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2 years ago

How do kinetic and potential energy transfer to one throughout a roller coaster ride?

Physics

1 answer:

mojhsa [17]2 years ago

6 0

Answer:

As the cars ascend the next hill, some kinetic energy is transformed back into potential energy. Then, when the cars descend this hill, potential energy is again changed to kinetic energy. This conversion between potential and kinetic energy continues throughout the ride.

Explanation:

hope it helps U

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Say you want to make a sling by swinging a mass M of 2.3 kg in a horizontal circle of radius 0.034 m, using a string of length 0

ycow [4]

Answer:

T = 764.41 N

Explanation:

In this case the tension of the string is determined by the centripetal force. The formula to calculate the centripetal force is given by:

$F_c=m\frac{v^2}{r}$ (1)

m: mass object = 2.3 kg

r: radius of the circular orbit = 0.034 m

v: tangential speed of the object

However, it is necessary to calculate the velocity v first. To find v you use the formula for the kinetic energy:

$K=\frac{1}{2}mv^2$

You have the value of the kinetic energy (13.0 J), then, you replace the values of K and m, and solve for v^2:

$v^2=\frac{2K}{m}=\frac{2(13.0J)}{2.3kg}=11.3\frac{m^2}{s^2}$

you replace this value of v in the equation (1). Also, you replace the values of r and m:

$F_c=(2.3kg)(\frac{11.3m^2/s^2}{0.034})=764.41N$

hence, the tension in the string must be T = Fc = 764.41 N

5 0

3 years ago

The mass of a star is 1.61·1031 kg and its angular velocity is 1.60E-7 rad/s. Find its new angular velocity if the diameter sudd

Tcecarenko [31]

Answer:

ω₂ = 1.9025 x 10⁻⁶ rad/s

Explanation:

given,

mass of star = 1.61 x 10³¹ kg

angular velocity = 1.60 x 10⁻⁷ rad/s

diameter suddenly shrinks = 0.29 x present size

r₂ = 0.29 r₁

using conservation of angular momentum

I₁ ω₁ = I₂ ω₂

$(\dfrac{2}{5}mr_1^2)\omega_1=(\dfrac{2}{5}mr2^2)\omega_2$

$r_1^2\times \omega_1=r_2^2\times \omega_2$

$r_1^2\times 1.60\times 10^{-7}=(0.29r_1)^2\times \omega_2$

$1.60\times 10^{-7}=0.0841\times \omega_2$

$\omega_2=\dfrac{1.60\times 10^{-7}}{0.0841}$

ω₂ = 1.9025 x 10⁻⁶ rad/s

5 0

3 years ago

Explain a charged by neutral body negatively by induction

Tpy6a [65]

If a negatively charged object is used to charge a neutral object by induction, then the neutral object will acquire a positive charge. And if a positively charged object is used to charge a neutral object by induction, then the neutral object will acquire a negative charge.

6 0

3 years ago

2. What type of motion objects that fall under the pull of gravity?

vivado [14]

Kinetic Energy Pulls Any object to the ground.This Energy is a part of Gravity.
Wish you happy timez!

4 0

3 years ago

A gymnast with mass 46.0 kg stands on the end of a uniform balance beam as shown. The beam is 5.00 m long and has a mass of 250

worty [1.4K]

Sum up the moments of the right and the left support:
∑ M2 = 0
F1 · 3.92 m = 46 kg · 9,81 m/s² · 4.46 m + 250 kg · 9.81 m/s² · 1.96 m
F 1 · 3.92 = 2012.6 + 4806.9
F 1 = 6819.5 : 3.92
F 1 = 1739.67 N
∑ M 1 = 0
F 2 · 3.92 m + 46 kg · 9.81 m/s² · 0.54 m = 250 kg · 9.81 m/s² · 1.96 m
F 2 · 3.92 + 243.68 = 4806.9
F 2 · 3.92 = 4563.22
F 2 = 4563.22 : 3.92
F 2 = 1164 N
Answer: The forces on the beam are: F 1 = 1739.67 N and F 2 = 1164 N.

3 0

4 years ago