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2 years ago

Please help me out (will give brainliest)

Physics

2 answers:

alisha [4.7K]2 years ago

3 0

Answer:

its black

Explanation:

because thats the color it is

almond37 [142]2 years ago

3 0

Answer:

Constants: 1 atm = 760 mm Hg = 760 torr = 1.01325 bar = 101.325 kPa R = 8.314 J·mol-1K-1= 0.08206 L·atm·mol-1K-11 Calorie = 1 kcal = 4.184 kJ Data:Properties of liquid water: b.p. (at 1 atm) = 100.00°CCH2O(l)= 4.184 J·g-1·K-1ΔHovap= 40.7 kJ·mol-1dH2O(l)(at 25°C) = 1.00 g·mL-1 KfH2O= 1.86 °C·kg·mol-1KbH2O= 0.52 °C·kg·mol-1Po(298 K)= 23.8 mm Hg Properties of ice: m.p. (at 1 atm) = 0.00°CCH2O(s) = 2.06 J·g-1·K-1 dH2O(s)= 0.917 g·mL-1 ΔHofus= 6.02 kJ·mol-1Formulae: ΔGo= ΔHo-TΔSo ΔG = ΔGo+ RT lnQΔGo= -RT lnKeqC = k P (or, S = k P) P = χPoΔT = KmπV = nRT [A]t= -k t + [A]oln[A]t= -k t + ln[A]o1/[A]t= k t + 1/[A]o

I am not sure If that helps im sorry I'm only in middle school :P.

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The centers of two 15.0 kg spheres are separated by 3.00 m. The magnitude of the gravitational force between the two spheres is

kompoz [17]

we have to use newtons law of gravitation which is
F=GMm/r^2
G=6.67 x 10^-11N kg^2/m^2
M=(15kg)
m=15 kg
r=(3.0m)^2
putting values we have
=(6.67 x 10^-11N kg^2/m^2)(15kg)(15kg)/(3.0m)^2 
=1.67 x 10^-9N

7 0

3 years ago

How do I solve this

Svetradugi [14.3K]

Answer:

W = 8.01 × 10^(-17) [J]

Explanation:

To solve this problem we need to know the electron is a subatomic particle with a negative elementary electrical charge (-1,602 × 10-19 C), The expression to calculate the work is given by:

W = q*V

where:

q = charge = 1,602 × 10^(-19) [C]

V = voltage = 500 [V]

W = work [J]

W = 1,602 × 10^(-19) * 500

W = 8.01 × 10^(-17) [J]

8 0

3 years ago

It took a crew 9 h 36 min to row 8 km upstream and back again. If the rate of flow of the stream was 2 km/h, what was the rowing

babunello [35]

Answer:

3 km/h

Explanation:

Let's call the rowing speed in still water x, in km/h.

Rowing speed in upstream is: x - 2 km/h

Rowing speed in downstream is: x + 2 km/h

It took a crew 9 h 36 min ( = 9 3/5 = 48/5) to row 8 km upstream and back again. Therefore:

8/(x - 2) + 8/(x + 2) = 48/5 (notice that: time = distance/speed)

Multiplying by x² - 2², which is equivalent to (x-2)*(x+2)

8*(x+2) + 8*(x-2) = (48/5)*(x² - 4)

Dividing by 8

(x+2) + (x-2) = (6/5)*(x² - 4)

2*x = (6/5)*x² - 24/5

0 = (6/5)*x² - 2*x - 24/5

Using quadratic formula

$x = \frac{2 \pm \sqrt{(-2)^2 - 4(6/5)(-24/5)}}{2(6/5)}$

$x = \frac{2 \pm 5.2}{2.4}$

$x_1 = \frac{2 + 5.2}{2.4}$

$x_1 = 3$

$x_2 = \frac{2 - 5.2}{2.4}$

$x_2 = -1\; 1/3$

A negative result has no sense, therefore the rowing speed in still water was 3 km/h

7 0

3 years ago

Is physical science part of physics?

telo118 [61]

Yes, it is. Physical science, the systematic study of the inorganic world

6 0

3 years ago

Read 2 more answers

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$