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lilavasa [31]
3 years ago
13

A snowboarder is sliding back and forth on a half pipe at one point she leaves the top of the half pipe and slides to the other

side choose when kinetic energy increases during the snowboarders ride
Physics
2 answers:
andrew11 [14]3 years ago
5 0

Answer:

what are you asking?

Explanation:

r-ruslan [8.4K]3 years ago
3 0

Answer:

he kinetic energy increases on the descent, being maximum at the lowest point of the trajectory.

Explanation:

In these semicircular sections the skaters slide from one side to the other, in the downward path their kinetic energy increases and their potential energy decreases; When it leaves the ramp and is in the air, the kinetic energy decreases rapidly, up to the point of maximum height where the kinetic energy is zero.

Consequently, the kinetic energy increases on the descent, being maximum at the lowest point of the trajectory.

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A concrete slab of mass 400 kg accelerates down a concrete slope inclined at 35°. The kinetic between the slab and slope is 0.60
belka [17]

I got you

Explanation:

normal force = 400 g cos 35

friction force up slope = .6 (400 g) cos 35

weight component down slope = 400 g sin 35

400 a = 400 g sin 35 - .6 (400 g cos 35)

a = g (sin 35 - .6 cos 35) = .082 g

I hope this helps you

8 0
2 years ago
A charge -353e is uniformly distributed along a circular arc of radius 5.30 cm, which subtends an angle of 48°. What is the line
vladimir2022 [97]

Answer:

- 1.3 x 10⁻¹⁵ C/m

Explanation:

Q = Total charge on the circular arc = - 353 e = - 353 (1.6 x 10⁻¹⁹) C = - 564.8 x 10⁻¹⁹ C

r = Radius of the arc = 5.30 cm = 0.053 m

θ = Angle subtended by the arc = 48° deg = 48 x 0.0175 rad = 0.84 rad        (Since 1 deg = 0.0175 rad)

L = length of the arc

length of the arc is given as

L = r θ

L = (0.053) (0.84)

L = 0.045 m

λ = Linear charge density

Linear charge density is given as

\lambda =\frac{Q}{L}

Inserting the values

\lambda =\frac{-564.8\times 10^{-19}}{0.045}

λ = - 1.3 x 10⁻¹⁵ C/m

4 0
3 years ago
A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object mo
Zinaida [17]

Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2

E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2

E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

3 0
3 years ago
The velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in
vivado [14]

The force the escaping gas exerts of the rocket is 10.42 N.

<h3>Force escaping gas exerts</h3>

The force the escaping gas exerts of the rocket is calculated as follows;

F = m(v - u)/t

where;

  • m is mass of the rocket
  • v is the final velocity of the rocket
  • u is the initial velocity of the rocket
  • t is time of motion

F = (0.25)(40 - 15)/0.6

F = 10.42 N

Thus, the force the escaping gas exerts of the rocket is 10.42 N.

Learn more about force here: brainly.com/question/12970081

#SPJ1

7 0
1 year ago
An ideal spring with spring constant k is hung from the ceiling. The initial length of the spring, with nothing attached to the
hram777 [196]

The mass m of the object = 5.25 kg

<h3>Further explanation</h3>

Given

k = spring constant = 3.5 N/cm

Δx= 30 cm - 15 cm = 15 cm

Required

the mass m

Solution

F=m.g

Hooke's Law

F = k.Δx

\tt m.g=k.\Delta x\\\\m.10=3.5\times 15\\\\m=5.25~kg

7 0
2 years ago
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