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Olegator [25]
3 years ago
12

These steps are followed when using the half-life of carbon-14 to determine the age of an object that contains carbon. what is t

he correct order of these steps

Physics
2 answers:
enyata [817]3 years ago
7 0

Answer:

the answer is b

Explanation:

a p e x

allsm [11]3 years ago
4 0
The Answer to this question is C
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A net force of –8750N is used to stop of 1250.kg car travelling 25m/s. What braking distance is needed to bring the car to a hal
Karolina [17]

Answer:

d = 44.64 m

Explanation:

Given that,

Net force acting on the car, F = -8750 N

The mass of the car, m = 1250 kg

Initial speed of the car, u = 25 m/s

Final speed, v = 0 (it stops)

The formula for the net force is :

F = ma

a is acceleration of the car

a=\dfrac{F}{m}\\\\a=\dfrac{-8750}{1250}\\\\a=-7\ m/s^2

Let d be the breaking distance. It can be calculated using third equation of motion as :

v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-7)}\\\\d=44.64\ m

So, the required distance covered by the car is 44.64 m.

4 0
3 years ago
What is the equation for the force of a spring
Komok [63]

Answer:

F=-kx

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yan lang po ang alam ko

7 0
3 years ago
Two flat surfaces are exposed to a uniform, horizontal magnetic field of magnitude 0.47 T. When viewed edge-on, the first surfac
alexgriva [62]

Answer:

a. A = 0.0859 m^2

b. A = 0.0178 m^2

Explanation:

Two flat surfaces are exposed to a uniform, horizontal magnetic field of magnitude 0.47 T. When viewed edge-on, the first surface is tilted at an angle of from the horizontal, and a net magnetic flux of 8.4 103 Wb passes through it. The same net magnetic flux passes through the second surface. (a) Determine the area of the first surface. (b) Find the smallest possible value for the area of the second surface.

take note that the question has not specified th angle which the surface is tilted so i assume the angle is at 12^{0} to the horizontal

flux = BAcos(\alpha)

B=magnetic flux in Weber

A=area of the flat surface in m^2

\alpha=the angle to the horizontal

a) 8.4 x10^-3= (.47)Acos(78)

alpha has to be the angle from the normal and not the horizontal so 90-12=78,

 8.4 x10^-3

/(.47)cos(78)

A = 0.0859 m^2

b) If flux remains the same then for it to be the smallest possible area it needs to be perpendicular to the magnetic field so alpha would be 0.

8.4 x10^-3 = (.47)Acos(0)

A = 0.0178 m^2

7 0
4 years ago
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