A car travels down a road at a certain velocity, vcar. The driver slows down so that the car is traveling only a third as fast as before. Which of the following is the correct expression for the resulting velocity?

a. 2vcar
b. 1/3vcar
c. -1/2vcar
d. -2vcar

Answers

Answer 1

The correct expression for the resulting velocity is -1/2 vcar. The correct option is c.

What is velocity?

The directional speed of an item in motion, as measured by a specific unit of time and observed from a certain point of reference, is what is referred to as velocity.

A car travels down a road at a certain velocity, is Vcar.

Let the initial velocity of the car travel downward be -V car

It is also, given, that the driver slows down so that the car is traveling only half as fast as before.

Resulting velocity = -1/2 vcar

Therefore, the correct option is c. -1/2vcar.

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Related Questions

The diagram below represents a wave.
What is the speed of the wave if its wavelength is 3.0 m?

Answers

The speed of the wave is determined as 18 m/s.

option D is the correct answer.

What is the speed of the wave?

The speed of the wave is the rate of change of wave's displacement with time.

The speed of the wave is calculated by applying the formula relating speed, wavelength and frequency of the wave as shown below.

v = fλ

where;

v is the speed of the wavef is the frequency of the waveλ is the wavelength of the wave

The frequency of the wave is calculated as follows;

f = 1 / T

where;

T is the period of the wave

The period of a wave is the time taken for the wave to complete one cycle.

In the picture given, 3 cycles of the wave = 0.5 s

1 cycle of the wave = ?

= 0.5 s / 3

= 0.16667 s

f = 1 / T

f = 1 / 0.16667

f = 6 Hz

The speed of the wave is calculated as follows;

v = fλ

v = 6 Hz  x  3 m

v = 18 m/s

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A physics student walks 6 meters east, 4 meters north, then 9 meters
west. What is the student's magnitude of displacement from starting
position?

Answers

Answer:

5 m

Explanation:

Displacement is the straight-line distance from the starting point to the end point.  Student ends up 4m north and 3m west from where he started. Use pythagorean theorem to solve for displacement (d).  

d² = 3² + 4² = 25

d = [tex]\sqrt{25}[/tex] = 5 m

Or, if you remember from geometry, a right triangle with legs of 3 and 4 has a hypotenuse of 5 (3-4-5 right triangle)

a vertical spring scale can measure weights up to 210 n . the scale extends by an amount of 13.5 cm from its equilibrium position at 0 n to the 210 n mark. a fish hanging from the bottom of the spring oscillates vertically at a frequency of 2.15 hz . part a ignoring the mass of the spring, what is the mass m of the fish?

Answers

sheesh sheesh                                                    

Explanation:

1. a 1.4 x 103 kg car is westbound at a velocity of 37.0 km/h when it collides with a 2.0 x 103 kg truck northbound at a velocity of 35 km/h. if these two vehicles lock together upon collision, what is the initial velocity of the vehicles after collision? (7.2 m/s 37o w of n) 2. a 6.2 kg object heading north at 3.0 m/s collides with an 8.0 kg object heading west at 3.5 m/s. if these two masses stick together upon collision, what is their velocity after collision? (2.4 m/s 56o w of n) 3. a 4.0 x 104 n truck moving west at a velocity of 8.0 m/s collides with a 3.0x104 n truck heading south at a velocity of 5.0 m/s. if these two vehicles lock together upon impact, what is their velocity?(5.0 m/s 25o s of w)

Answers

The velocity of vehicles after the collision is 15.82m/sec if car mass is 1.4 × 10³kg and truck mass is 2×10³kg

We know that we need to conserve the momentum of watermelon in a certain direction. Also,we know that initial momentum of watermelon is zero,it means that according to law of conservation of momentum final momentum should be zero.

We know that momentum =mass × velocity

So,initial momentum of car is =1.4 × 10³kg × 37km/hr

Similarly,,initial momentum of truck=2×10³kg × 35km/hr

Now,it is given that both vehicles lock together,so total mass of both vehicles is =1.4 × 10³kg +  2×10³kg=3.4 ×10³kg

Assume final velocity of both vehicle is v

So,following the law of conservation,we get

=>1.4 × 10³kg × 37 + 2×10³kg × 35 = 3.4 ×10³ ×v

=>51.8 ×10³ + 2×10³=3.4×10³ ×v

=>53.8 ×10³ kg-m/sec = 3.4×10³ ×v

=>v = 53.8/3.4

=>v=15.82m/sec

Hence,final velocity of both vehicle after collision is 15.82m/sec.

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which sentence correctly describes when the triple-alpha process occurs in the life cycle of an average-mass star

Answers

Three helium-4 nuclei (alpha particles) are turned into carbon by a series of nuclear fusion processes known as the triple-alpha process.

What results from the triple-alpha procedure as the finished product?

The triple-alpha process and the alpha process are two classes of nuclear fusion reactions that stars use to change helium into heavier elements. The alpha process is also referred to as the alpha ladder. [1] Only helium is used in the triple-alpha process, which also yields carbon.

Which statement concerning alpha particles in an atom is accurate?

Two protons and two neutrons make up alpha particles, which are identical to helium nuclei and have a positive charge.

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Maher is trying to find the density of a plastic toy. The toy’s mass is 160 g.
Maher placed the toy in a graduated cylinder that has 70 ml of water, the water level increased
till 150 ml.
a) Find the toy’s volume.

Answers

The volume of the toy is 80 ml.

To find the volume of the plastic toy, Maher can use the principle of buoyancy. When an object is placed in a fluid, it will displace an amount of fluid equal to its own volume. The volume of the displaced fluid can be measured and used to calculate the volume of the object.

In this case, Maher has placed the toy in a graduated cylinder filled with water, and he has observed that the water level increased from 70 ml to 150 ml. This means that the toy displaced 150 - 70 = 80 ml of water.

The volume of the toy is equal to the volume of the displaced water, so the toy's volume is 80 ml. This is the volume of the toy when it is completely submerged in water.

It's important to note that the volume of an object can change depending on its temperature, pressure, and other factors. To get an accurate measurement of the volume of the toy, Maher should make sure to measure the volume of the displaced water carefully and under controlled conditions.

Answer:

Volume of the toy: [tex]80\; {\rm mL}[/tex].

Average density of the toy: [tex]2\; {\rm g\cdot mL^{-1}}[/tex] (or equivalently, [tex]2\; {\rm g \cdot cm^{-3}}[/tex].)

Explanation:

The graduated cylinder initially measures the volume of water in this cylinder:

[tex]V(\text{water}) = 70\; {\rm mL}[/tex].

Assume that the toy is submerged in the water. The graduated cylinder would then measure the volume of the water and the toy, combined:

[tex]V(\text{water}) + V(\text{toy}) = 150\; {\rm mL}[/tex].

Rearrange to find the volume of the toy:

[tex]\begin{aligned}V(\text{toy}) &= 150\; {\rm mL} - V(\text{water}) \\ &= 150\; {\rm mL} - 70\; {\rm mL} \\ &= 80\; {\rm mL}\end{aligned}[/tex].

To find the average density of this toy, divide mass by volume:

[tex]\begin{aligned}(\text{average density}) &= \frac{(\text{mass})}{(\text{volume})} \\ &= \frac{160\; {\rm g}}{80\; {\rm mL}} \\ &= 2\; {\rm g\cdot mL^{-1}}\end{aligned}[/tex].

a simple harmonic oscillator takes 11.5 s to undergo five complete vibrations. (a) find the period of its motion. s (b) find the frequency in hertz. hz (c) find the angular frequency in radians per second. rad/s

Answers

The required values are a) T = 2.3 seconds, b)  f = 0.434 Hz, c) ω = 2.7 rd/s.

Which of the following motions is simply harmonic?

Simple harmonic motion, a particular kind of periodic motion in which a particle repeatedly oscillates around a mean location, In U-tube oscillating liquid column motion is hence simple harmonic.

According to question:

Given,

Time = 11.5 seconds to five complete vibrations.

a) Time period is time taken to complete one vibration,

So T = 11.5/5 = 2.3 seconds

b) Frequency(f) = 1/T

f = 1/2.3 = 0.434 second inverse.

c) By using formula of angular frequency ω = 2π/T

ω = 2π/2.3 = 2.7 hz

Thus, final values are a) T = 2.3 seconds, b)  f = 0.434 second inverse, c) ω = 2.7 rd/s.

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based on its surface temperature of 6,000 k, most photons that leave the sun's surface lie in which region of the electromagnetic spectrum?

Answers

Most photons that leave the sun's surface have a surface temperature of 6,000 K, which corresponds to the visible light region of the electromagnetic spectrum.

The electromagnetic spectrum is a continuous range of wavelengths and frequencies that includes radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays. The wavelengths and frequencies of these different types of electromagnetic radiation are all different and correspond to different regions of the spectrum.

The visible light region of the electromagnetic spectrum is the portion of the spectrum that can be seen by the human eye. It ranges from about 400 nanometers (nm) to about 700 nm in wavelength, and corresponds to frequencies of about 7.5 x 10^14 Hz to about 4.3 x 10^14 Hz. Photons with wavelengths and frequencies in this range have enough energy to excite the photoreceptors in the human eye, allowing us to see them.

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5. if the star sirius emits 23 times more energy than the sun, why does the sun appear brighter in the sky? include the terms luminosity and apparent brightness in your response.

Answers

The star Sirius is indeed brighter but our Sun appears to be brighter because it is closer to us.

The intrinsic luminosity of Sirius is exactly 25.4 times more than the sun but if we compare the distances of these stars, Sirius is 8.6 light years away from us and the Sun is just 8 light minutes away from us. This marks the obvious reason why our Sun appears more bright.

Even mathematically, it can be justified as the apparent brightness of any luminous object depends on the inverse square of the distance of that luminous object. Therefore the sun's light energy dominates our sky and we can't see any traces of Sirius from the earth.

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