

$U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}$

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$U = 1.15\times 10^{-30}$

Now we know

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2$

$KE = 4.55 \times 10^{-27} kg$

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

4 0

2 years ago

Two speedboats are traveling at the same speed relative to the water in opposite directions in a moving river. An observer on th

DENIUS [597]

Answer:

a) vboat = 5.95 m/s b) vriver= 1.05 m/s

Explanation:

a) As observed from the shore, the speed of the boats can be expressed as the vector sum, of the boat speed relative to the water and the river speed relative to the shore, as follows:

vb₁s = vb₁w + vrs

In one case, the boat is moving in the same direction as the water:

vb₁s = vb₁w + vrs = 7.0 m/s (1)

For the other boat, it is clear that is moving in an opposite direction:

vb₂s = vb₂w - vrs = 4.9 m/s (2)

As we know that vb₁w = vb₂w, adding both sides, we can remove the river speed from the equation, as follows:

vb₁w = vb₂w = $\frac{7.0 m/s + 4.9 m/s}{2} =5.95 m/s$

b) Replacing this value in (1) and solving for vriver, we have:

vriver = 7.0 m/s - 5.95 m/s = 1.05 m/s

(we could have arrived to the same result subtracting both sides in (1), and (2))

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2 years ago