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Maru [420]
3 years ago
8

NEED IT ASAP PLEASE 3 examples of how we use physics in our everyday life. Please explain throughly.

Physics
1 answer:
Flauer [41]3 years ago
3 0

Answer:

  1. Alarm Clock. The buzzing sound of an alarm clock helps you wake up in the morning as per your schedule. The sound is something that you can’t see, but hear or experience.
  2. Cell Phones Cellphones have become like Oxygen gas in modern social life. Hardly, anyone would have been untouched by the effects of a cell phone. Whether conveying any urgent message or doing incessant gossips, cellphones are everywhere. But do you know how does a cell phone work? It works on the principle of electricity and the electromagnetic spectrum, undulating patterns of electricity and magnetism.
  3. Walking.Now, when you get ready for your office/school, whatever medium of commutation is, you certainly have to walk up to a certain distance. You can easily walk is just because of Physics. While you have a walk in a park or on a tar road, you have a good grip without slipping because of a sort of roughness or resistance between the soles of your shoes and the surface of the road.

Explanation:

physical is related to things perceived through the senses as opposed to the mind; tangible or concrete.

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50 g of liquid Y at 10 Celcius and 200 g of liquid Y at 40Celcius are mixed. Final temperature of the mixture is measured as 15
Sholpan [36]
50 g of liquid X at 10 Celcius and 200 g of liquid Y

mx*cx*(t-tx)+my*cy*(t-ty)=0
cx/cy = - my*(t-ty) : mx*(t-tx) = (my/mx)  * (ty - t) / (t-tx)
cx/cy  = 200/50*(40-15)/(15-10) = 20
cx/cy   = 20



5 0
3 years ago
What is the mass of an object travelling at 25 m/s with a kinetic energy of 3775 J?
Arte-miy333 [17]

Answer:

<h2>You can do 6516-3775=2741</h2><h2 /><h2>The difference is 27141 kilometers</h2><h2 /><h2>Did this help?</h2><h2>⇒  Yes or No?</h2>
4 0
2 years ago
A 1kg ball is dropped (from rest) 100m onto a spring with spring constant 125N/m. How much does the spring compress?
Anastasy [175]

Answer:

3.95 m

Explanation:

m = 1 kg, h = 100 m, k = 125 N/m

Let the spring is compressed by y.

Use the conservation of energy

potential energy of the mass is equal to the energy stored in the spring

m x g x h = 1/2 x ky^2

1 x 9.8 x 100 = 0.5 x 125 x y^2

y^2 = 15.68

y = 3.95 m

7 0
3 years ago
A man jogs at a speed of 1.6 m/s. His dog waits 1.8 s and then takes off running at a speed of 3 m/s to catch the man. How far w
inessss [21]

Answer:

The dog catches up with the man 6.1714m later.

Explanation:

The first thing to take into account is the speed formula. It is v=\frac{d}{t}, where v is speed, d is distance and t is time. From this formula, we can get the distance formula by finding d, it is d=v\cdot t

Now, the distance equation for the man would be:

d_{man}=v_{man}\cdot t=1.6\cdot t

The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.

d_{dog}=v_{dog}\cdot (t-1.8)=3\cdot (t-1.8)

For a better understanding, at t=1.8 the dog must be in d=0. Let's verify:

d_{dog}=v_{dog}\cdot (1.8-1.8)=3\cdot (0)=0

Now, for finding how far they have each traveled when the dog catches up with the man we must match the equations of each one.

d_{man}=d_{dog}

1.6\cdot t=3\cdot (t-1.8)

1.6\cdot t=3\cdot t-5.4

1.4\cdot t=5.4

t=\frac{5.4}{1.4}

t=3.8571s

The result obtained previously means that the dog catches up with the man 3.8571s after the man started running.

That value is used in the man's distance equation.

d_{man}=1.6\cdot t=1.6\cdot (3.8571)

d_{man}=6.1714m

Finally, the dog catches up with the man 6.1714m later.

6 0
3 years ago
Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m
aev [14]

Answer:

Explanation:

Given

charge of first body q_1=2.5\ mu C

charge of second body q_2=-3.5\ mu C

Particle 1 is at origin and particle 2 is at x=0.6\ m

third Particle which charge +q must be placed left of 2.5\mu C because it will repel the q charge while -3.5\mu C will attract it

suppose it is placed at a distance of x m

F_{1q}=\frac{kq(2.5)}{x^2}

F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}

F_{1q}+F_{2q}=0

\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0

\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}

\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}

0.6+x=1.1832x

x=3.27\ m

5 0
3 years ago
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