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Sophie [7]
3 years ago
5

HELP I NEED THIS ANSWERED AS FAST AS POSSIBLE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Physics
1 answer:
WITCHER [35]3 years ago
8 0

Displacement is d  


Vf² = Vi² + 2 g d  


(-20²) = (+10²) + 2 (-9.8) d  


-19.6 d = 300  


d = -15.3 m  


negative means lower


time is t  


d = Vi t + 1/2 g t²



-15.3 = 10 t + (-4.9) t²



4.9 t² - 10 t -15.3 = 0  


t = 3.06 s

Hope this helps -John

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un movil que parte del reposo alcanza una velocidad de 75 m/s en 13 segundos ¿cual su aceleracion y el espacio que recorrio en l
Dmitriy789 [7]

Answer:

Acceleration = 5.77 m/s²

Distance cover in 13 seconds = 487.56 meter

Explanation:

Given:

Final velocity of mobile device = 75 m/s

initial velocity of mobile device = 0 m/s

Time taken = 13 seconds

Find:

Acceleration

Distance cover in 13 seconds

Computation:

v = u + at

75 = 0 + (a)(13)

13a = 75

a = 5.77

Acceleration = 5.77 m/s²

s = ut + (1/2)(a)(t²)

s = (0)(t) + (1/2)(5.77)(13²)

Distance cover in 13 seconds = 487.56 meter

8 0
3 years ago
There is a naturally occurring vertical electric field near the Earth’s surface that points toward the ground. In fair weather c
trasher [3.6K]

Answer:

\frac{F}{W} = 9.37 \times 10^{-4}

Explanation:

Radius of the pollen is given as

r = 12.0 \mu m

Volume of the pollen is given as

V = \frac{4}{3}\pi r^3

V = \frac{4}{3}\pi (12\mu m)^3

V = 7.24 \times 10^{-15} m^3

mass of the pollen is given as

m = \rho V

m = 7.24 \times 10^{-12}

so weight of the pollen is given as

W = mg

W = (7.24 \times 10^{-12})(9.81)

W = 7.1 \times 10^{-11}

Now electric force on the pollen is given

F = qE

F = (-0.700\times 10^{-15})(95)

F = 6.65 \times 10^{-14} N

now ratio of electric force and weight is given as

\frac{F}{W} = \frac{6.65 \times 10^{-14}}{7.1 \times 10^{-11}}

\frac{F}{W} = 9.37 \times 10^{-4}

7 0
3 years ago
Bài 4: Một vật dao động điều hòa với phương trình x = 4cos(4πt - π/3) cm.
ra1l [238]

Answer:

g

Explanation:

6 0
3 years ago
Which sequence shows all the spectral colors of visible light arranged in an increasing order of their frequency? A. red, yellow
Ad libitum [116K]

Answer:

The answer is B. red, orange, yellow, green, blue, indigo, violet

Explanation:

Most textbooks have the acronym ROYGBV to express the order in which colors appear on the spectrum of light, indigo is included in your list, and that's not a problem, although it's not typical. This spectrum of light is the same order in which colors appear in rainbows.

8 0
3 years ago
Read 2 more answers
A 150 g copper bowl contains 210 g of water, both at 24.0°C. A very hot 430 g copper cylinder is dropped into the water, causing
Dahasolnce [82]

Answer:

A. 15969.22 cal

B. 1052,22 cal

C. 528,87 °C

Explanation:

To solve this kind of question, a proper method is to work from the data that you have towards the data that you need. Also, it is recommended to analyze related equations as they could give us clues on how to find the missing information or the information that the problem is asking us.

Let us start with Question A. It is important to remember that energy transfers with the environment are being neglected; this means that all the energy that the cylinder lose is picked up by the water and the copper bowl. To find the amount of energy transferred to the water, we first find the amount of energy necessary to raise the water’s temperature to 100°C and then we find the amount of energy necessary to evaporate the 17.1 g of water indicated by the question. This would be:

Q = m_water * CP_water *∆T =210g *1 cal/(g K) * (100°C-24°C) = 15960 cal

Q_evap = m_wat * L = 17,1 g * 539 cal/kg* (1 kg)/(1000 g) =9.2169 cal

Therefore, the total energy that was transferred to the water is the sum of these components, that would be Q_tot = 15960 cal + 9.2159 cal = 15969.22 cal.  Let´s also remember that a temperature difference in K is equal to a temperature difference in ° C

To solve Question B, we use the same method. We must find the amount of energy necessary to raise the temperature from its initial temperature to the one stated by the problem to be the equilibrium temperature of the system (100°C):

Q= m_copper *CP_copper *∆T = 150g * 0.0923 cal/(g K) * (100°C-24°C) = 1052,22 cal

If we add the components we just found in questions A and B, we can find the amount of energy than the Copper cylinder lost, this would be: Q_tot = 15969.22 cal + 1052.22 cal = 17021.44 cal.

The question C asks us to find the initial temperature of the cylinder and Q_tot will help us to find it.

We know that Q_tot is the energy lost by the cylinder and we also know that Q_tot = m_cylinder * CP_copper * ∆T. Therefore, what we need to do  is clear the last term of the equation and find the initial temperature.

Q_tot = m_cylinder *CP_copper *∆T → T_fin-T_initial = Q_tot/(m_cylinder*CP_copper ) = (-17021.44 cal)/(430g*0.0923 cal/(g K))

→ T_initial = 100°C + (-17021.44 cal)/(430g * 0.0923 cal/(g K)) = 528,87 °C

If we convert the 100°C to K before we do the calculation, the result would be the same one, You would only need to add 273,15 to the final result to check it out.  

Hope everything was clear. If you have any further question, I'll be happy to help :D

5 0
3 years ago
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