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kozerog [31]
2 years ago
11

A rock is thrown down from the top of a cliff with a velocity of 3.61 m/s (down). The cliff is 28.4 m above the ground. Determin

e the velocity of the rock just before it hits the ground.
Physics
1 answer:
Katarina [22]2 years ago
4 0
I’m not sure look it up in the internet!
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A car speed off around a bend at a constant 10m/s explain why it's velocity is not constant
VARVARA [1.3K]
It's velocity is not constant as direction is changing.
 
We know, velocity is speed with direction, so if direction is changing, velocity can't be constant, doesn't matter that speed is constant.

Hope this helps!
8 0
3 years ago
In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) s
sladkih [1.3K]

Answer:

  • quality factor (Q) = 69.99
  • inductor = 1.591 x 10⁻⁴ H
  • capacitor = 3.248 x 10⁻¹⁰ F

Explanation:

Given;

resonance frequency (F₀) = 700 kHz

resistor, R =  10 Ohm

bandwidth (BW) = 10 kHz

bandwidth (BW)  = \frac{R}{2\pi L}

BW = \frac{R}{2\pi L}

make L (inductor) the subject of the formula

L = \frac{R}{2\pi *BW}  =  \frac{10}{2\pi *10,000} =1.591 *10^{-4} \ H = \ 0.1591\ mH

F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}

make C (capacitor)  the subject of the formula

C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F

quality factor (Q) = \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99

quality factor (Q) =  69.99

5 0
3 years ago
An astronaut has a mass of 90 kg. The astronaut travels to Mars, where gravity is weaker. The acceleration of free fall on the s
marin [14]

Answer:

w = mg

w = 90 * 3.7

w = 333 N

hope it helps you

3 0
2 years ago
Read 2 more answers
WILL GIVE BRAINLIEST!!<br> What is a way to transfer charge in which an object becomes polarized?
Andreyy89

Answer:

I answered Number 4 (Solids and Elasticity)

Explanation:

solids and elasticity

4 0
3 years ago
Calculate the potential energy of a +1.0μC point charge sitting 0.1m from a -5.0μC point charge.
AfilCa [17]

Answer:

P.E.      =   -0.449 J

Explanation:

Potential energy of a charge particle in any electrostatic field is defined as the amount of work done ( in negative ) to bring that charge particle from any position to a new position r.

Now Potential energy is defined by this formula,

P.E. = k q₁ q₂/ r

where P.E. is the potential energy.

k = 1/( 4πε₀) = 8.99 × 10⁹ C²/ ( Nm²)

q₁ = charge of one particle = +1.0μC

q₂ = charge of another particle = -5.0μC

r = distance = 0.1 m

Now , P.E. = 8.99 × 10⁹C²/ ( Nm²) * ( -5.0 × 10⁻⁶ C ) × ( 1 × 10⁻⁶ C ) / 0.1 m

          P.E.      =  -0.449 J

8 0
3 years ago
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