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Ilya [14]
3 years ago
5

rickey approaches third base. He dives head first, hitting the ground at 6.75 m/s and reaching the base at 5.91 m/s in 2.5 secon

ds. Determine the distance Rickey slides across the ground before touching the base.
Physics
1 answer:
Gekata [30.6K]3 years ago
4 0

Answer:

15.825 m

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 6.75 m/s

v = Final velocity = 5.91 m/s

s = Displacement

a = Acceleration

Equation of motion

v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{5.91-6.75}{2.5}\\\Rightarrow a=-0.336\ m/s^2

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.91^2-6.75^2}{2\times -0.336}\\\Rightarrow s=15.825\ m

The distance Rickey slides across the ground before touching the base is 15.825 m

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What kinds of bonds can happen between the elements of a compound
Anit [1.1K]
What kinds of bonds can happen between the elements of a compound?

<span>Covalent and Ionic</span>

4 0
3 years ago
Sabiendo que el indice de refracción dela gua es 1,33 calcula el ángulo de refracción resultante para un rayo de luz que incide
Kaylis [27]

Answer:

35.16 degrees

Explanation:

Knowing that the index of refraction of the guide is 1.33, calculate the resulting angle of refraction for a ray of light that falls on a pool with an angle of incidence of 50º

Refractive index, n = 1.33

The angle of incidence, i = 50°

We need to find the angle of refraction. let it is r. It can be calculated using Snells law as follows:

n=\dfrac{\sin i}{\sin r}\\\\\sin r=\dfrac{\sin i}{n}\\\\r=\sin^{-1}(\dfrac{\sin i}{n})\\\\r=\sin^{-1}(\dfrac{\sin 50}{1.33})\\\\r=35.16^{\circ}

So, the angle of refraction is 35.16 degrees.

3 0
2 years ago
A 5.00-V battery charges the parallel plates in a capacitor, with a plate area of 865 mm2 and an air-filled separation of 3.00 m
Westkost [7]

Answer:

W = 3.21x10⁻¹¹ J

Explanation:

The work required to separate the plates can be calculated using the following equation:

W = U_{2} - U_{1} = \frac{1}{2}(C_{2}V_{2}^{2} - C_{1}V_{1}^{2})

<u>Where</u>:

U₂: is the final stored energy

U₁: is the initial stored energy

C₂: is the final capacitance

C₁: is the initial capacitance

V₁: is the initial potential difference = 5.00 V

V₂: is the final potential difference

The initial and final capacitance is:

C_{1} = \epsilon_{0}*\frac{A}{d_{1}}

<u>Where</u>:

ε₀: is the vacuum permittivity = 8.85x10⁻¹² C²/(N*m²)

d: is the initial distance = 3.00 mm = 3.00x10⁻³ m    

A: is the plate area = 865 mm² =  8.65x10⁻⁴ m²

C_{1} = \epsilon_{0}*\frac{A}{d_{1}} = 8.85 \cdot 10^{-12} C^{2}/(N*m^{2})*\frac{8.65 \cdot 10^{-4} m^{2}}{3.00 \cdot 10^{-3} m} = 2.55 \cdot 10^{-12} F      

Similarly, C₂ is:

C_{2} = \epsilon_{0}*\frac{A}{d_{2}} = 8.85 \cdot 10^{-12} C^{2}/(N*m^{2})*\frac{8.65 \cdot 10^{-4} m^{2}}{3.00 + 3.00 \cdot 10^{-3} m} = 1.28 \cdot 10^{-12} F

Now, V₂ can be calculated by finding the initial charge (q₁):

q_{1} = C_{1}V_{1} = 2.55 \cdot 10^{-12} F*5.00 V = 1.28 \cdot 10^{-11} C

Since, q₁ is equal to q₂, V₂ is:

V_{2} = \frac{q_{2}}{C_{2}} = \frac{1.28 \cdot 10^{-11} C}{1.28 \cdot 10^{-12} F} = 10 V

Finally, we can find the work:

W = \frac{1}{2}(C_{2}V_{2}^{2} - C_{1}V_{1}^{2}) = \frac{1}{2}(1.28 \cdot 10^{-12} F*(10 V)^{2} - 2.55 \cdot 10^{-12} F(5.00 V)^{2}) = 3.21 \cdot 10^{-11} J

Therefore, the work required to separate the plates is 3.21x10⁻¹¹ J.

I hope it helps you!

4 0
3 years ago
Find the velocity in m/s of a swimmer who swims 110m toward the shore in 72 s
Sladkaya [172]
1.5 m/s toward shore 

8 0
3 years ago
Which of the following is a scalar quantity?
neonofarm [45]

Answer:

55

Explanation:

I hope this answer help u

4 0
3 years ago
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