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omeli [17]
3 years ago
15

Mr. Llama walked from his house to the bus stop. The bus stop is 2 miles from his house. He returned back to his house from the

bus stop. What is his displacement? Question 3 options: A. 4 feet B. 2 miles C. 4 miles D. 0 miles PLZ ANSWER I DONT HAVE MUCH TIME! HURRY!
Physics
1 answer:
Hoochie [10]3 years ago
7 0

Answer:

Displacement of Mr. Llama: Option D. 0 miles.

Explanation:

The magnitude of the displacement of an object is equal to the distance between its final position and its initial position. In other words, as long as the initial and final positions of the object stay unchanged, the path that this object took will not affect its displacement.

For Mr. Llama:

  • Final position: Mr. Llama's house;
  • Initial position: Mr. Llama's house.

The distance between the final and initial position of Mr. Llama is equal to zero. As a result, the magnitude of Mr. Llama's displacement in the entire process will also be equal to zero.

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A mover loads a 100 kg box into the back of a moving truck by
NeX [460]

Answer:

2.7

Explanation:

The following data were obtained from the question:

Mass (m) of box = 100 Kg

Length (L) of ramp = 4 m

Height (H) of ramp = 1.5 m

Mechanical advantage (MA) of ramp =?

Mechanical advantage of a ramp is simply defined as the ratio of the length of the ramp to the height of the ramp. Mathematically, it is given by:

Mechanical Advantage = Lenght / height

MA= L/H

With the above formula, we can obtain the mechanical advantage of the ramp as follow:

Length (L) of ramp = 4 m

Height (H) of ramp = 1.5 m

Mechanical advantage (MA) of ramp =?

MA = 4/1.5

MA = 2.7

Therefore, the mechanical advantage of the ramp is 2.7

3 0
3 years ago
Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150
Phoenix [80]

Answer:

The temperature of the core raises by 2.8^{o}C every second.

Explanation:

Since the average specific heat of the reactor core is 0.3349 kJ/kgC

It means that we require 0.3349 kJ of heat to raise the temperature of 1 kg of core material by 1 degree Celsius

Thus reactor core whose mass is 1.60\times 10^{5}kg will require

0.3349\times 1.60\times 10^{5}kJ\\\\=0.53584\times 10^{5}kJ

energy to raise it's temperature by 1 degree Celsius in 1 second

Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by

\frac{150\times 10^{6}}{0.53584\times 10^{8}}=2.8^{o}C/s

5 0
3 years ago
What is a force that attracts all matter to each other?
Temka [501]
The answer is gravity. I hope this helps. 
5 0
3 years ago
Read 2 more answers
Which of the following is an incorrect statement of one of newtons laws of motion?
weqwewe [10]

Explanation:

The newton's laws of motion are:

First law:

  "A body will remain in its state of rest or of uniform motion along a path unless it is acted upon by an external force. ".

This is popularly called the law of inertia.

Second law:

 "the acceleration of a body is produced by a net force that is inversely proportional to the mass of the body".

Third law:

  "action and reaction forces are equal and opposite in direction".

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

5 0
3 years ago
How much heat is needed to raise the temperature of a 100.0g of water by 85.0 c?
KatRina [158]

Answer:

35.7kJ

Explanation:

we can calculate the amount of heat energy required , using this formula

Q = mcθ

where.

Q = heat energy (Joules, J)  

m = mass of a substance (kg)  

c = specific heat capacity (units Jkg^{-1} C^{-1})

θ  = change in temperature (Celcius,C or Kelvin K)

Assume Specific heat capacity (c) of water =4200Jkg^{-1} C^{-1}

mass =0.1 kg

Q=0.1 kg*4200Jkg^{-1} C^{-1}*85C\\=35700J\\=35.7kJ

3 0
3 years ago
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