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stiks02 [169]
3 years ago
15

a car travelling at 50m/h on a horizontal highway (a) if the coefficient of static friction between road and tyres on a rainy da

y is 0.100, what is the minimum distance in which the car will stop?

Physics
1 answer:
aleksandrvk [35]3 years ago
3 0
Hope this helps!

-Lilly

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Jake calculates that the frequency of a wave is 500 hertz and that the wave is moving at 1,250 m/s. What is the wavelength of th
Neko [114]
Frequency (f) = 500 hz (SI)
Velocity (V) = 1250 m/s (SI)
Wavelength (Lambda) = ? meters

v =  \lambda \times f
1250 =  \lambda \times 500 \\ \lambda = 1250 \div 500 \\ \lambda = 2.5 \: meters
6 0
3 years ago
The Moon's center is 3.9x10 m from Earth's center. The Moon is 1.5x10^8 km from the Sun's center. If the mass of the Moon is 7.3
nika2105 [10]

Explanation:

It is given that The Moon's center is 3.9x10⁸ m from Earth's center. The moon 1.5x10⁸ km from the Sun's center. We need to find the ratio of the gravitational forces exerted by Earth and the Sun on the Moon.

The gravitational force is given by :

F=\dfrac{Gm_em_m}{r^2}

It means F\propto \dfrac{1}{r^2}

So,

\dfrac{F_1}{F_2}=\dfrac{r_2}{r_1}

r₁ = 3.9x10⁸ km

r₂= 1.5x10⁸ km

So,

\dfrac{F_1}{F_2}=\dfrac{1.5\times 10^8}{3.9\times 10^8}\\\\\dfrac{F_1}{F_2}=\dfrac{5}{13}

Hence, the ratio of the gravitational forces exerted by Earth and the Sun on the Moon is 5:13.

3 0
3 years ago
When the force acting on the body equal to acceleration?
topjm [15]

Answer:

Acceleration and velocity Newton's second law says that when a constant force acts on a massive body, it causes it to accelerate, i.e., to change its velocity, at a constant rate. In the simplest case, a force applied to an object at rest causes it to accelerate in the direction of the force.

5 0
2 years ago
A battery-operated car utilizes a 12.0 V system. Initially, the car is at rest at the base of a 195 m high hill. Some time later
ale4655 [162]

Answer:

140265.8 C = 1.403 × 10⁵ C

Explanation:

The battery's electric potential energy is used to account for the kinetic and potential work done in moving the car up this hill.

Potential work required to move the 757 kg car up a vertical height of 195 m = mgh

P.E = 757 × 9.8 × 195 = 1446627 J

Kinetic work done = (1/2)(m)(v²)

K.E = (1/2)(757)(25²) = 236562.5 J

Total work done in moving the car up that height = 1446627 + 236562.5 = 1683189.5 J

And this would be equal to the potential of the battery.

For the battery, potential difference = (electric potential energy)/(charges moved)

ΔV = ΔU/q

q = ΔU/ΔV

ΔU = 1683189.5 J

ΔV = 12.0 V

q = 1683189.5/12 = 140265.8 C

7 0
3 years ago
Read 2 more answers
The si unit of measurement is a way for scientists to?
Vera_Pavlovna [14]
Have a universal record base. Everyone is able to understand the data compiled since the same measurement systems are being used around the world. This is just to simplify all of the information.
7 0
3 years ago
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