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NeX [460]
3 years ago
12

How much power is developed in a car that can pull a 5000 N at a constant speed of 20 m/s?

Physics
1 answer:
egoroff_w [7]3 years ago
5 0

Answer:

YAYA MAN

Explanation:

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How do i do this??????????
Sliva [168]
Power = I^2 x R
Energy = Power x Time
4 0
3 years ago
Una barra de aluminio que esta a 78 GRADOS CENTIGRADOS entra en contacto con una barra de cobre de la misma longitud y área que
stiks02 [169]

Answer:

Al llegar a su equilibrio térmico ambas barran tendrán una temperatura de 53 grados centígrados.

Explanation:

Dado que una barra de aluminio que está a 78 grados centígrados entra en contacto con una barra de cobre de la misma longitud y área que esta a 28 grados centígrados, y posteriormente se lleva acabo la transferencia de energía entre ambas barras llegando a su equilibrio térmico, para determinar la temperatura a la que ambas barras llegarán se debe realizar el siguiente cálculo:

(78 + 28) / 2 = X

106 / 2 = X

53 = X

Por lo tanto, al llegar a su equilibrio térmico ambas barran tendrán una temperatura de 53 grados centígrados.

8 0
3 years ago
A 1200 Kg car rounds a corner of radius r = 45m. If the coefficient of static friction between the ties and the road is us = 0.8
Vaselesa [24]

Answer:

The greatest speed of the car is 19.36m/s

Explanation:

The maximum speed the car will attain without skidding is given by:

F= uN = umg ...eq1

But F = mv^2/r

mv^2/r = umg

Dividing both sides by m, leaves you with:

V= Sqrt(ugr)

Where u = coefficient of static friction

g = acceleration due to gravity

r = raduis

Given:

U = 0.82

r=0.82

g= 9.8m/s

V = Sqrt(0.82 × 9.8 × 45)

V = Sqrt(374.85)

V = 19.36m/s

5 0
3 years ago
Why is density used to determine the identity of matter?
dedylja [7]
 a substance's density is the same at a certain pressure and temperature, and the density of one substance is usually different than another substance.
4 0
3 years ago
A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 691 nm. Determine the angle
expeople1 [14]

Explanation:

Given that,

Wavelength of the light, \lambda=691\ nm=691\times 10^{-9}\ m

(a) Slit width, a=3.8\times 10^{-4}\ m

The angle that locates the first dark fringe is given by :

sin\theta=\dfrac{\lambda}{a}

sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-4}}

\theta=0.104^{\circ}

(b) Slit width, a=3.8\times 10^{-6}\ m

The angle that locates the first dark fringe is given by :

sin\theta=\dfrac{\lambda}{a}

sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-6}}

\theta=10.47^{\circ}

Hence, this is the required solution.

7 0
3 years ago
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