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2 years ago

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The Lamborghini Huracan has an initial acceleration of 0.80g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode al

ong, what would the car's acceleration be?

1 answer:

irina [24]2 years ago

6 0

Answer:

a = 15.1 g

Explanation:

The relation between mass and acceleration is given by :

$m\propto \dfrac{1}{a}$

If a₁ = 0.80g, m₁ = 1510 kg, m₂ = 80 kg, we need to find a₂

So,

$\dfrac{m_1}{m_2}=\dfrac{a_2}{a_1}\\\\a_2=\dfrac{a_1m_1}{m_2}\\\\a_2=\dfrac{0.8g\times 1510}{80}\\\\a_2=15.1g$

So, the car's acceleration would be 15.1 g.

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Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the

astraxan [27]

Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

Ratio of the current = I₁ : I₂ = 3 : 1

Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

$B = \frac{\mu_oI}{2\pi R}$

Where $\mu_o$ is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

$B_1 = \frac{\mu_oI_1}{2\pi R}$

$B_2 = \frac{\mu_oI_2}{2\pi R}$

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T = $\frac{\mu_oI_1}{2\pi R}$ - $\frac{\mu_oI_2}{2\pi R}$

or

4.0×10⁻⁶T = $\frac{\mu_o}{2\pi R}\times (I_1-I_2 )$

also

$\frac{I_1}{I_2} = \frac{3}{1}$

⇒ $I_1 = 3\times I_2$

substituting the values in the above equation we get

4.0×10⁻⁶T = $\frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)$

⇒ $I_2 = 1A$

also

$I_1 = 3\times I_2$

⇒ $I_1 = 3\times 1A$

⇒ $I_1 = 3A$

Hence, the larger of the two currents is 3A

3 0

2 years ago

A metal, M, forms an oxide having the formula M2O3 containing 52.92% metal by mass. Determine the atomic weight in g/mole of the

irina [24]

Answer:

The atomic weight in g/mole of the metal (molar mass) is 8.87.

Explanation:

To begin, it is possible to assume that, as a sample, it has 100 g of the compound. This means that:

52.92% metal: 52.92 g M
47.80% oxygen: 47.80 g O

Using the molar mass of oxygen, which is 16 g / mol, it is possible to calculate the amount of moles of oxygen present in the sample using the rule of three:

$moles of oxygen=\frac{47.8g*1mol}{16g}$

moles of oxygen=2.9875

The chemical formula of metal oxide tells you that:

2 M⁺³ + 3 O²⁻ ⇒ M₂O₃

In the previous equation you can see that you need 3 oxygen anions to react with two metal cations. Then:

$2.9875 moles of oxygen*\frac{2 moles of metal M}{1 mol of oxygen} = 5.975 moles of metal M$

You have 52.92 g of metal in the sample, then the molar mass of the metal is:

$molar mass=\frac{52.92 g}{5.975 mol}$

molar mass≅ 8.87 g/mol

<u><em> The atomic weight in g/mole of the metal (molar mass) is 8.87.</em></u>

The closest match to this value is Beryllium (Be), which has an atomic mass of 9.0122 g / mol.

3 0

2 years ago

A 25kg child sits on one end of a 2m see saw. How far from the pivot point should a rock of 50kg be placed on the other side of

ivann1987 [24]

Answer:

a rock of 50kg should be placed =drock=0.5m from the pivot point of see saw

Explanation:

τchild=τrock

Use the equation for torque in this equation.

(F)child(d)child)=(F)rock(d)rock)

The force of each object will be equal to the force of gravity.

(m)childg(d)child)=(m)rockg(d)rock)

Gravity can be canceled from each side of the equation. for simplicity.

(m)child(d)child)=(m)rock(d)rock)

Now we can use the mass of the rock and the mass of the child. The total length of the seesaw is two meters, and the child sits at one end. The child's distance from the center of the seesaw will be one meter.

(25kg)(1m)=(50kg)drock

Solve for the distance between the rock and the center of the seesaw.

drock=25kg⋅m50kg

drock=0.5m

6 0

2 years ago

Lidia plans an experimental investigation to see how the thickness of a lens affects the point where a beam of light is focused.

alexandr1967 [171]

Answer:

The type of light and the material of lenz.

Explanation:

1) As the investigation is based on how the thickness of a lens effect the other variable. Thickness of the lenz is independent variable. So Lidia has to experiment with the different thicknesses in order to find the effect on dependent variable.

2) As the investigation is based to find the point where the beam of light is focused. It's a dependent variable and Lidia has no control over it. So the only thing she can do is to measure and observe how it respond to the changes in independent variable.

3) For conclusion, she has to make sure that the other variables are not effecting the output or results that is the beam point where the light is focused. So she must have to kept constant the type of light and material of lenz otherwise she won't be able to discriminate the effect of thickness of lenz from other causes.

8 0

2 years ago

Compared to the resistivity of a 0.4-meter length of 1-millimeter-diameter copper wire at 0 degrees Celsius, the resistivity of

Mazyrski [523]

Answer:

Resistivity of both wires are same

Explanation:

Length of one wire, $l_1=0.4 m$

Diameter, $d_1=1mm$

Radius, $r_1=\frac{d_1}{2}=\frac{1}{2}mm=0.5\times 10^{-3} m$

$1mm=10^{-3} m$

$l_2=0.8 m$

$d_2=1mm$

$r_2=0.5\times 10^{-3} m$

Temperature in each case is same.

Area of each wire, $A_1=A_2=A=\pi r^2=\pi (0.5\times 10^{-3})^2m^2$

Resistivity is the property of material due to which it offers resistance to the flow of current.

Resistivity of material depends upon the temperature and material by which it is made.

It does not depends upon the length of object.

Therefore, the resistivity of both wires of different length are same.

3 0

2 years ago

Read 2 more answers