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Aleksandr-060686 [28]
2 years ago
12

A ball is dropped from the top of a building. When does the ball have the most potential energy?

Physics
1 answer:
exis [7]2 years ago
5 0

Answer:

When it hits the ground

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A Shaolin monk of mass 60 kg is able to do a ‘finger stand’: he supports his whole weight on his two index fingers, giving him a
ss7ja [257]

Answer:

P = 1471500 [Pa]

Explanation:

We must remember that pressure is defined as the relationship between Force over the area.

P=F/A

where:

P = pressure [Pa] (units of pascals)

F = force [N] (units of Newtons)

A = area of contact = 4 [cm²]

But first we must convert from cm² to m²

A = 4[cm^{2}]*\frac{1^{2} m^{2} }{100^{2} cm^{2} }

A = 0.0004 [m²]

Also, the weight should be calculated as follows:

w = m*g

where:

m = mass = 60 [kg]

g = gravity acceleration = 9.81 [m/s²]

Now replacing:

w = 60*9.81\\w = 588.6[N]

And the pressure:

P=588.6/0.0004\\P=1471500 [Pa]

Because 1 [Pa] = 1 [N/m²]

8 0
3 years ago
shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at const
Contact [7]

Answer:

a) T=0.40 N

b) T=1.9 s

Explanation:

Let's find the radius of the circumference first. We know that bob follows a circular path of circumference 0.94 m, it means that the perimeter is 0.94 m.

The perimeter of a circunference is:

P=2\pi r=0.94

r=\frac{0.94}{2\pi}=0.15 m

Now, we need to find the angle of the pendulum from vertical.

tan(\alpha)=\frac{r}{L}=\frac{0.15}{0.90}=0.17

\alpha=9.44 ^{\circ}

Let's apply Newton's second law to find the tension.

\sum F=ma_{c}=m\omega^{2}r

We use centripetal acceleration here, because we have a circular motion.

The vertical equation of motion will be:

Tcos(\alpha)=mg (1)

The horizontal equation of motion will be:

Tsin(\alpha)=m\omega^{2}r (2)

a) We can find T usinf the equation (1):

T=\frac {mg}{cos(\alpha)}=\frac{0.04*9.81}{cos(9.44)}=0.40 N

We can find the angular velocity (ω) from the equation (2):

\omega=\sqrt{\frac{Tsin(\alpha)}{mr}}=3.31 rad/s

b) We know that the period is T=2π/ω, therefore:

T=\frac{2\pi}{\omega}=\frac{2\pi}{3.31}=1.9 s

I hope it helps you!

8 0
2 years ago
the force of a bag of potatoes on a vegetable stand is 22.5 N down. what is the force of the vegetable stand on the bag of potat
natima [27]

As per Newton's III law we can say that

Force applied by object 1 on object 2 is always equal in magnitude and opposite in direction of the force that object 2 apply on object 1.

So we can say it as

F_{12} = -F_{21}

now here above question is based upon the same

if a bag of vegetables applied a force F = 22.5 N of the surface stand the the same surface will apply same magnitude of force in opposite direction on the vegetables bag

So our answer will be F = 22.5 N (upwards).

4 0
2 years ago
Please help me out !!!!!!
nikdorinn [45]

Answer:

89

Explanation:

8 0
3 years ago
A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 19
Hunter-Best [27]

The car will take 300 m before it stops due to applying break.

<h3>What's the relation between initial velocity, final velocity, acceleration and distance?</h3>
  • As per Newton's equation of motion, V² - U² = 2aS
  • V= final velocity velocity of the object, U = initial velocity velocity of the object, a= acceleration, S = distance covered by the object
  • Here, U = 60 ft/sec, V = 0 m/s, a= -6 ft/sec²
  • So, 0² - 60² = 2×6× S

=> -3600 = -12S

=> S = 3600/12 = 300 m

Thus, we can conclude that the distance covered by the car is 300 m before it stopped.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 6 ft/sec². How long will it take before the car stops?

Learn more about the Newton's equation of motion here:

brainly.com/question/8898885

#SPJ1

7 0
1 year ago
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