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3 years ago

15

A 1200kg car is moving at 17.3 m/s when a force is applied the direction of the car's motion. The car speeds up to 29.4 m/s. If

the force is applied for 58s what the magnitude of the force?

1 answer:

slavikrds [6]3 years ago

7 0

The car undergoes an average acceleration <em>a</em> of

<em>a</em> = (29.4 m/s - 17.3 m/s) / (58 s) ≈ 0.209 m/s²

so the force has a magnitude <em>F</em> of

<em>F</em> = (1200 kg) <em>a</em> ≈ 250 N

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Un cuerpo de 480 g de masa es atraído con una fuerza de 3.9 E-6 N por otro cuerpo de 196 g de masa. Calcula la distancia a la qu

Soloha48 [4]

Answer:

La distancia entre los dos cuerpos es aproximadamente 1.269 milímetros.

Explanation:

Asumamos que ambos cuerpos son partículas, la fuerza de atracción ( $F$ ), en newtons, entre ambos cuerpos se define mediante la Ley de Newton de la Gravitación Universal, cuya ecuación es:

$F = G \cdot \frac{m_{1}\cdot m_{2}}{r^{2}}$ (1)

Donde:

$G$ - Constante de la gravitación universal, en metros cúbicos por kilogramo-segundo cuadrado.

$m_{1}, m_{2}$ - Masas de los cuerpos, en kilogramos.

$r$ - Distancia entre los cuerpos, en metros.

Si sabemos que $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $m_{1} = 0.48\,kg$ , $m_{2} = 0.196\,kg$ y $F = 3.9\times 10^{-6}\,N$ , entonces la distancia entre los dos cuerpos es:

$r = \sqrt{\frac{G\cdot m_{1}\cdot m_{2}}{F} }$

$r \approx 1.269\times 10^{-3}\,m$

Es decir, la distancia entre los dos cuerpos es aproximadamente 1.269 milímetros.

5 0

3 years ago

A charge if 2 nC is placed 10cm to the right of a conducting sphere with a diameter of 2 cm. A charge of 5 nC is placed 10 nC to

SCORPION-xisa [38]

Answer:

The charge at the center of the conducting sphere is zero.

Explanation:

A principle of conductor materials is that the electric field inside a conductor in electrostatic state is always zero. The gauss law says that the flux of a electric field in a closed surface is proportional to the charge enclosed by the surface. Then if the Electric field inside of a conductor is zero, imperatively the charge anywhere inside the conductor is zero too, so the charge at the center of the sphere is zero.

3 0

4 years ago

A U-tube open at both ends is partially filled withwater. Oil

ArbitrLikvidat [17]

Answer:

a) h_w = 0.02139 m , b) v₁ = 9.74 m / s

Explanation:

For this exercise we use the pascal principle that states that the pressure at one point is the same regardless of body shape.

At the initial moment (before emptying the oil), we fix the point on the surface of the liquid, in this case the left and right sides are in balance.

P₁ = P₂ = P₀

Now we add the 5 cm (h₂ = 0.05 m) of oil, in this case the weight of the oil creates an extra pressure that pushes the water, let's look for how much the water moved (h_w). The weight of oil added is equal to the weight of displaced water

W_w = W_oil

m_w g= m_oil g

Density is defined

ρ = m / V

we replace

ρ_w V_w = ρ_oil W_oil

V = A h

ρ_w A h_w = ρ_oil A h_oil

h_w = h_oil ρ_oil / ρ_w

Now let's analyze the pressure at the initial reference height for both sides two had in U

Right side

We have the atmospheric pressure, with its decrease due to the lower height, plus the oil pressure above the reference level

h’= 0.05 cm - h_w

P = (P₀ - ρ_air g (0.05-h_w)) + ρ_oil g (0.05-h_w)

Left side

We have at the same point, the atmospheric pressure with its reduction due to the height change plus the water pressure

P = (P₀ - ρ_air h h_w) + ρ_w g h_w

As we have the same point we can equalize the pressure

(P₀ -ρ_air g (0.05-h_w)) +ρ_oil g (0.05-h_w) = (Po -ρ_air h h_w) +ρ_w g h_w

ρ_air g (h_w - (0.05-h_w)) = ρ_w g h_w - ρ_oil g (0.05-h_w)

- ρ_air g 0.05 = h_w g ( ρ_w + ρ _oil) - rho_oil g 0.05

h_w g ( ρ_w + ρ_oil) = g 0.05 ( ρ_air - ρ_oil)

calculate

h_w = 0.05 (ρ_ oil- ρ_air) / ( ρ_w + ρ_oil)

h_w = 0.05 (750 - 1.29) / (1000-750)

h_w = 0.02139 m

The amount that decreases the height on one side is equal to the amount that increases the other

b) cover the right side and blow the air on the left side, let's use Bernoulli's equation, where index 1 will be for the left side and index 2 for the right side

P₁ + 1/2 ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Since they indicate that both sides are at the same height y₁ = y₂, the right side is protected from the wind speed therefore v₂ = 0, let's write the equation

P₁ + ½ ρ_air v₁² = P₂

v₁² = (P₂-P₁) 2 / ρ_air

Let's analyze the pressure on each side of the tube, since the new equilibrium height is the height that was added of oil distributed between the two tubes, bone

h’= 2.5 cm = 0.025 m

P₁ = P₀ - ρ_w g h’

P₂ = P₀ - ρ_oil g h ’

P₂-P₁ = g h’ (rho_w - Rho_oil)

We replace

v₁² = 2g h’ (ρ_w –ρ_oil) / ρ_air

calculate

v₁² = 2 9.8 0.025 (1000 - 750) /1.29

v₁ = √ 94.96

v₁ = 9.74 m / s

3 0

4 years ago

Which of these attributes is generally attributed to Kepler

kotegsom [21]

Kepler's first "law":
The orbits of the planets are ellipses, with the sun at one focus.
This would include Mars.

3 0

4 years ago

What is kinematics?<br>explain!!~<br><br>thankyou ~

artcher [175]

Answer:

Kinematics is the branch of mechanics concerned with the motion of objects without reference to the forces which cause the motion.

-the features or properties of motion in an object.

Explanation:

Hope this helps <3

Have a great day!

7 0

2 years ago

Read 2 more answers