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sdas [7]
2 years ago
8

An athlete prepares to throw a 2.0-kilogram discus. His arm is 0.75 meters long. He spins around several times with the discus a

t the end of his out-stretched arm so that the discus reaches a velocity of 5.0 m/s. What is the centripetal force acting on the discus?
9.4 N
14 N
27 N
66 N
Physics
2 answers:
wlad13 [49]2 years ago
7 0
Centripetal Force (Fcp) = ?

His arm length = Radius (R) = 0.75 m

Discus velocity = Linear Velocity (V) = 5 m/s

Discus mass (m) = 2 kg

Centripetal Acceleration (Acp) = V^2/R or W^2 x R
In this case i will use the V^2/R formula, because it uses the discus velocity (V).

fcp = m \times acp \\ fcp = m \times {v}^{2} \div r \\ fcp = 2 \times {5}^{2} \div 0.75 \\ fcp = 2 \times 25 \div 0.75 \\ fcp = 50 \div 0.75
fcp = 66.666... = 66 \: newtons

Answer: Last option, 66 N.
Masteriza [31]2 years ago
7 0

Answer:

the answer is 66 N

Explanation:

Centripetal Force (Fcp) = ?

His arm length = Radius (R) = 0.75 m

Discus velocity = Linear Velocity (V) = 5 m/s

Discus mass (m) = 2 kg

Centripetal Acceleration (Acp) = V^2/R or W^2 x R

In this case i will use the V^2/R formula, because it uses the discus velocity (V).

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loris [4]

Answer:

<em>1) 1.56 m/s</em>

<em>2) 1.11 m/s</em>

<em></em>

Explanation:

The bus travels 6 blocks east and 8 blocks west

Each block is 100 m long

time taken to travel through this distance = 15 min

average speed of the bus = ?

the total blocks traveled = 6 + 8 = 14 blocks

total distance traveled = 14 x 100 m = 1400 m

time taken = 15 min = 15 x 60 sec = 900 sec

average speed of the bus = distance traveled/time taken

==> 1400/900 = <em>1.56 m/s</em>

<em></em>

2) velocity = displacement/time taken

displacement is the shortest distance between the starting position and the final position of the bus.

The displacement of this bus will be the hypotenuse of the triangle formed by the motion of the bus

the distance traveled east = 6 x 100 = 600 m

the distance traveled north = 8 x 100 = 800 m

displacement = \sqrt{800^2 + 600^2} = 1000 m

The velocity of the bus = 1000/900 = <em>1.11 m/s</em>

4 0
2 years ago
One of the great dangers to mountain climbers is an avalanche, in which a large mass of snow and ice breaks loose and goes on an
sammy [17]

Answer:

Explanation:

The acceleration of an object down a slope (neglecting friction, µ = 0) is:

a = g × sin θ

Where,

g is the acceleration due to gravity and θ is the angle of the slope.

a = (9.8 × sin (21.5º)

= 3.592 m/s²

Using equations of motion,

S = ut + 1/2at²

Since, u = 0,

S = 1/2at²

347 = 1/2 × (3.592)t²

t² = 193.21

= sqrt(193.21)

= 13.9 s.

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The diagram shows movement of thermal energy. At bottom a fire has red curved lines labeled Y with arrowheads pointing upward to
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Answer:

X and Z

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6 0
2 years ago
Read 2 more answers
answers Collision derivation problem. If the car has a mass of 0.2 kg, the ratio of height to width of the ramp is 12/75, the in
Natasha2012 [34]

Answer:

4.8967m

Explanation:

Given the following data;

M = 0.2kg

∆p = 0.58kgm/s

S(i) = 2.25m

Ratio h/w = 12/75

Firstly, we use conservation of momentum to find the velocity

Therefore, ∆p = MV

0.58kgm/s = 0.2V

V = 0.58/2

V = 2.9m/s

Then, we can use the conservation of energy to solve for maximum height the car can go

E(i) = E(f)

1/2mV² = mgh

Mass cancels out

1/2V² = gh

h = 1/2V²/g = V²/2g

h = (2.9)²/2(9.8)

h = 8.41/19.6 = 0.429m

Since we have gotten the heigh, the next thing is to solve for actual slant of the ramp and initial displacement using similar triangles.

h/w = 0.429/x

X = 0.429×75/12

X = 2.6815

Therefore, by Pythagoreans rule

S(ramp) = √2.68125²+0.429²

S(ramp) = 2.64671

Finally, S(t) = S(ramp) + S(i)

= 2.64671+2.25

= 4.8967m

3 0
2 years ago
If vx=9.80 units and vy=-6.40 units, determine the magnitude and direction of v
dexar [7]
The resultant vector can be determined by the component vectors. The component vectors are vector lying along the x and y-axes. The equation for the resultant vector, v is:

v = √(vx² + vy²)
v = √[(9.80)² + (-6.40)²]
v = √137 or 11.7 units
5 0
2 years ago
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