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Ronch [10]
3 years ago
11

What does the magnitude of centripetal acceleration depend on ?

Physics
1 answer:
MariettaO [177]3 years ago
3 0

Explanation:

Centripetal acceleration ac is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. It is perpendicular to the linear velocity v and has the magnitude ac=v2r;ac=rω2 a c = v 2 r ; a c = r ω 2 .

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PLS HELP Do you believe this relationship between incident and reflected angles would occur even if the medium interface were cu
Anna35 [415]
The Law of reflection would still hold even off a curved surface. Since the angles are measured from the normal, which is perpendicular to the surface, curved surfaces don't matter. This is basis of curved mirrors such as concave and convex 
4 0
3 years ago
What is the increase in pressure required to decrease volume of mercury by 0.001%
REY [17]

Answer:

Explanation:

Using Boyles law

Boyle's law states that, the volume of a given gas is inversely proportional to it's pressure, provided that temperature is constant

V ∝ 1 / P

V = k / P

VP = k

Then,

V_1 • P_1 = V_2 • P_2

So, if we want an increase in pressure that will decrease volume of mercury by 0.001%

Then, let initial volume be V_1 = V

New volume is V_2 = 0.001% of V

V_2 = 0.00001•V

Let initial pressure be P_1 = P

So,

Using the equation above

V_1•P_1 = V_2•P_2

V × P = 0.00001•V × P_2

Make P_2 subject of formula by dividing be 0.00001•V

P_2 = V × P / 0.00001 × V

Then,

P_2 = 100000 P

So, the new pressure has to be 10^5 times of the old pressure

Now, using bulk modulus

Bulk modulus of mercury=2.8x10¹⁰N/m²

bulk modulus = P/(-∆V/V)

-∆V = 0.001% of V

-∆V = 0.00001•V

-∆V = 10^-5•V

-∆V/V = 10^-5

Them,

Bulk modulus = P / (-∆V/V)

2.8 × 10^10 = P / 10^-5

P = 2.8 × 10^10 × 10^-5

P = 2.8 × 10^5 N/m²

3 0
3 years ago
What sort of fitness component would dance be<br> categorized under
zalisa [80]

Answer:

aerobic fitness

Explanation:

i just looked it up lol

6 0
3 years ago
Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiem
statuscvo [17]

Answer:

t = 0.33h = 1200s

x = 18.33 km

Explanation:

If the origin of coordinates is at the second car, you can write the following equations for both cars:

car 1:

x=x_o+v_1t    (1)

xo = 10 km

v1 = 55km/h

car 2:

x'=v_2t    (2)

v2 = 85km/h

For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:

x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}

t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s

The position in which both cars coincides is:

x=(55km/h)(0.33h)=18.33km

6 0
3 years ago
HELP!
Angelina_Jolie [31]

Answer:

speed is increasing.

this is because the graph has a positive slope (due to it going up ), if it were decreasing, the graph would be going down

4 0
3 years ago
Read 2 more answers
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