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WITCHER [35]
3 years ago
6

Jump rope with a double jump. As you bring the jump rope in front of you, cross your right arm over your left. For the next jump

, cross your left arm over your right. Describes which type of jump?
Single bounce

Straddle

Cross-Over

Double jump
Physics
2 answers:
timama [110]3 years ago
8 0

Answer:

cross over

Explanation:

Andrews [41]3 years ago
6 0
It was a cross over.
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What is reflection of light
AnnZ [28]

Answer: When a ray of light approaches a smooth polished surface and the light ray bounces back, it is called the reflection of light. The incident light ray which lands upon the surface is said to be reflected off the surface. The ray that bounces back is called the reflected ray.<u> </u>

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4 0
3 years ago
A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s . Assume in this problem that air drag
Readme [11.4K]

Answer:

(a)0.0675  J

(b)0.0675 J

(c)0.0675 J

(d)0.0675 J

(e)-0.0675 J

(f)0.459 m

Explanation:

15g = 0.015 kg

(a) Kinetic energy as it leaves the hand

E_k = \frac{mv^2}{2} = \frac{0.015*3^2}{2} = 0.0675  J

(b) By the law of energy conservation, the work done by gravitational energy as it rises to its peak is the same as the kinetic energy as the ball leave the hand, which is 0.0675 J

(c) The change in potential energy would also be the same as 0.0675J in accordance with conservation law of energy.

(d) The gravitational energy at peak point would also be the same as 0.0675J

(e) In this case as the reference point is reversed, we would have to negate the original potential energy. So the potential energy as the ball leaves hand is -0.0675J

(f) Since at the maximum height the ball has potential energy of 0.0675J. This means:

mgh = 0.0675

0.015*9.81h = 0.0675

h = 0.459 m

The ball would reach a maximum height of 0.459 m

4 0
3 years ago
1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp
Tanzania [10]

Answer with explanation:

We are given that  

Mass of ball,m_1=75 g=\frac{75}{1000}=0075kg

1 kg=1000 g

Height,h_1=1.6 m

h_2=0.6 m

Horizontal velocity,v_x=2 m/s

Mass of platem_2=400 g=\frac{400}{1000}=0.4 kg

a.Initial velocity of plate,u_2=0

Velocity before impact=u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s

Where g=9.8 m/s^2

Velocity after impact,v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s

According to law of conservation of momentum  

m_1u_1+m_2u_1=-m_1v_1+m_2v_2

Substitute the values  

0.075\times 5.6+0=-0.075\times 3.4+0.4v_2

0.4v_2=0.075\times 5.6+0.075\times 3.4

v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s

Velocity of plate=1.69 m/s

b.Initial energy=\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J

Final energy=\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2

Final energy=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

6 0
2 years ago
Speed=100m/sec<br> Frequency=10 Hz<br> Wavelength=?
elena-14-01-66 [18.8K]

Answer:

Wavelength = 10 m

Explanation:

Given:

Speed = 100 ms^{-1}

Frequency = 10 Hz = 10 s^{-1}

To find : Wavelength = ?

We know that the relationship between wavelength λ, frequency f and speed v is given by the equation

    v = fλ

Therefore wavelength λ = v/f

                                        = 100 ms^{-1} / 10 ms^{-1}

                                        = 10 m

Hence wavelength = 10 m

4 0
3 years ago
In the equation F = Kq1 q2/r2 solve for q2. Solve for r.
Tpy6a [65]

Answer:

r = √(k q₁ q₂ / F)

Explanation:

F = k q₁ q₂ / r²

Multiply both sides by r²:

F r² = k q₁ q₂

Divide both sides by F:

r² = k q₁ q₂ / F

Take the square root of both sides:

r = √(k q₁ q₂ / F)

3 0
3 years ago
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