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GuDViN [60]
2 years ago
15

jesse is swinging miguel in a circle at a tangential speed of 3.50 m/s. if the radius of the circle is 0.600 m and miguel has a

mass of 11.0 kg, what is the centripetal force on miguel? round to the nearest whole number.
Physics
2 answers:
Morgarella [4.7K]2 years ago
6 0
Centripetal acceleration = (speed)² / (radius) .

Force = (mass) · (acceleration)

Centripetal force = (mass) · (speed)² / (radius) .

                             = (11 kg) · (3.5 m/s)² / (0.6 m)

                             = (11 kg) · (12.25 m²/s²) / (0.6 m)

                             =  (11 · 12.25) / 0.6  kg-m/s²

                             =      224.58 newtons.    (about 50.5 pounds)

That's the tension in Miguel's arm or leg or whatever part of his body
Jesse is swinging him by.  It's the centripetal force that's needed in
order to swing 11 kg in a circle with a radius of 0.6 meter, at 3.5
meters/second.  If the force is less than that, then the mass has to
either swing slower or else move out to follow a bigger circle.
Dimas [21]2 years ago
3 0

the answer is 225 n , got it right

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A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th
ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = 9.8\ m/s^2

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

                Kinetic energy K.E = \frac{1}{2}mv^2.

From above we know that -

   Kinetic energy at the bottom of the cliff = potential energy at a height h

                 \frac{1}{2}mv^2=\ mgh

                ⇒ v^2=\ 2gh

                ⇒ h=\ \frac{v^2}{2g}

                ⇒ h=\ \frac{(24.2)^2}{2\times 9.8}

                ⇒ h=\ 29.88\ m

Hence, the height of the cliff is 29.88 m

             


5 0
1 year ago
A confused dragonfly flies forward and backward in a straight line. Its motion is shown on the following graph of horizontal pos
sertanlavr [38]

Answer:

  0

Explanation:

Assuming your graph and question match the attachment, the average speed is 0. The bug ends up where it started, so its displacement is zero.

  average speed = displacement/time = 0/(8 s)

  average speed = 0

7 0
2 years ago
An airplane flying parallel to the ground undergoes two consecutive dis- placements. The first is 75 km 30.0° west of north, and
torisob [31]

Answer: displacement of airplane is 172 km in direction 34.2 degrees East of North

Explanation:

In constructing the two displacements it is noticed that the angle between the 75 km vector and the 155 km vector is a right angle (90 degrees).

 

Hence if the plane starts out at A, it travels to B, 75 km away, then turns 90 degrees to the right (clockwise) and travels to C, 155 km away from B. Angle ABC is 90 degrees, hence we can use Pythagoras theorem to solve for AC

 

AC2 = AB2 + BC2 ; AC^2 = 752 + 1552  ; from this we get AC = 172 km (3 significant figures)

 

Angle BAC = Tan-1(155/75) ; giving angle BAC = 64.2 degrees

 

Hence AC is in a direction (64.2 - 30) = 34.2 degrees East of North

 

Therefore the displacement of the airplane is 172 km in a direction 34.2 degrees East of North

5 0
1 year ago
You and your friend Peter are putting new shingles on a roof pitched at 20degrees . You're sitting on the very top of the roof w
Anit [1.1K]

Answer:

v₀ =3.8 m/s

Explanation:

Newton's second law of the box:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Known data

m=2.1 kg  mass of the box

d= 5.4m  length of the roof

θ = 20° angle θ of the roof with respect to the horizontal direction

μk= 0.51 : coefficient of kinetic friction between the box and the roof  

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the box on the roof  and the y-axis in the direction perpendicular to it.

W: Weight of the box  : In vertical direction

N : Normal force : perpendicular to the direction the  roof

fk : Friction force: parallel to the direction to the roof

Calculated of the weight  of the box

W= m*g  =  (2.1 kg)*(9.8 m/s²)= 20.58 N

x-y weight components

Wx= Wsin θ= (20.58)*sin(20)° =7.039 N

Wy= Wcos θ =(20.58)*cos(20)°= 19.34 N

Calculated of the Normal force

∑Fy = m*ay    ay = 0

N-Wy= 0

N=Wy =19.34 N

Calculated of the Friction force:

fk=μk*N= 0.51* 19.34 N = 9.86 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx-f = ( 2.1)*a

7.039 - 9.86  = ( 2.1)*a

-2.821 = ( 2.1)*a

a=(-2.821) /( 2.1)

a= -1.34  m/s²

Kinematics of the box

Because the box moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d Formula (2)

Where:  

d:displacement  = 5.4 m

v₀: initial speed  

vf: final speed  = 0

a : acceleration of the box = -1.34  m/s²

We replace data in the formula (2)

0²=v₀²+2*(-1.34)*(5.4)

2*(1.34)*(5.4)= v₀²

v_{o} =\sqrt{14.472}

v₀ = 3.8 m/s

7 0
2 years ago
You have a suspect on the run in a dry, wooded area; they’ve ditched the gun but are hiding from you. how long do you have to fi
erma4kov [3.2K]

Answer:

C. a few days

Explanation:

The evidence that we will have to investigate is the gunshot residue on the body of the suspect. When a person fires a gun there will be a deposition of material from the components of the chemicals present in the gun. This deposit is called gunshot residue and consists of particles of the propellant or explosive primer.

Experimentation has shown that the period of time that is person will test possible for gunshot residue is 5.27 days.

7 0
1 year ago
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