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velikii [3]
3 years ago
14

if you run around a circle at 4.5 m/s and the circle has a radius of 7.7 m, what is your centripetal acceleration?

Physics
1 answer:
madreJ [45]3 years ago
7 0

Answer:

Centripetal acceleration,

a_{c} =2.63\ m/s^{2} }

Explanation:

Centripetal acceleration:

Centripetal acceleration is the idea that any object moving in a circle, in something called circular motion, will have an acceleration vector pointed towards the center of that circle.

Centripetal means towards the center.

Examples of centripetal acceleration (acceleration pointing towards the center of rotation) include such situations as cars moving on the cicular part of the road.

An acceleration is a change in velocity.

Formula for Centripetal acceleration:

a_{c} =\frac{(velocity)^{2} }{radius}

Given here,

Velocity = 4.5 m/s

radius = 7.7 m

To Find :

a_{c} = ?

Solution:

We have,

a_{c} =\frac{(velocity)^{2} }{radius}

Substituting  given value in it we get

a_{c} =\frac{(4.5)^{2}}{7.7} \\\\a_{c} =\frac{20.25}{7.7}\\\\a_{c} =2.629\ m/s^{2} \\\\\therefore a_{c} =2.63\ m/s^{2

Centripetal acceleration,

a_{c} =2.63\ m/s^{2} }

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salantis [7]

Answer:

1.05m or 105cm

Explanation:

Using the hooke's law equation as follows;

F = –k.x

Where;

F = force (N)

x = extension length (m)

k = constant of proportionality (N/m)

According to the information given in this question;

Displacement (x) = 85cm = 85/100 = 0.85m

Force = 12500N

Using F = kx, we find the proportionality constant

k = F/x

K = 12500/0.85

K = 14705.8N/m.

Also, since K = 14705.8N/m, the displacement (x), when the force increases to 15500N is;

F = kx

x = F/k

x = 15500/14705.8

x = 1.05m or 105cm

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3 years ago
The diagram shows particles that make up an atom. Which label best completes the diagram?
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The right answer is "Strong nuclear force"
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A pendulum at position A is released and swings through position B to position Con the other side.
zaharov [31]

Explanation:

Given the conditions A,B and C when the pendulum is released, at point A the initial velocity of the pendulum is zero(0), the potential energy stored is maximum(P.E= max),

the conditions can be summarized bellow

point A

initial velocity= 0

final velocity=0

P.E= Max

K.E= 0

point B

initial velocity= maximum

final velocity=maximum

P.E=K.E

point C

initial velocity= min

final velocity=min

P.E= 0

K.E= max

3 0
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Objects A and B are separated by 2m and the force of attraction between them is 8 x 10^-10 N . If the mass of A is 8kg what is t
Jobisdone [24]

Answer:

Only object a and b???????

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2 years ago
Three blocks are placed in contact on a horizontal frictionless surface. A constant force of magnitude F is applied to the box o
Lina20 [59]

Answer:

A) M

Explanation:

The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:

Box with mass M

\Sigma F = F - F' = M\cdot a

Box with mass 2M

\Sigma F = F' - F'' = 2\cdot M \cdot a

Box with mass 3M

\Sigma F = F'' = 3\cdot M \cdot a

On the third equation, acceleration can be modelled in terms of F'':

a = \frac{F''}{3\cdot M}

An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.

F' = 2\cdot M \cdot a + F''

F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''

F' = \frac{5}{3}\cdot F''

Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:

F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)

F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''

F = 2\cdot F''

F'' = \frac{1}{2}\cdot F

Afterwards, F' as function of the external force can be obtained by direct substitution:

F' = \frac{5}{6}\cdot F

The net forces of each block are now calculated:

Box with mass M

M\cdot a = F - \frac{5}{6}\cdot F

M\cdot a = \frac{1}{6}\cdot F

Box with mass 2M

2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F

2\cdot M \cdot a = \frac{1}{3}\cdot F

Box with mass 3M

3\cdot M \cdot a = \frac{1}{2}\cdot F

As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.

8 0
3 years ago
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