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Bas_tet [7]
3 years ago
5

If a 990 kg car is on the road and the Ff is 360 n what is the normal force

Physics
2 answers:
Veseljchak [2.6K]3 years ago
8 0
Given that a car is in the road, there is only movement in the x-direction. There is no movement in the y-direction.

Looking at the y-direction for the normal force:

F = N - mg
0 = N - mg, (no movement in y-dir.)
N = mg
N = (990)(9.8)
N = 9702 newtons

The normal force exerted on the car by the road is 9702 newtons.
Arturiano [62]3 years ago
8 0

Answer:

9700 N upward

Explanation:

Got right on my test.

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What is 60 kilometers to meters (1 km = 1000 m)
mestny [16]
60,000 meters. no explanation
5 0
3 years ago
Read 2 more answers
The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.750 W/m2. (a) Determine the intensit
sveticcg [70]

Answer:

a)   I = 3.63 W / m² , b)   I = 0.750 W / m²

Explanation:

The intensity of a sound wave is given by the relation

         I = P / A = ½ ρ v (2π f s_{max})²

         I = (½ ρ v 4π² s_{max}²) f²

a) with the initial condition let's call the intensity Io

        cte = (½ ρ v 4π² s_{max}²)

         I₀ = cte s² f₀²

        I₀ = cte 10 6

If frequency is increase f = 2.20 10³ Hz

         I = constant (2.20 10³) 2

         I = cte 4.84 10⁶

let's find the relationship of the two quantities

        I / Io = 4.84

        I = 4.84 Io

        I = 4.84 0.750

        I = 3.63 W / m²

b) in this case the frequency is reduced to f = 0.250 10³ Hz and the displacement s = 4 s or

        I = cte (f s)²

        I = constant (0.250 10³ 4)²

 

        I = cte 1 10⁶

         

the relationship

        I / Io = 1

        I = Io

        I = 0.750 W / m²

6 0
2 years ago
A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with
Julli [10]

Explanation:

Given that,

Capacitor C=1.66\ \mu F

Resistor R=80.0\ \Omega

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

f_{c}=\dfrac{1}{2\pi R C}

Where, R = resistor

C = capacitor

Put the value into the formula

f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}

f_{c}=1198.45\ Hz

(B). We need to calculate the V_{R} when f = \dfrac{1}{2f_{c}}

Using formula of  V_{R}

V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})

Put the value into the formula

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=2.280\ Volt

(C). We need to calculate the V_{R} when f = f_{c}

Using formula of  V_{R}

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=3.606\ Volt

(D). We need to calculate the V_{R} when f = 2f_{c}

Using formula of  V_{R}

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=4.561\ Volt

Hence, This is the required solution.

8 0
3 years ago
How do i find the weight of the suitcase<br>pls help asap​
Citrus2011 [14]
F=W=mg
12N(given)= m*9.8
m= 1.22 kg
3 0
3 years ago
I'm not sure if the answer is A or B... someone help
Talja [164]
Its b i literally have had this exact question
8 0
3 years ago
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