2 years ago

Which term describes the image of an object located between a concave mirror and its focal point?

Physics

1 answer:

labwork [276]2 years ago

6 0

Answer: I think it is b.

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The energy generated by the sun travels to Earth as electromagnetic waves of varying lengths. Which statement describes an elect

Paha777 [63]

Answer: 4(45)

Explanation:

It’s not equal

3 0

2 years ago

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The density of nuclear matter is about 1018 kg/m3. Given that 1 mL is equal in volume to 1 cm3, what is the density of nuclear m

Sonbull [250]

Answer:

density is $10^{6}$ Mg/µL

Explanation:

given data

density of nuclear = $10^{18}$ kg/m³

1 ml = 1 cm³

to find out

density of nuclear matter in Mg/µL

solution

we know here

1 Mg = 1000 kg

1 m³ is equal to $10^{6}$ cm³

and here 1 cm³ is equal to 1 mL

so we can say 1 mL is equal to 10³ µL

so by these we can convert density

density = $10^{18}$ kg/m³

density = $10^{18}$ kg/m³ × $\frac{10^{-3} }{10^{6} }$ Mg/µL

density = $10^{6}$ Mg/µL

8 0

3 years ago

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Calculated the measurement uncertainty for Kinetic Energy when :mass = 1.3[kg] +/- 0.4[kg]velocity= 5.2 [m/s] +/- 0.2 [m/s]KE= 1

andriy [413]

Answer:

$\rm KE\pm \Delta KE = 17.6\pm 6.8\ J.$

Explanation:

<u>Given:</u>

Mass, $\rm m\pm\Delta m = 1.3\pm 0.4\ kg.$

Velocity, $\rm v\pm \Delta v = 5.2\pm 0.2\ m/s.$

where,

$\rm \Delta m,\ \Delta v$ are the uncertainties in mass and velocity respectively.

The kinetic energy is given by

$\rm KE = \dfrac 12 mv^2 = \dfrac 12 \times 1.3\times 5.2^2=17.576\approx 17.6\ J.$

The uncertainty in kinetic energy is given as:

$\rm \dfrac{\Delta KE}{KE}=\dfrac{\Delta m}{m}+\dfrac{2\Delta v}{v}\\\dfrac{\Delta KE}{17.6}=\dfrac{0.4}{1.3}+\dfrac{2\times 0.2}{5.2}\\\dfrac{\Delta KE}{17.6}=0.384\\\Rightarrow \Delta KE = 17.6\times 0.384 = 6.7854\ J\approx6.8\ J\\\\Thus,\\\\KE\pm \Delta KE = 17.6\pm 6.8\ J.$