A particle traveling around a circle at constant speed will not experience an acceleration.

A 1100 kg automobile is at rest at a traffic signal. At the instant the light turns green, the automobile starts to move with a

Viktor [21]

Answer:

a ) 21 m b) 7 m/s

Explanation:

For this case, we can consider for our reference system, that point zero (this is, at the rest of the traffic signal) is when X₀ = 0, and direction of movement is from left to right (positive sign)

For the car:

a= acceleration is constant = 5 m/s² m1 , mass = 1100 Kg , V = not constant

This is an uniformly accelerated rectilinear movement, and applicable formulas are:

V = V₀ + at , X = X₀ + V₀t + 1/2at²

For the truck:

V = speed is constant = 7 m/s , mass = 2000 Kg, a = 0

This is an uniform rectilinear movement, and the applicable formula is:

X = X₀ + Vt

a) At t = 3.0 sec, we can use the formula for the track, considering that X₀ = 0 (when it pass the rest of the traffic signal)

X = 0 + (7 m/s)x(3 s) = 21 m

So, at 3 sec, truck will be at 21 m from the rest of traffic light and, as truck has a constant speed; at that exact second; truck and car will be together as a com, so both will be 21 m away from point zero

b) At the exact time (3 sec, or in distance words, 21 m) car and truck will be together as a com, so they will both have the exact speed, hence, speed of car at that point will be 7 m/s

6 0

3 years ago

What element is used for lightbulb

Artemon [7]

<span>Lightbulbs are filled with the element argon and the filaments are made of metallic element tungsten. The argon in the bulb protects the filament from the surrounding air, which would otherwise oxidize the filament. </span>

5 0

3 years ago

What are the basic si units for the wavelength of light?

noname [10]

Answer:

Meter (m)

Explanation:

The wavelenght of a light wave is a measure of the distance between two successive crests (or two successive troughs) of a light wave.

Since the SI units for the distance is the meter (m), then the SI unit for the wavelength is also the meter (m).

Wavelength is related to the frequency of the light wave by:

$\lambda=\frac{c}{f}$

where

c is the speed of light

f is the frequency of the light

7 0

3 years ago

explain how you can determine if a chemical reaction a single replacement reaction or a double replacement reaction

Kobotan [32]

However, there is one type of double-replacement reaction that we can predict: the precipitation reaction. A precipitation reaction occurs when two ionic compounds are dissolved in water and form a new ionic compound that does not dissolve; this new compound falls out of solution as a solid precipitate.

8 0

3 years ago

A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= $\int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx$

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*0.8(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = $\frac{36}{11}$ kg/m

Mass of rope = weight of rope × change in distance

= $\frac{36}{11}$ kg/m × (11-x)m = 3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*3.27(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

8 0

4 years ago