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3 years ago

How much mass energy could be obtained from the complete conversion of a 235 g hamburger?

Physics

1 answer:

Radda [10]3 years ago

8 0

E = mc²
E = 0.235 kg · (3×10⁸ m/s)² = 0.235 · 9×10¹⁶ kg·m/s²
E = 2.115×10¹⁶ J
The answer is d) 2.12×10¹⁶ J

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You spy on an anchor on the bottom of a lake. What is the direction and amount of force the water exerts on it: a) zero

vaieri [72.5K]

Answer:

c) up, less than the weight of the anchor

Explanation:

An object floats when the weight of the water the object displaces is equal to the weight of the object. In this case the weight of the anchor and the weight of the water it displaces is not the same so it sinks. Buoyancy is the force that acts on body when it is in water.

When buoyancy is less than the force that the object is generating due to gravity the object sinks. Buoyancy always acts up.

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3 years ago

A quantity of N2 occupies a volume of 1.4 L at 290 K and 1.0 atm. The gas expands to a volume of 3.3 L as the result of a change

lions [1.4K]

Answer:

$\rho = 0.50 g/L$

Explanation:

As we know that

PV = nRT

here we have

$P = 1.0 atm$

$P = 1.013 \times 10^5 Pa$

so we have

$V = 1.4 \times 10^{-3} m^3$

T = 290 K

now we have

$(1.013 \times 10^5)(1.4 \times 10^{-3}) = n(8.31)(290)$

$n = 0.06$

now the mass of gas is given as

$m = n M$

$m = (0.06)(28)$

$m = 1.65 g$

now density of gas when its volume is increased to 3.3 L

so we will have

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

$\rho = \frac{1.65 g}{3.3 L}$

$\rho = 0.50 g/L$

5 0

3 years ago

How can a small human retina detect objects larger than itself?

muminat

The correct answer would be A) the lenses of the eye change the size of images

5 0

3 years ago

In this problem you will consider the balance of thermal energy radiated and absorbed by a person.Assume that the person is wear

Viefleur [7K]

Answer: $P=14.6 W$

Explanation:

According to the Stefan-Boltzmann law for real radiating bodies:

$P=\sigma A \epsilon T^{4}$ (1)

Where:

$P$ is the energy radiated (in Watts)

$\sigma=5.67(10)^{-8}\frac{W}{m^{2} K^{4}}$ is the Stefan-Boltzmann's constant.

$A$ is the Surface area of the body

$T=30\°C + 273.15= 303.15 K$ is the effective temperature of the body (its surface absolute temperature) in Kelvin

$\epsilon=0.6$ is the body's emissivity

On the other hand, we are told the human body is roughly approximated to a cylinder of length $L=2.0m$ and circumference $C=0.8m$ .

The circumference of a circle is: $C=0.8m=2 \pi r$ where $r$ is the radius. Hence $r=\frac{0.8m}{2 \pi}=0.1273 m$ .

Now we have to input this value for $r$ in the Area of a cylinder formula:

$A=\pi r^{2}L$

$A=\pi (0.1273 m)^{2}(2 m)$

$A=0.0509 m^{2}$ (2)

Substituting (2) in (1):

$P=(5.67(10)^{-8}\frac{W}{m^{2} K^{4}}) (0.0509 m^{2}) (0.6) (303.15 K)^{4}$ (3)

Finally:

$P=14.62 W \approx 14.6 W$

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3 years ago

A softball of mass 0.220 kg that is moving with a speed of 5.5 m/s (in the positive direction) collides head-on and elastically

Elanso [62]

Answer:

The velocity and mass of the target ball are 1.6 m/s and 1.29 kg.

Explanation:

Given that,

Mass of softball = 0.220 kg

Speed = 5.5 m/s

(a). We need to calculate the velocity of the target ball

Using conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$0.220\times5.5+m_{2}\times0=0.220\times(-3.9)+m_{2}v_{2}$

$1.21=-0.858+m_{2}v_{2}$

$m_{2}v_{2}=2.068$ ....(I)

The velocity approach is equal to the separation of velocity

$u_{1}-u_{2}=v_{2}-v_{1}$

$5.5-0=v_{2}-(-3.9)$

$v_{2}=1.6\ m/s$

(b). We need to calculate the mass of the target ball

Now, Put the value of v₂ in equation (I)

$m_{2}\times1.6=2.068$

$m_{2}=\dfrac{2.068}{1.6}$

$m_{2}=1.29\ kg$

Hence, The velocity and mass of the target ball are 1.6 m/s and 1.29 kg.

3 0

3 years ago