• 500 waves pass by in 2 second. These waves have a wavelength of 6

What is the speed of a bobsled whose distance-time graph indicates that it traveled 113m in 29s?

Elza [17]

In this item, we are asked to determine the speed of the bobsled given the distance traveled and the time it takes to cover the certain distance. This can mathematically be expressed as,
speed = distance / time

Substituting the given values in this item,
speed = (113 m) / (29 s)
speed = 3.90 m/s

ANSWER: 3.90 m/s

3 0

3 years ago

A truck traveling at a velocity of 33m/s comes to a halt by decelerating at 11m/s^2. How far does the truck travel in the proces

snow_lady [41]

Answer:

The truck was moving 16.5 m/s during the time it took to stop, which was 3 seconds.

Initial velocity = 33 m/s

Final velocity = 0 m/s

Average velocity = (33 + 0) / 2 m/s = 16.5 m/s

Explanation:

First, how long does it take the truck to come to a complete stop?

( 33 m/s ) / ( 11 m / s^2 ) = 3 seconds

Then we can look at the average velocity between when the truck started decelerating and when it came to a complete stop. Because the deceleration is constant (always 11m/s^2) we can use this trick.

4 0

3 years ago

IBM has a fast computer that it calls the Blue Gene/L that can do '136.8

dmitriy555 [2]

Answer:

138.6 megacalculations

Explanation:

This is a pretty straightforward one.

All it needs is to convert the degree of measurement.

Dimensions in physics are attributed names, which state the power to which they're are raised. Just as how

Kilo and Mega means the numbers are raised to the power of 3 and 6 respectively. There also exists the ones that indicates how small, such as milli and micro, which are to the powers of -3 & -6.

The question says the IBM computer calculates at an astonishing 136.8 teracalculations.

Tera in physics means it's raised to the power of 12. Thus, the IBM calculates at an astonishing rate of

136.8*10^12 calculations per second.

We're then asked how many calculations it does in 1 micro second. Like I had highlighted earlier, 1 micro second is 1 raised to the power of -6. Or succinctly put,

1 micro second = 1*10^-6.

If the IBM does

138.6*10^12 = 1 second,

Then it does

x = 1*10^-6 second.

When we cross multiply, we have

138.6*10^12 * 1*10^-6, and that is

138.6*10^6 calculations, or say, 138.6 megacalculations.

The IBM does 138.6 megacalculations in 1 micro second, which is still astonishing, by the way

6 0

3 years ago

A body of volume 2000 cm3 has a mass 4 kg. Find the density of the body. Will the body sink or float in water, if the density of

Brums [2.3K]

Answer:

$2 g/cm^3$ , it will sink

Explanation:

The density of an object is given by

$d=\frac{m}{V}$

where

m is the mass of the object

V is its volume

For the body in the problem, we have

m = 4 kg = 4000 g

$V=2000 cm^3$

Therefore, its density is

$d=\frac{4000}{2000}=2 g/cm^3$

And the object will sink in water, because its density is larger than that of water, which is $1 g/cm^3$ . (an object sinks when its density is larger than that of water, otherwise it floats).

6 0

3 years ago

A pendulum is lifted to a certain height and then released, which causes the

borishaifa [10]

Answer:

Answer is C

Explanation:

Let's say the pendulum starts swinging from its max height from the left. It then will go down and reach the equilibrium position, this will make it lose GPE while gaining KE (the loss in GPE = gain in KE). At the equilibrium position it has the max KE (max velocity) and minimum GPE. After passing the equilibrium it then starts to head up to the max height on the right, the pendulum gains GPE while losing KE and at the top will have minimum KE while having max GPE. Meaning throughout its joruney the total energy remains constant as
Total energy = KE + GPE

I have attached a simple diagram below, the y axis is the energy and x axis being the time (where t = 0 is the pendulum starting from max height left of the equilibrium). The green curve the the GPE and blue curve is KE. Red line shows that at all times the energy is constant.

8 0

2 years ago