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gogolik [260]
1 year ago
10

If you jog for 1 hr and travel 10 km, the calculation of 10 km/hr describes your.

Physics
1 answer:
zloy xaker [14]1 year ago
8 0

Answer:

[Speed] or [Velocity (if there is a direction)]

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. Suppose that the displacement of an object is related to time according to the expression: x(t) = B t3 . Using dimensional ana
-Dominant- [34]

Answer:

The dimensionality of B is <em>length</em> per cubic <em>time</em>.

Explanation:

Units for displacement and time are <em>length</em> [L] and <em>time</em> [T], respectively. Then, formula can be tested for dimensional analysis as follows:

[L] = B\cdot [T]^{3}

Now, let is clear B to determine its units:

B = \frac{[L]}{[T]^{3}}

The dimensionality of B is <em>length</em> per cubic <em>time</em>.

6 0
3 years ago
#1- How much Magnesium is consumed in this reaction? *input # only.
____ [38]

Answer:

use the mole concept

Explanation:

the mole to mole ratio

4 0
3 years ago
Sounds travel faster in Question 1 options: warmer air. cooler air. Temperature does not influence the speed of sound. a vacuum.
olya-2409 [2.1K]

Answer:

warmer air

Explanation:

the particles are more excited which increases the probability that the particles will bump into each other

8 0
2 years ago
A person wishes to heat pot of fresh water from 20°C to 100°C in order to boil water for pasta. They calculate that their pot ho
nasty-shy [4]
Given:
Water, 2 kilograms
T1 = 20 degrees Celsius, T2 = 100 degrees Celsius.  

Required:
Heat produced  

Solution:
Q (heat) = nRT = nR(T2 = T1)
Q (heat) = 2 kilograms (4.184 kiloJoules per kilogram Celsius) (100 degrees Celsius – 20 degrees Celsius)
<u>Q (heat) = 669.42 Joules
</u>This is the amount of heat produced in boiling 2 kg of water.
6 0
3 years ago
Water falls without splashing at a rate of 0.200 L/s from a height of 3.60 m into a 0.730 kg bucket on a scale. If the bucket is
dimaraw [331]

Answer:

15.106 N

Explanation:

From the given information,

The weight of the bucket can be calculated as:

W_b = m_bg =  \\ \\  W_b = (0.730 \  kg) ( 9.80 \ m/s^2) \\ \\ W_b = 7.154 \ N

The mass of the water accumulated in the bucket after 3.20s is:

m_w= (0.20 \ L/s) ( 3.20)s

m _w=0.64 \ kg

To determine the weight of the water accumulated in the bucket, we have:

W_w = m_w g

W_w = ( 0.64  \ kg )(9.80\  m  \  /s^2)

W_w = 6.272 \ N

For the speed of the water before hitting the bucket; we have:

v = \sqrt{2gh}

v = \sqrt{2*9.80 \ m/s^2 * 3.60 \ m}

v = 8.4 m/s

Now, the force required to stop the water later when it already hit the bucket is:

F = v ( \dfrac {dm}{dt} )

F = (8.4 \ m/s)( 0.200 \ L/s)

F = 1.68 N

Finally, the reading scale is:

F_{scale = 7.154 N + 6.272 N + 1.68 N

= 15.106 N

6 0
2 years ago
Read 2 more answers
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