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3 years ago

What is the height of a 50 kg person that has a potential energy of 4900J?

Physics

1 answer:

ASHA 777 [7]3 years ago

7 0

Answer:

10 m

Explanation:

PE = mgh

m = 50kg

g= 9.8 m/s^2

h=?

so,

4900= 50×9.8× h

50h = 4900/9.8

50h = 500

h = 10 m

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For atomic hydrogen, the Paschen series of lines occurs when nf = 3, whereas the Brackett series occurs when nf = 4 in the equat

zvonat [6]

Answer:

$(\lambda_{max} )_{brackett} < (\lambda_{min} )_{paschen}$

So the two wavelength range will over lap

Explanation:

The Rydberg equation is given by

$\frac{1}{\lambda} =\frac{2\pi mk^2e^4}{h^3c} t (\frac{1}{n^2_f}-\frac{1}{n^2_i} )$

m is the mass of electron

k = 1/4π∈₀

∈₀ = is the permitivity of free space

e is the charge of electron

h is the plank constant

c is the speed of light in vaccum

z is the atomic number = 1

$\frac{1}{\lambda} =R (\frac{1}{n^2_f}-\frac{1}{n^2_i} )$

where R is the Rydberg constant = 1.097373 × 10⁷m⁻¹

For Paschen series of H spectrum

$n_f = 3$

$n_i = 5,6,7 ...$

in Paschen series of H spectrum

The maximum wavelength occur for $n_i = 4$

$\frac{1}{\lambda_m_a_x } =(1.097373 \times 10^7)(\frac{1}{9} - \frac{1}{16} )\\\\\lambda_m_a_x=1874.6nm$

The minimum wavelength occur for $n_i$ = ∞

$\frac{1}{\lambda_m_i_n } =(1.097373 \times 10^7)(\frac{1}{9} - \frac{1}{_o_o} )\\\\\lambda_m_a_x=820.14nm$

The brackett series of H spectrum

The maximum wavelength occur for $n_i = 4$

$\frac{1}{\lambda_m_a_x } =(1.097373 \times 10^7)(\frac{1}{16} - \frac{1}{25} )\\\\\lambda_m_a_x=4050.05nm$

The minimum wavelength occur for $n_i$ = ∞

$\frac{1}{\lambda_m_i_n } =(1.097373 \times 10^7)(\frac{1}{16} - \frac{1}{_o_o} )\\\\\lambda_m_a_x=1458.03nm$

$(\lambda_{max} )_{brackett} < (\lambda_{min} )_{paschen}$

So the two wavelength range will over lap

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3 years ago

Q2.

zzz [600]

Answer:

800W

Explanation:

Given the following data;

Voltage = 230 V

Model = VSO Ha

Power = 800 W

Power is calculated by the multiplication of voltage and current flowing through an electric circuit.

Mathematically, power is given by the formula;

Power = current * voltage

The S.I unit for power is Watts and it is a measure of the energy consumption of an electronic device.

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3 years ago

A metal ring 4.00 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpend

liq [111]

Answer:

a) 0.21N/C

b) counterclockwise

Explanation:

a) to find the magnitude of the electric field you can use the following formula:

$\int Eds=-\frac{\Delta \Phi_B}{\Delta t}=-\frac{\Delta AB}{\Delta t}$

A: area of the ring = pi*r^2

E: electric field

Ф_B: magnetic flux

In the line integral you can assume E as constant. Furthermore, you calculate the change in the magnetic flux by taking into account that the time interval is 1.12/0.21=5.33s. By replacing in the formula you obtain:

$\frac{\Delta \Phi_B}{\Delta t}=\frac{A(B_f-B_i)}{5.33s}=\frac{\pi(0.04m)^2(1.12T)}{5.33}=1.056*10^{-3}W/s$

$E\int ds=E(2\pi r)=1.056*10^{-3}W/s\\\\E=\frac{1.056*10^{-3}W/s}{\pi(0.04m)^2}=0.21\frac{N}{C}$

the magnitude if the induced electric field is 0.21N/C

b) By the Lenz's law you can conclude that the current has a direction in a counterclockwise

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3 years ago

I) Friction depends on the........ of surface.

zhuklara [117]

Answer:

material of the surface

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3 years ago

A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

marta [7]

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

The current I is any motion of charge from one region to another, so this is given by:

 $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$

Finally, for the drift velocity magnitude vd, we find:

 $v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

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3 years ago