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3 years ago

.. As you increase the

Physics

1 answer:

Llana [10]3 years ago

3 0

Answer:

air pressure increases and temperature decreases

Explanation:

Hope this helps

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A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points

mr Goodwill [35]

Answer:

Approximately $11.0\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81 \; \rm N \cdot kg^{-1}$ , and that the tabletop is level.)

Explanation:

Weight of the book:

$W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N$ .

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, $F(\text{normal force}) \approx 10.202\; \rm N$ .

As a side note, the $F_N$ and $W$ on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction $F(\text{kinetic friction})$ on it will be equal to

$\mu_{\rm k}$ , the coefficient of kinetic friction, times
$F(\text{normal force})$ , the normal force that's acting on it.

That is:

$\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}$ .

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that $15.0\; \rm N$ applied force. The net force on the book shall be:

$\begin{aligned}& F(\text{net force}) \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}$ .

Apply Newton's Second Law to find the acceleration of this book:

$\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}$ .

6 0

2 years ago

The mass of Jupiter is 1.9 × 1027 and that of the sun is 1.99 × 1030. The

n200080 [17]

Answer:

F = 4.147 × 10^23

v = 1.31 × 10^4

Explanation:

Given the following :

mass of Jupiter (m1) = 1.9 × 10^27

Mass of sun (m2) = 1.99 × 10^30

Distance between sun and jupiter (r) = 7.8 × 10^11m

Gravitational force (F) :

(Gm1m2) / r^2

Where ; G = 6.673×10^-11 ( Gravitational constant)

F = [(6.673×10^-11) × (1.9 × 10^27) × (1.99 × 10^30)] / (7.8 × 10^11)^2

F = [25.231 × 10^(-11+27+30)] / (60.84 × 10^22)

F = (25.231 × 10^46) / (60.84 × 10^22)

F = 3.235 × 10^(46 - 22)

F = 0.4147 × 10^24

F = 4.147 × 10^23

Speed of Jupiter (v) :

v = √(Fr) / m1

v = √[(4.147 × 10^23) × (7.8 × 10^11) / (1.9 × 10^27)

v = √32.3466 × 10^(23+11) / 1.9 × 10^27

v = √32.3466× 10^34 / 1.9 × 10^27

v = √17. 023 × 10^34-27

v = √17.023 × 10^7

v = 13047.221

v = 1.31 × 10^4

4 0

3 years ago

An insect 5.00 mm tall is placed 20.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ

jenyasd209 [6]

Answer:

a) i = -9.63 cm , h ’= .0.24075 cm erect

b) i = 259.74 cm ,

Explanation:

For this exercise let's start by finding the focal length of the lens

1 / f = (n-1) (1 / R₁ - 1 / R₂)

1 / f = (1.70 -1)) 1 / ∞ - 1/13)

1 / f = 0.0538

f = - 18.57 cm

Now we can use the constructor equation

1 / f = 1 / o + 1 / i

1 / i = 1 / f - 1 / o

1 / i = -1 / 18.57 -1/20

1 / i = -0.1038 cm

I = -9.63 cm

For the height of the

image let's use magnification

m = h '/ h = - i / o

h ’= -h i / o

h ’= - 0.5 (-9.63) / 20

h ’= .0.24075 cm

b) we invert the lens

The focal length is

1 / f = (1.70 -1) (1/13 - 1 / int)

1 / f = 0.0538

f = 18.57 cm

1 / i = 1 / f -1 / o

1 / I = 1 / 18.57 - 1/20

1 / I = 3.85 10-3

i = 259.74 cm

h ’= - 0.5 259.74 / 20

h ’= 6.4935 cm

7 0

2 years ago

What kind of model is shown below?

Rudiy27

D. a foot model

btw this is a joke right cuz there ain’t no picture lol

7 0

3 years ago

1 point

Svetllana [295]

Answer:

Beta 17,000K, bc warmer is more blue

5 0

2 years ago