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3 years ago

The distance between Jupiter and the Sun is 5.2 AU. What is the distance in millions of kilometers?

Physics

1 answer:

miss Akunina [59]3 years ago

8 0

Answer:

777.9 Mkm

Explanation:

(M = Mega = $10^{6}$ )

AU is the Astronomical Unit for distance (1AU=149,597,870.691 km)

The total distance in km then is:

$5.2 * 149,597,870.691 km = 777908927.593 km$

To give the result in millions, we use the prefix Mega:

$(\frac{1M}{10^{6} } ) * 777908927.593 km = 777.9 Mkm$

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If the car ha a mass of 1000 kilograms, what is its momentum (v=35m/s)

Black_prince [1.1K]

Answer:

The momentum of the car is 35,000 kg.m/s

Explanation:

Momentum

Momentum is often defined as mass in motion.

Since all objects have mass, if it's moving, then it has momentum. It can be calculated as the product of the mass by the velocity of the object:

$\vec p = m\vec v$

If only magnitudes are considered:

p = mv

The car has a mass of m=1,000 kg and travels at v=35 m/s. Calculating its momentum:

p = 1,000 kg * 35 m/s

p = 35,000 kg.m/s

The momentum of the car is 35,000 kg.m/s

4 0

3 years ago

Which of the following is equal to impulse? A. Ft B. m/s C. pt D. mv

tankabanditka [31]

Answer: A (Ft)

Explanation: The impulse experienced by the object equals the change in momentum of the object. In equation form, F • t = m • Δ v

5 0

3 years ago

From a set of graphed data the slope of the best fit line is found to be 1.35 m/s and the slope of the worst fit line is 1.29m/s

Svetradugi [14.3K]

Solution:

Let the slope of the best fit line be represented by ' $m_{best}$ '

and the slope of the worst fit line be represented by ' $m_{worst}$ '

Given that:

$m_{best}$ = 1.35 m/s

$m_{worst}$ = 1.29 m/s

Then the uncertainity in the slope of the line is given by the formula:

$\Delta m = \frac{m_{best}-m_{worst}}{2}$ (1)

Substituting values in eqn (1), we get

$\Delta m = \frac{1.35 - 1.29}{2}$ = 0.03 m/s

8 0

3 years ago

A hiker determines the length of a lake by listening for the echo of her shout reflected by a cliff at the far end of the lake.

ArbitrLikvidat [17]

Answer:

L = 499 m

Explanation:

If we assume that the speed of sound is constant, that travels along a straight line, and that the echo is instantaneous, we can find the total distance travelled by the sound, as follows, just applying the definition of average velocity:

$\Delta x = v_{s} * t = 343 m/s* 2.91 s = 998 m$

If we assume that the time needed to reach to the cliff, is the same used for the return travel, the length of the lake will be exactly half of the total distance calculated:

$l_{lake} = \frac{\Delta x}{2} = \frac{998m}{2} = 499 m$