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Svetach [21]
2 years ago
15

2. A car accelerates down the road. What is the "reaction" to the tires pushing on the road?

Physics
1 answer:
Artist 52 [7]2 years ago
7 0
The answer would be friction
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He distance between two consecutive crests is 2.5 meters. Which characteristic of the wave does this distance represent? amplitu
kupik [55]

Frequency is measured in units of reciprocal time.
Period is measured in units of time.
Phase is a number without units that represents a fraction of a wave.

None of these is measured in meters, so none of them can be the answer.
It must be either amplitude or wavelength.

Amplitude is a quantity that's measured at one or two points in the same wave.
The question is talking about points on consecutive waves.

<em>Wavelength is</em> the only choice left.  That must be it.

3 0
2 years ago
Read 2 more answers
A 3-kg block, attached to a spring, executes simple harmonic motion according to x= 2cos(50t) where x is in meters and t is in s
saw5 [17]

Explanation:

w=50=√k/m=√k/3

SO

(50²)*3 is 7500N/m

4 0
3 years ago
Separating the electron from the proton in a hydrogen atom takes 2.18 ✕ 10−18 j of work. through what electric potential differe
My name is Ann [436]
Electric potential = work done/charge of electron = 2.18×10⁻¹⁸/1.6×10⁻¹⁹
                                                                              =  13.625 V
6 0
2 years ago
Suppose a car approaches a hill and has an initial speed of
kvv77 [185]

Answer:

a) 1.73*10^5 J

b) 3645 N

Explanation:

106 km/h = 106 * 1000/3600 = 29.4 m/s

If KE = PE, then

mgh = 1/2mv²

gh = 1/2v²

h = v²/2g

h = 29.4² / 2 * 9.81

h = 864.36 / 19.62

h = 44.06 m

Loss of energy = mgΔh

E = 780 * 9.81 * (44.06 - 21.5)

E = 7651.8 * 22.56

E = 172624.6 J

Thus, the amount if energy lost is 1.73*10^5 J

Work done = Force * distance

Force = work done / distance

Force = 172624.6 / (21.5/sin27°)

Force = 172624.6 / 47.36

Force = 3645 N

5 0
2 years ago
You have two square metal plates with side lengths of (6.50 C) cm. You want to make a parallel-plate capacitor that will hold a
gtnhenbr [62]

Answer:

The necessary separation between  the two parallel plates is 0.104 mm

Explanation:

Given;

length of each side of the square plate, L = 6.5 cm = 0.065 m

charge on each plate, Q = 12.5 nC

potential difference across the plates, V = 34.8 V

Potential difference across parallel plates is given as;

V = \frac{Qd}{L^2 \epsilon_o} \\\\d = \frac{V L^2 \epsilon_o}{Q}

Where;

d is the separation or distance between the two parallel plates;

d = \frac{VL^2 \epsilon_o}{Q} \\\\d =  \frac{34.8*(0.065)^2 *8.854*10^{-12}}{12.5*10^{-9}} \\\\d = 0.000104 \ m\\\\d = 0.104 \ mm

Therefore, the necessary separation between  the two parallel plates is 0.104 mm

6 0
3 years ago
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