Because of churning rock and ash, the largest volcanic eruptions are sometimes surrounded by

An exoplanet with one half of Earth's mass and 50% of Earth's radius is discovered.

Georgia [21]

Answer:

The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet

Explanation:

The given parameters are;

The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M

The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R

The weight of the cadet on Earth = 800 N

$The \ weight, W =G\dfrac{M \times m}{R^{2}} = 800 \ N$

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

$W_1 =G\dfrac{\dfrac{M}{2} \times m}{ \left ( \dfrac{R}{2} \right ) ^{2}} = G\dfrac{\dfrac{M}{2} \times m \times 4}{ R ^{2}} = 2 \times G \times \dfrac{M \times m}{R^{2}} = 2 \times 800 \, N = 1,600 \, N$

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.

7 0

3 years ago

A race car has a centripetal acceleration of 13.33 m/s^2 as it goes around a curve. if the curve is a circle with a radius 30 m

anzhelika [568]

Answer:

The speed of the car, v = 19.997 m/s

Explanation:

Given,

The centripetal acceleration of the car, a = 13.33 m/s²

The radius of the curve, r = 30 m

The centripetal force acting on the car is given by the formula

F = mv²/r

Where v²/r is the acceleration component of the force

a = v²/r

Substituting the values in the above equation

13.33 = v²/30

v² = 13.33 x 30

v² = 399.9

v = 19.997 m/s

Hence, the speed of the car, v = 19.997 m/s

3 0

3 years ago

A student rides her bike to school. Her school is 5 miles from home. She travels at an average rate of 15 miles per hour. How mu

Elden [556K]

Answer:

.3 repeating hours

Explanation:

3 0

2 years ago

Unpolarized light is incident upon two ideal polarizing filters that do not have their transmission axes aligned. If 19% of the

Zepler [3.9K]

Answer:

51.94°

Explanation:

$I_0$ = Unpolarized light

$I_2$ = Light after passing though second filter = $0.19I_0$

Polarized light passing through first filter

$I_1=\frac{I_0}{2}$

Polarized light passing through second filter

$I_2=\frac{I_0}{2}cos^2\theta\\\Rightarrow 0.19I_0=\frac{I_0}{2}cos^2\theta\\\Rightarrow cos^2\theta=\frac{0.19I_0}{\frac{I_0}{2}}\\\Rightarrow cos\theta=\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{0.19\times 2}\\\Rightarrow \theta=cos^{-1}\sqrt{0.38}\\\Rightarrow \theta=51.94^{\circ}$

The angle between the two filters is 51.94°

5 0

3 years ago

answers At the beginning of 2019, Robotics Inc. acquired a manufacturing facility for $12.3 million. $9.3 million of the purchas

pshichka [43]

Answer:

$613,077

Explanation:

Calculation for the depreciation on the building for 2021

First step is to calculate the depreciation expense for the year 2019 and 2020 using this formula

Depreciation expense = (Purchase price - Residual value) ÷ (Useful life)

Let plug in the formula

Depreciation expense= ($9,300,000 - $1,300,000) ÷ (20 years)

Depreciation expense= ($8,000,000) ÷ (20 years)

Depreciation expense= $400,000

Second step is to calculate the book value for the year 2021 using this formula

Book value = Purchase price - Depreciation expenses *2 years

Let plug in the formula

Book value= $9,300,000 - $400,000 × 2

Book value= $8,500,000

Last step is to calculate the depreciation for 2021 using this formula

2021 Depreciation=(Book value-Residual value)÷Useful life

2021 Depreciation= ($8,500,000 - $530,000) ÷ (15 years - 2 years)

2021 Depreciation=$7,970,000÷13 years

2021 Depreciation= $613,077

Therefore the depreciation on the building for 2021 will be $613,077

8 0

3 years ago