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deff fn [24]
4 years ago
15

1. The speaker uses needles from two

Physics
1 answer:
GuDViN [60]4 years ago
6 0

Answer:

1. Georgia

2. Ten

3. Repeating

Explanation:

kmkn

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what is the net force required to give an automobile with a mass of 1,600 kg an acceleration of 4.5m/s
Dmitrij [34]

Hi there!

Recall Newton's Second Law:

\large\boxed{\Sigma F = ma}}

∑F = Net force (N)

m = mass (kg)

a = acceleration (m/s²)

We are already given the mass and acceleration, so we can plug these values into the equation:

∑F = 1600 · 4.5 = 7200 N

7 0
3 years ago
(a) two protons in a molecule are 4.50 10-10m apart. find the magnitude of the electric force exerted by one proton on the other
Anuta_ua [19.1K]

Answer:

a)  1. 1365 × 10⁻⁹N

b) 9.1862 × 10⁻⁴⁶N

c) 8.61 × 10⁻¹¹C/Kg

Explanation:

Ke = 8.99× 109 N.m2 / C2 ,      G = 6.67 × 10-11 N. m2 / kg2

a)  F = Ke . q₁.q₂/r²  

=  (8.99× 109 N.m2 / C2 ) ×(1.60×10⁻¹⁹C)²/(4.50×10⁻¹⁰C)²

= 1. 1365 × 10⁻⁹N

b)  

F = Gm₁m₂/r²

= (6.67× 10⁻¹¹)×(1.67×10⁻²⁷)²/(4.50×10⁻¹⁰)²

= 9.1862 × 10⁻⁴⁶N

The electric force is larger by 8.0497 ×10³⁷ times

c)

if Keq₁q₂/r² Gm₁m₂/r²,

with q₁=q₂ = q,  and m₁ =m₂ = m

Then q/m =\sqrt{G/M} = \sqrt{(6.67 X 10^{-11} } /8.89 X 10^{9}

= 8.61 × 10⁻¹¹C/Kg

8 0
3 years ago
Which of the following describes a displacement vs. time graph that looks like this?
lubasha [3.4K]

constant non zero acceleration

6 0
3 years ago
Air friction damps a tuning fork so the amplitude decreases by 1/10 per second. Resonances] 10 Hz. what % chnge in frequency res
aleksandr82 [10.1K]

Answer:

The percentage change in frequency is 10%.

Explanation:

Given that,

Amplitude decreases by 1/10 per second.

Resonance = 10 Hz

We need to calculate the change in frequency

Using formula of change in frequency

\Delta f=\Delta A\times R

Put the value into the formula

\Delta f=\dfrac{1}{10}\times10

\Delta f=1\ Hz

We need to calculate the resonant frequency

Using formula of resonant frequency

f_{r}=R-\Delta f

Put the value into the formula

f_{r}=10-1

f_{r}=9\ Hz

We need to calculate the percentage change in frequency

Using formula of percentage change in frequency

f=\dfrac{10-9}{10}

f=10\%

Hence, The percentage change in frequency is 10%.

7 0
4 years ago
If a CFOC was launched and travels 65 meters and is in the air for 3 seconds, what is the launch velocity and angle?
labwork [276]

Answer:

Lauch velocity (u) = 26.15 m/s

Lauch Angle (θ) = 35°

Explanation:

From the question given above, the following data were obtained:

Range (R) = 65 m

Time of flight (T) = 3 s

Acceleration due to gravity (g) = 10 m/s²

Lauch velocity (u) =?

Lauch Angle (θ) =?

R = u²Sin2θ /g

65 = u² × Sin2θ /10

Recall:

Sin2θ = 2SinθCosθ

65 = u² × 2SinθCosθ / 10

65 = u² × SinθCosθ / 5

Cross multiply

65 × 5 = u² × SinθCosθ

325 = u² × SinθCosθ .....(1)

T = 2uSinθ / g

3 = 2uSinθ / 10

3 = uSinθ / 5

Cross multiply

3 × 5 = uSinθ

15 = u × Sinθ

Divide both side by Sinθ

u = 15 / Sinθ....... (2)

Substitute the value of u in equation (2) into equation (1)

325 = u² × SinθCosθ

u = 15 / Sinθ

325 = (15 / Sinθ)² × SinθCosθ

325 = 225 / Sin²θ × SinθCosθ

325 = 225 × SinθCosθ / Sin²θ

325 = 225 × Cosθ / Sinθ

Cross multiply

325 × Sineθ = 225 × Cosθ

Divide both side by Cosθ

325 × Sineθ / Cosθ = 225

Divide both side by 325

Sineθ / Cosθ = 225 / 325

Sineθ / Cosθ = 0.6923

Recall:

Sineθ / Cosθ = Tanθ

Tanθ = 0.6923

Take the inverse of Tan

θ = Tan¯¹ 0.6923

θ = 35°

Substitute the value of θ into equation (2) to obtain the value of u.

u = 15 / Sinθ

θ = 35°

u = 15 / Sin 35

u = 15 / 0.5736

u = 26.15 m/s

Summary:

Lauch velocity (u) = 26.15 m/s

Lauch Angle (θ) = 35°

8 0
3 years ago
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