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olasank [31]
3 years ago
11

A bucket tied to a rope is moving at a constant speed of 5.0 m/s in a circle of radius

Physics
1 answer:
jeyben [28]3 years ago
6 0

Answer:

a=12.5\ m/s^2

Explanation:

Given that,

The speed of the bucket tied to a rope, v = 5 m/s

The radius of the circle, r =2 m

We need to find the magnitude of the centripetal acceleration of the bucket. The formula for the centripetal acceleration is given by :

a=\dfrac{v^2}{r}\\\\a=\dfrac{(5)^2}{2}\\\\a=12.5\ m/s^2

So, the centripetal acceleration of the bucket is 12.5\ m/s^2.

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The electrical resistance of an element in a platinum resistance thermometer at 100°c, 0°c and room temperature are 75.00Ω, 63.0
Dominik [7]

Answer:

16.6 °C

Explanation:

From the question given above, the following data were obtained:

Temperature at upper fixed point (Tᵤ) = 100 °C

Resistance at upper fixed point (Rᵤ) = 75 Ω

Temperature at lower fixed point (Tₗ) = 0 °C

Resistance at lower fixed point (Rₗ) = 63.00Ω

Resistance at room temperature (R) = 64.992 Ω

Room temperature (T) =?

T – Tₗ / Tᵤ – Tₗ = R – Rₗ / Rᵤ – Rₗ

T – 0 / 100 – 0 = 64.992 – 63  / 75 – 63

T / 100 = 1.992 / 12

Cross multiply

T × 12 = 100 × 1.992

T × 12 = 199.2

Divide both side by 12

T = 199.2 / 12

T = 16.6 °C

Thus, the room temperature is 16.6 °C

6 0
2 years ago
Look at a photograph of a fault. notice how the right side appears lower than the left side. this happens when pieces of crust a
dybincka [34]
I believe it’s divergent boundary but I might be wrong
5 0
3 years ago
Read 2 more answers
Select the words that make the sentence a true statement.
ZanzabumX [31]

Answer:

1.  force

2. greater

Explanation:

4 0
3 years ago
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A fish scale, consisting of a spring with spring constant k=200N/m, is hung vertically from the ceiling. A 2.6 kg fish is attach
Olegator [25]

Answer:

Explanation:

The fish is initially at rest and it is also at rest when the spring is fully stretched at the maximum distance.

Change in gravity potential energy = change in spring potential energy

mgh = 1/2kh^2

Assume gravity constant g is 10m/s^2

2.6*10*h = 1/2*200*h^2

100h^2 - 26h = 0

2h(50h - 13) = 0

h = 0 or h = 13/50 = 0.65m

h = 0 is before the spring is stretched

So the maximum distance is 0.65m.

3 0
2 years ago
Read 2 more answers
Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

5 0
3 years ago
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