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3 years ago

A car accelerates from 25 km/hr in 30 seconds. what Is the acceleration?

2 answers:

7 0

Acceleration is units of distance per time squared, take speed divided by time:

(25 km/hr)/(30 sec) = 0.833 km/hr/sec

these units aren't very helpful, so convert to units more likely to represent what your measuring. probably meters per seconds squared:

(0.833 km/hr/sec)*(1 hr/ 3600sec)*(1000m/km) = 0.23 m/s^2

kaheart [24]3 years ago

6 0

25 km/hr = 6,9 m/s
acceleration = 6,9 / 30 = 0,23

acceleration = 0,23 m / s²

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Un vas plin cu lichid cântăreşte 175kg. Ceea ce reprezintă de 5 ori masa vasului gol. Ştiind că volumul interior al vasului este

Mariana [72]

a) Density of the liquid: $823.5kg/m^3$

b) Weight of the liquid: 1372 N

Explanation:

Translation of the text:

"A full tank with liquid weighs 175kg. Which is 5 times the mass of the empty vessel. Knowing that the inside volume of the vessel is 0.17kl, calculate:

a) the density of the liquid;

b) the weight of the liquid."

We know that the full tank with liquid has a total mass of M = 175 kg. We can write the total mass as

$M=m_L + m_V$ (1)

where

$m_L$ is the mass of the liquid

$m_V$ is the mass of the vessel

We also know that the total mass M is 5 times the mass of the empty vessel, so we have:

$M=5m_V\\m_V=\frac{M}{5}=\frac{175}{5}=35 kg$

which is the mass of the empty vessel.

Therefore, we can find the mass of the liquid only using (1):

$m_L=M-m_V=175-35=140 kg$

The density of the liquid is given by

$d=\frac{m}{V}$

where

m = 140 kg (mass of the liquid)

V = 0.170 kL = 170 L = $0.170 m^3$ (volume of the liquid, which is equal to the volume of the vessel)

So we get

$d=\frac{140}{0.170}=823.5kg/m^3$

The weight of a body is given by

$F=mg$

where

m is its mass

g is the acceleration due to gravity

For the liquid in this problem, we have

m = 140 kg (mass)

$g=9.8 m/s^2$ (acceleration due to gravity)

Therefore, its weight is

$F=(140)(9.8)=1372 N$

Learn more about density:

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6 0

3 years ago

Unpolarized light with an average intensity of 845 W/m2 moves along the x-axis when it enters a Polarizer A with a vertical tran

horsena [70]

Answer:

θ = 36.2º

Explanation:

When light passes through a polarizer it becomes polarized and if it then passes through a second polarizer, it must comply with Malus's law

I = I₀ cos² tea

The non-polarized light between the first polarized of this leaves half the intensity, with vertical polarization

I₁ = I₀ / 2

I₁ = 845/2

I₁ = 422.5 W / m²

In this case, the incident light in the second polarizer has an intensity of I₁ = 422.5 W / m² and the light that passes through the polarizer has a value of

I = 275 W / m ²

Cos² θ = I / I₁

Cos θ = √ I / I₁

Cos θ = √ (275 / 422.5)

Cos θ = 0.80678

θ = cos⁻¹ 0.80678

θ = 36.2º

This is the angle between the two polarizers

8 0

3 years ago

On August 10, 1972, a large meteorite skipped across the atmosphere above the western United States and western Canada, much lik

Anon25 [30]

a) $4.62\cdot 10^{14} J$

b) 0.110 megatons

c) 8.46 bombs

Explanation:

The energy lost by the meteorite is equal to the difference between its final kinetic energy and its initial kinetic energy:

$\Delta K=K_f-K_i$

Which can be rewritten as:

$\Delta K=\frac{1}{2}mv^2-\frac{1}{2}mu^2$

where:

$m=3.2\cdot 10^6 kg$ is the mass of the meteorite

$v=0$ is the final speed of the meteorite

$u=17 km/s = 17,000 m/s$ is the initial speed of the meteorite

Substituting the values into the equation, we found the loss in energy of the meteorite:

$\Delta K=0-\frac{1}{2}(3.2\cdot 10^6)(17000)^2=-4.62\cdot 10^{14} J$

So, the energy lost by the meteorite is $4.62\cdot 10^{14} J$

The energy equivalent to 1 megaton of TNT is

$E_{TNT}=4.2\cdot 10^{15} J$

Here the energy lost by the meteorite is

$E=4.62\cdot 10^{14} J$

Therefore, in order to write the energy lost by the meteorite as a multiple of the energy of 1 megaton of TNT, we have to divide the energy lost by the meteorite by the energy equivalent to 1 TNT; we find:

$\frac{E}{E_{TNT}}=\frac{4.62\cdot 10^{14}}{4.2\cdot 10^{15}}=0.110$