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lord [1]
3 years ago
13

Why is science always changing

Physics
2 answers:
Anastaziya [24]3 years ago
8 0

Answer:

Science is changing because little by little scientists descover new things.

Explanation:

Discovery

Hope this helps:)

kykrilka [37]3 years ago
8 0
Because scientists discover new things, new ways to think about a singular method or principle
You might be interested in
Please help on this one
mixas84 [53]
Ep=mgh
h= Ep/mg
h=57÷(3.3×9.8)
h= 57÷32.34
h= 1.8m
So; the answer is B. 1.8m
6 0
3 years ago
Why is the ocean warmer than the air at night
Sliva [168]

You see, during the day the ocean collects heat from the sun. So the air above the ocean get warm at night, but the rest of the air on the land gets cooler because water has the ability to collect energy from the Sun.

4 0
2 years ago
On earth, two parts of a space probe weigh 14500 N and 4800 N. These parts are separated by a center-to-center distance of 18 m
Nastasia [14]

Answer:

F = 1.489*10^{-7}  N

Explanation: Weight of space probes on earth is given by:W= m*g

W= weight of the object( in N)

m= mass of the object (in kg)

g=acceleration due to gravity(9.81 \frac{m}{s^{2} })

Therefore,

m_{1} = \frac{14500}{9.81}

m_{1} = 1478.08  kg

Similarly,

m_{2} = \frac{4800}{9.81}

m_{2} = 489.29  kg

Now, considering these two parts as uniform spherical objects

Also, according to Superposition principle, gravitational net force experienced by an object is sum of all individual forces on the object.

Force between these two objects is given by:

F =  \frac{Gm_{1} m_{2}}{R^{2} }

G= gravitational constant (6.67 * 10^{-11} m^{3} kg^{-1} s^{-2})

m_{1} , m_{2}= masses of the object

R= distance between their centres (in m)(18 m)

Substituiting all these values into the above formula

F = 1.489*10^{-7}  N

This is the magnitude of force experienced by each part in the direction towards the other part, i.e the gravitational force is attractive in nature.

7 0
3 years ago
Is a neutron star also a black hole?
coldgirl [10]

No.  A neutron star is the weird remains of a star that blew its outer layers off
in a nova event, and then had enough mass left so that gravity crushed its
electrons into its protons, and then what was left of it shrank down to a sphere
of unimaginably dense neutron soup.  But it didn't have enough mass to go
any farther than that.

A black hole is the remains of a star that had enough mass to go even farther
than that.  No force in the universe was able to stop it from contracting, so it
kept contracting until its mass occupied no volume ... zero.  It became even
more weird, and is composed of a substance that we don't know anything about
and can't describe, and occupies zero volume.

Contrary to popular fairy tales, a black hole doesn't reach out and "suck things in".
It's just so small (zero) that things can get very close to it.  You know that gravity
gets stronger as you get closer to an object, so if the object has no size at all, you
can get really really close to it, and THAT's where the gravity gets really strong.
You may weigh, let's say, 100 pounds on the Earth.  But you're like 4,000 miles
from the center of the Earth.  What if all of the earth's mass was crammed into
the size of a bean.  Then you could get 1 inch from it, and at that distance from
the mass of the Earth, you would weigh 25,344,000,000 pounds. 
But Earth's mass is not enough to make a black hole.  That takes a minimum
of about 3 times the mass of the sun, which is right about 1 million times the
Earth's mass.   THEN you can get a lightweight black hole.
Do you see how it works now ?

I know.  It all seems too fantastic to be true. 
It sure does.

8 0
3 years ago
Two identical strings, of identical lengths of 2.00 m and linear mass density of μ=0.0065kg/m, are fixed on both ends. String A
kolezko [41]

Answer:

beat frequency = 13.87 Hz

Explanation:

given data

lengths l = 2.00 m

linear mass density μ = 0.0065 kg/m

String A is under a tension T1 = 120.00 N

String B is under a tension T2 = 130.00 N

n = 10 mode

to find out

beat frequency

solution

we know here that length L is

L = n × \frac{ \lambda }{2}      ........1

so  λ = \frac{2L}{10}  

and velocity is express as

V = \sqrt{\frac{T}{\mu } }    .................2

so

frequency for string A = f1 = \frac{V1}{\lambda}

f1 = \frac{\sqrt{\frac{T}{\mu } }}{\frac{2L}{10}}

f1 = \frac{10}{2L} \sqrt{\frac{T1}{\mu } }      

and

f2 = \frac{10}{2L} \sqrt{\frac{T2}{\mu } }

so

beat frequency is = f2 - f1

put here value

beat frequency = \frac{10}{2*2} \sqrt{\frac{130}{0.0065}}  - \frac{10}{2*2} \sqrt{\frac{120}{0.0065} }

beat frequency = 13.87 Hz

6 0
3 years ago
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