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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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A car moving with constant acceleration covers the distance between two points 60 m apart in 6.0 s. Its speed as it passes the s
BlackZzzverrR [31]

Answer:

The speed in the first point is: 4.98m/s

The acceleration is: 1.67m/s^2

The prior distance from the first point is: 7.42m

Explanation:

For part a and b:

We have a system with two equations and two variables.

We have these data:

X = distance = 60m

t = time = 6.0s

Sf = Final speed = 15m/s

And We need to find:

So = Inicial speed

a = aceleration

We are going to use these equation:

Sf^2=So^2+(2*a*x)

Sf=So+(a*t)

We are going to put our data:

(15m/s)^2=So^2+(2*a*60m)

15m/s=So+(a*6s)

With these equation, you can decide a method for solve. In this case, We are going to use an egualiazation method.

\sqrt{(15m/s)^2-(2*a*60m)}=So

15m/s-(a*6s)=So

\sqrt{(15m/s)^2-(2*a*60m)}=15m/s-(a*6s)

[\sqrt{(15m/s)^2-(2*a*60m)}]^{2}=[15m/s-(a*6s)]^{2}

(15m/s)^2-(2*a*60m)}=(15m/s)^{2}-2*(a*6s)*(15m/s)+(a*6s)^{2}

-120m*a=-180m*a+36s^{2}*a^{2}

0=120m*a-180m*a+36s^{2}*a^{2}

0=-60m*a+36s^{2}*a^{2}

0=(-60m+36s^{2}*a)*a

0=a1

\frac{60m}{36s^{2}} = a2

1.67m/s^{2}=a2

If we analyze the situation, we need to have an aceleretarion  greater than cero. We are going to choose a = 1.67m/s^2

After, we are going to determine the speed in the first point:

Sf=So+(a*t)

15m/s=So+1.67m/s^2*6s

15m/s-(1.67m/s^2*6s)=So

4.98m/s=So

For part c:

We are going to use:

Sf^2=So^2+(2*a*x)

(4.98m/s)^2=0^2+(2*(1.67m/s^2)*x)

\frac{24.80m^2/s^2}{3.34m/s^2}=x

7.42m=x

5 0
2 years ago
Suppose 1 kg of Hydrogen is converted into Helium. a) What is the mass of the He produced? b) How much energy is released in thi
morpeh [17]

Answer:

a) m = 993 g

b) E = 6.50 × 10¹⁴ J

Explanation:

atomic mass of hydrogen = 1.00794

4 hydrogen atom will make a helium atom = 4 × 1.00794 = 4.03176

we know atomic mass of helium = 4.002602

difference in the atomic mass of helium = 4.03176-4.002602 = 0.029158

fraction of mass lost = \dfrac{0.029158}{4.03176}= 0.00723

loss of mass for 1000 g = 1000 × 0.00723 = 7.23

a) mass of helium produced = 1000-7.23 = 993 g (approx.)

b) energy released in the process

E = m c²

E = 0.00723 × (3× 10⁸)²

E = 6.50 × 10¹⁴ J

4 0
2 years ago
Read 2 more answers
Simone is walking her dog on a leash. The dog is pulling with a force of 32 N to the right and Simone is pulling backward with a
sammy [17]

Answer:

Net force acting on them is 16 N and it is acting to the right side.

Explanation:

It is given that,

Force acting by the dog, F_1 = 32\ N (right side)

Force acting by Simone , F_2 = -16\ N (backward)

Let backward direction is taken to be negative while right side is taken to be positive.

The net force will act in the direction where the magnitude of force is maximum. Net force is given by :

F=F_1+F_2

F=32+(-16)    

F = 16 N

So, the net force is 16 N and it is acting to the right side.

6 0
2 years ago
Read 2 more answers
7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh
pentagon [3]

Answer:

The frictional force needed to overcome the cart is 4.83N

Explanation:

The frictional force can be obtained using the following formula:

F= \mu R

where \mu is the coefficient of friction = 0.02

R = Normal reaction of the load = mgcos\theta = 25 \times 9.81 \times cos 10 = 241.52N

Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

F=0.02 \times 241.52N

F = 4.83 N

Hence, the frictional force needed to overcome the cart is 4.83N

4 0
2 years ago
A cubical shell with edges of length a is positioned so that two adjacent sides of one face are coincident with the +x and +y ax
Bingel [31]

Answer:

Q = ba⁴ * ε₀

Explanation:

From Gauss's Law, we know that

flux Φ = Q / ε₀

where ε₀ = 8.85e-12 C²/N·m²

and also,

Φ = EAcosθ

The field is directed along the x-axis, so that all of the flux passes through the side of the cube at x = a. This means that θ = 0º, and thus

Φ = EAcos0

Φ = EA

E = bx² meanwhile, we are interested in the point where x = a, so we substitute and then

E = ba²

Since A = a² for the cube face, we have

Q / ε₀ = E * A

Q / ε₀ = ba² * a²

so that

Q = ba⁴ * ε₀

5 0
2 years ago
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