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Kaylis [27]
3 years ago
9

3. Calculate the momentum of a 1200kg car with a velocity of 25m/s.

Physics
1 answer:
antiseptic1488 [7]3 years ago
4 0

momm=massxvelocity

momm=1200x2.5=120x25=600x5=3000kgm/s

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Un cuerpo se desplaza desde el punto 5,7 m al 7,5 en linea recta. cual es el valor del desplazamiento de la coordenada x?
adoni [48]

Answer:

Δx = 2m

Explanation:

From the given information:

De la información dada:

El desplazamiento ocurrió en un vector que se mueve desde una posición inicial particular hasta el final.

Entonces, supongamos que el desplazamiento es D;

Luego:

D = (7,5 ) - (5,7)

D = (2, -2)

Por lo tanto, el valor del desplazamiento de la coordenada x; Δx = 2m

8 0
3 years ago
You are at a furniture store and notice that a Grandfather clock has its time regulated by a physical pendulum that consists of
artcher [175]

Answer:

The distance is 1.026 m.

Explanation:

mass of rod, M = 1.23 kg

Length, L = 1.25 m

mass, m = 10 kg

Time period, T = 2 s

Let the distance is d.

The formula of the time period is given by

T = 2\pi\sqrt\frac{\frac{1}{3}ML^2+md^2}{(M +m)g}\\\\2\times 2 = 4\pi^2\times \frac{\frac{1}{3}\times1.23\times1.25\times 1.25+ 10d^2}{(1.23 + 10)\times9.8}\\\\11.16  = 0.64 + 10d^2\\\\d= 1.026 m

3 0
3 years ago
Taking the density of air to be 1.29 kg/m3, what is the magnitude of the angular momentum (in kg · m2/s) of a cubic meter of air
Svet_ta [14]

The magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

The rotating equivalent of linear momentum in physics is called angular momentum. Because it is a conserved quantity—the total angular momentum of a closed system stays constant—it is significant in physics. Both the direction and the amplitude of angular momentum are preserved.

Given the density of air to be 1.29 kg/m3 and a wind speed of 73.0 mi/h

We have to find the magnitude of the angular momentum

Let,

ρ = Density of air = 1.29 kg/m^3

v = Speed of wind = 73.0 mi/h = 0.032 km/s

M = angular momentum of air

Let the volume of air be 1 m^3

Mass = Volume x ρ = 1 x 1.29 = 1.29 kg

Momentum = M = mass x velocity

Momentum = 1.29 x 0.0032

Momentum = 4.128 x 10^(-3) kg·m^2/s

Hence the magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

Learn more about angular momentum here:

brainly.com/question/7538238

#SPJ10

8 0
2 years ago
How does a rubber rod become negatively charged through friction?
stira [4]
I think it is c I'm only in 7th grade but I'm pretty sure that the answer is c
5 0
4 years ago
Read 2 more answers
The motion of spinning a hula hoop around one's hips can be modeled as a hoop rotating around an axis not through the center, bu
Citrus2011 [14]

Answer and Explanation:

Based on the given information, the formula and the computation is given below:

a. The rotational inertia of the hoop is shown below:

I_H = I_R + Mh^2

= MR^2 + Mh^2

= 0.73 \times (0.60^2 + 0.38^2)

= 0.73 × (0.36 +  0.1444)

= 0.368 kg\ mg^2

b. Now the rotational kinetic energy is

= Half \times Inertia \times omega^2

= 0.5 \times 0.368 \times 14.1^2

= 36.58 J

We simply applied the above formula for rotational inertia and rotational kinetic energy in order to reach with the correct answer

6 0
3 years ago
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