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Softa [21]
3 years ago
5

a ball is thrown horizontally from a 20 m high building with a speed of 5.0 m/s. How far from the base of the building does the

ball hit the ground?

Physics
1 answer:
kipiarov [429]3 years ago
7 0
Given:
Dy= 20 m
Vi = 5.0 m/s horizontally
A=9.81 m/s^2

Find:
Horizontal displacement

Solution:
D=ViT+(1/2)AT^2
Dy=(1/2)AT^2
T^2=Dy/(1/2)A
T=sqrt(Dy/(1/2)A)
T=sqrt(20/4.905)
T=2.0s

Dx=ViT
Dx=(5.0)(2.0)
Dx=10. meters
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Humans have 46 chromosomes. The cells created by meiosis have
alukav5142 [94]
Meiosis creates the gamete cells, because these cells are used in reproduction they only have 26 chromosomes two of these cells join together to make a full 46 chromosomes.

Hope this helps! :)
7 0
3 years ago
This question relates to the practicality of searching for intelligent life in other solar systems by detecting their radio broa
Montano1993 [528]

Answer:

2.77287\times 10^{15}\ m

Explanation:

P = Power = 50 kW

n = Number of photons per second

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

\nu = Frequency = 781 kHz

r = Distance at which the photon intensity is i = 1 photon/m²

Power is given by

P=nh\nu\\\Rightarrow n=\dfrac{P}{h\nu}\\\Rightarrow n=\dfrac{50000}{6.626\times 10^{-34}\times 781000}\\\Rightarrow n=9.66201\times 10^{31}\ photons/s

Photon intensity is given by

i=\dfrac{n}{4\pi r^2}\\\Rightarrow 1=\dfrac{9.66201\times 10^{31}}{4\pi r^2}\\\Rightarrow r=\sqrt{\dfrac{9.66201\times 10^{31}}{4\pi}}\\\Rightarrow r=2.77287\times 10^{15}\ m

The distance is 2.77287\times 10^{15}\ m

3 0
3 years ago
During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away from another trolley, Y, of mass (2.561 ±
Alla [95]

Answer:

P = 1 (14,045 ± 0.03 )  k gm/s

Explanation:

In this exercise we are asked about the uncertainty of the momentum of the two carriages

            Δ (Pₓ / Py) =?

 Let's start by finding the momentum of each vehicle

car X

        Pₓ = m vₓ

        Pₓ = 2.34 2.5

        Pₓ = 5.85 kg m

car Y

        Py = 2,561 3.2

        Py = 8,195 kgm

How do we calculate the absolute uncertainty at the two moments?

          ΔPₓ = m Δv + v Δm

          ΔPₓ = 2.34 0.01 + 2.561 0.01

          ΔPₓ = 0.05 kg m

         ΔP_{y} = m Δv + v Δm

         ΔP_{y} = 2,561 0.01+ 3.2 0.001

         ΔP_{y} = 0.03 kg m

now we have the uncertainty of each moment

          P = Pₓ / P_{y}

          ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²

          ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²

          ΔP = 0.006 + 0.0026

          ΔP = 0.009 kg m

The result is

           P = 14,045 ± 0.039 = (14,045 ± 0.03 )  k gm/s

7 0
3 years ago
Please help i'm going to throw up from stress
Eddi Din [679]

Answer:

Explanation:

First of all, I used the specific heat of water as 4182 J/(kgC) and the specific heat of ethyl alcohol (EtOH) as 2440 J/(kgC); that means that we need the masses in kg, not g.

120.g = .1200 kg of ethyl alcohol. Now for the formula:

t_f=\frac{(m_{H2O}*spheat_{H2O}*temp_{H2O})+(m_{EtOH}*spheat_{EtOH}*temp_{EtOH})}{(m_{H2O}*spheat_{H2O})+(m_{EtOH}*spheat_{EtOH})} where spheat is specific heat.

Filling that horrifying-looking formula in with some values:

16.0=\frac{(x*4182*20.0)+(.1200*2440*10.0)}{(x*4182)+(.1200*2440)} and

16.0=\frac{83640x+2928}{4182x+292.8} and

16(4182x + 292.8) = 83640x + 2928 and

66912x + 4684.8 = 83640x + 2928 and

1756.8 = 16728x so

x = .105 kg and the amount of water added is 105 g

4 0
3 years ago
Why might surface mining be less risky for miners than underground mining?
aev [14]
Surface miners work aboveground. (Apex) ^-^

6 0
2 years ago
Read 2 more answers
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