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anzhelika [568]
3 years ago
5

An amusement park ride spins you around in a circle of radius 2.5 m with a speed of 8.5 m/s. If your mass is 75 kg, what is the

centripetal force acting on you? A. 1530 N B. 2168 N C. 255 N D. 2359 N
Physics
2 answers:
nataly862011 [7]3 years ago
8 0

B is the correct answer - APEX

zhannawk [14.2K]3 years ago
3 0
B is the answer to this truly did the math
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Determine the volume displaced and then calculate the density of this 54 g sample of brass.
inessss [21]

Answer:

DETAILS IN THE QUESTION INSUFFICIENT TO ANSWER

Explanation:

Assuming the liquid to be water ,

the density d_{w} of water is : 1000kgm^{-3}=1gcm^{-3}

Buoyant force exerted by a liquid on an object with V_{imm} of it's volume immersed is :

F_{B}=V_{imm}*d_{l}*g

where ,

  • F_{B} is the buoyant force
  • d_{l} is the density of the liquid
  • g is the acceleration due to gravity

Thus at equilibrium:

m_{brass}*g=V_{imm}*d_{l}*g\\m_{brass}=V_{imm}*d_{l}\\54=V_{imm}*1\\V_{imm}=54cm^{3}

from these , we get the density of brass to be 1gcm^{-3}

which is not possible

7 0
3 years ago
what is the acceleration of an object if the object has an initial speed of 230 m/s and speeds up to 650 m/s. The time it takes
adell [148]

Answer:

explain this

Explanation:

7 0
2 years ago
What is the relationship between the spring constant and the period in a mass hanging on a spring oscillation and why?
anygoal [31]

Explanation:

Period of a mass on a spring is:

T = 2π√(m/k)

T is inversely proportional with the square root of k.  So as the spring constant increases, the period decreases.

3 0
3 years ago
Read 2 more answers
Given that water at standard pressure freezes at 0∘c, which corresponds to 32∘f, and that it boils at 100∘c, which corresponds t
Zolol [24]
The temperature difference of 1 K is equivalent to the temperature difference of 1 °C. Therefore, we find the relationship between the change in °F and °C.
A change of 212 - 32 °F is the same as a change of 100 - 0 °C. Thus:
(212 - 32) °F = (100 - 0) °C
1 °C = 1.8 °F
1 K = 1.8 °F
6 0
3 years ago
A 0.500 kg rock is whirled in a vertical circle of a radius 0.60 m . the velocity of the rock at the bottom of the swing is 4.0
OLEGan [10]

Explanation:

Centripetal acceleration is:

a = v² / r

a = (4.0 m/s)² / 0.60 m

a = 26.6 m/s²

7 0
3 years ago
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