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3 years ago

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An object of mass 20 kg is raised vertically through a distance of 8 m above ground level. If g = 10 m/s2, what is the gravitati

onal potential energy of this object at this position?

1 answer:

IgorC [24]3 years ago

6 0

2400joules

Explanation:

P.E

m= 20kg h=8m g=10m

P.E= 20×8×10

=2400joules

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In a photoelectric effect experiment you illuminate a metal with light of an unknown wavelength and measure the maximum kinetic

grigory [225]

Answer:

(a). The first wavelength is 401.0 nm.

(b). The metal's work function is 2.55 eV.

Explanation:

Given that,

Maximum kinetic energy = 0.65 eV

Second wavelength $\lambda_{2}= \dfrac{2}{3}\times\lambda_{1}$

(a). We need to calculate the wavelength

Using equation of work function for first wavelength

$\dfrac{hc}{\lambda_{1}}-W_{0}=0.65\ eV$ .....(I)

For second wavelength,

$\dfrac{hc}{\lambda_{2}}-W_{0}=2.3\ eV$

Put the value of second wavelength

$\dfrac{1.5 hc}{\lambda_{1}}-W_{0}=2.3\ eV$ ....(II)

By subtraction equation (I) from (II)

$0.5\dfrac{hc}{\lambda_{1}}=1.55$

$\lambda_{1}=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{3.1\times1.6\times10^{-19}}$

$\lambda_{1}=4.010\times10^{-7}\ m$

$\lambda_{1}=401.0\times10^{-9}\ m$

$\lambda_{1}=401.0\ nm$

(b). We need to calculate the work function

Using formula of work function

$W_{0}=(\dfrac{hc}{\lambda}-0.55)$

Put the value into the formula

$W_{0}=(\dfrac{6.63\times10^{-34}\times3\times10^{8}}{1.6\times10^{-19}\times401.0\times10^{-9}}-0.55)$

$W_{0}=2.55\ eV$

Hence, (a). The first wavelength is 401.0 nm.

(b). The metal's work function is 2.55 eV.

4 0

3 years ago

Read 2 more answers

Crystals can have _____ bonds between their atoms.

Dmitry_Shevchenko [17]

<h3><u>Answer;</u></h3>

All types of

Crystals can have <em><u>all types of</u></em> bonds between their atoms.

<h3><u>Explanation</u>;</h3>

<em><u>Crystals are solid materials which contains atoms, molecules or ions that are arranged in a complex structure forming a crystal lattice.</u></em>
<u><em>Crystals may have all types of bonds</em></u>, these includes; <em><u>ionic bonds, formed as a result of transfer between a metallic bond and a non-metal atom, Covalent bond, formed as a result of sharing electrons between non metal atoms, hydrogen bonds, metallic bonds Van der Waals bonds, etc.</u></em>

8 0

4 years ago

Read 2 more answers

A cylindrical bucket, open at the top, is 28.0 cm high and 11.0 cm in diameter. A circular hole with a cross-sectional area 1.55

svlad2 [7]

Answer:

so height is 0.1283 m

Explanation:

given data

height = 28 cm

diameter = 11 cm

cross-sectional area = 1.55 cm2

water flow rate = 2.46×10^−4 m3/s

to find out

How high will the water in the bucket rise

solution

we know that here

potential energy = kinetic energy

mgh = 1/2 mv²

multiply both sides by the 2 and we get

2mgh=mv²

solve it we get

√(2gh) = v ....................1

h = v²/2g ...............2

and

flow rate = A V

2.46×10^−4 = V 1.55×10^−4

V = 1.5870 m/s

so from 2

h = v²/2g

h = 1.5870²/ 2(9.81)

h = 0.1283 m

so height is 0.1283 m

6 0

4 years ago

The concrete post (Ec = 3.6 × 106 psi and αc = 5.5 × 10-6/°F) is reinforced with six steel bars, each of 78-in. diameter (Es = 2

masha68 [24]

Answer:

the normal stress induced by the concrete post $\sigma_c = 67.26 psi$

the normal stress induced by the steel $\sigma_s = - 1795.84 psi$

Explanation:

Given that:

Modulus for elasticity for concrete post $E_c$ = $3.6 *10^6$ psi

Thermal coefficient for concrete post $\alpha _c$ = $5.5 *10^{-6}/^0F$

Modulus for elasticity of steel bar $E_s$ = $29*10^6psi$

Thermal coefficient of steel bar $\alpha _2$ = $6.5*10^{-6}/^0F$

Change in temperature ΔT = 80°F

Diameter of the steel rood = 7/8-in

Area of the steel rod $A_s$ = $6(\frac{ \pi}{4} )(d_s)^2$

= $6(\frac{ \pi}{4} )(\frac{7}{8} )^2$

= 3.61 in²

Area of concrete parts $A_c$ = (10)(10) - $A_s$

= (100 - 3.61) in²

= 96.39 in²

The total strain developed in the concrete post can be expressed as:

= $[\frac{1}{E_cA_c}+\frac{1}{E_sA_s}]P=(\alpha_s-\alpha_c)(\delta T)$

= $[\frac{1}{(3.6*10^6)(96.39)}+\frac{1}{(29*10^6)(3.61)}]P=(6.5*10^{-6}-5.5*10^{-6})(80)$

= $[(2.88*10^{-9}) +(9.55*10^{-9}]P = 8.0*10^{-5}$

= $1.234*10^{-8}P = 8.0*10^{-5}$

$P = \frac {8.0*10^{-5}}{1.234*10^{-8}}$

P = 6482.98 lb

Since, the normal stress in concrete is induced as a result of temperature rise; we have the expression :

$\sigma_c =\frac{P}{A_c}$

$\sigma_c = \frac{6482.98}{96.39}$

$\sigma_c = 67.26 psi$

Thus, the normal stress induced by the concrete post $\sigma_c = 67.26 psi$

Also; the normal stress in the steel bars induced as a result of temperature rise is as follows:

$\sigma_s = \frac{-P}{A_s}$

$\sigma_s =\frac{-6482.98}{3.61}$

$\sigma_s = - 1795.84 psi$

Thus, the normal stress induced by the steel $\sigma_s = - 1795.84 psi$

6 0

4 years ago

On a wave, the distance between two points that have identical status is termed by ______

Sedaia [141]

D wavelength

velocity deals w/ speed
frequency/amplitude deals with sound

5 0

3 years ago

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