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ankoles [38]
3 years ago
10

Q1.

Physics
1 answer:
Vesnalui [34]3 years ago
4 0

Answer:

P = 450 J

Explanation:

Given that,

Mass of a child, m = 18 kg

The vertical distance from the top to the bottom of the slide is 2.5 metres.

The Gravitational field strength = 10 N/kg

We need to find the decrease in gravitational potential energy of the child sliding from the  top to the bottom of the slide.

The formula for the gravitational potential energy is given by :

P = mgh

Substituting all the values,

P = 18 kg × 10 m/s² × 2.5 m

P = 450 J

Hence, the decrease in gravitational potential energy is 450 J.

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What is the magnitude of electrical force of attraction between an copper nucleus (29 protons) and its innermost electron if the
Agata [3.3K]
The charge of the copper nucleus is 29 times the charge of one proton:
Q=29 q= 29 \cdot 1.6 \cdot 10^{-19}C=4.64 \cdot 10^{-18}C
the charge of the electron is
e=-1.6 \cdot 10^{-19}C
and their separation is
r=1.0 \cdot 10^{-12} m

The magnitude of the electrostatic force between them is given by:
F=k_e  \frac{Qe}{r^2}
where k_e is the Coulomb's constant. If we substitute the numbers, we find (we can ignore the negative sign of the electron charge, since we are interested only in the magnitude of the force)
F=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(4.64 \cdot 10^{-18}C)(1.6 \cdot 10^{-19}C)}{(1.0 \cdot 10^{-12} m)^2}=6.68 \cdot 10^{-3} N
3 0
3 years ago
A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per un
Zarrin [17]

Answer:

steady state temperature =88.7deg C

t=time within  1 deg C of it steady state is 8.31s

Explanation:

A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per unit length of 0.01 Ω/m. If a current of 100 A flows through the wire and the convection coefficient is 500W/m2K, what is the steady state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature within 1-degC of the steady state value? The density of the wire is 8,000kg/m3, its heat capacity is 500 J/kgK and its thermal condu

The diameter of the wire is known to be=1mm

properties=

The density of the wire is 8,000 kg/m3,

heat capacity is 500 J/kgK

themal conductivity is 20W/m.K

electrical resistance per unit length of 0.01 Ω/m

from lump capavity method

B_{i} =\frac{hr/2}{k}

500*(2.5*10^-4)/20

0.006<0.1

we know also, to find steady state temperature

\piDh(T-Tinf)=I^{2} R_{e}

make T the subject of the equation , we have

T=25+\frac{100^2*0.01}{\pi*0.001*500 }

T=88.7 degC

rate of chnage in temperature

dT/dt=\frac{I^2*Re}{rho*c*\pi*D^2/4 } -\frac{4h}{rho*c*D} (T-Tinf)

at t=o and integrating both sides\frac{T-Tinf-(I^2*Re/\pi*Dh) }{Ti-Tinf-(I^2*Re/\pi*Dh } =exp\frac{-4ht}{rho*c*D}

we have

\frac{87.7-25-63.7}{25-25-63.7} =exp\frac{4*500t}{8000*500*0.001}

t=8.31s

steady state temperature =88.7deg C

t=time within  1 degC of it steady stae is 8.31s

7 0
3 years ago
The chemical formula for sodium fluoride is A) Na7F B) NaF C) NaF2 D) NaF7
ikadub [295]
The correct answer would be the letter B.) NaF this attracts opposite charges. The final formula of sodium fluoride would be NaF it’s <span>compound is formed by the complete transfer of electrons from a metal to a nonmetal.</span>
8 0
3 years ago
Read 2 more answers
Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu
Luba_88 [7]

Answer:

V_1=8 V_2

Explanation:

Given that:

  • Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
  • separation distance of capacitor 2, d_2=d
  • separation distance of capacitor 1, d_1=2d
  • quantity of charge on capacitor 2, Q_2=Q
  • quantity of charge on capacitor 1, Q_1=4Q

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

C=\frac{k.\epsilon_0.A}{d}.....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}

From eq. (1)

For capacitor 2:

C_2=\frac{k.\epsilon_0.A}{d}

For capacitor 1:

C_1=\frac{k.\epsilon_0.A}{2d}

C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]

We know, potential differences across a capacitor is given by:

V=\frac{Q}{C}..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}

V_2=\frac{Q.d}{k.\epsilon_0.A}

& for capacitor 1:

V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}

V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}

V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]

V_1=8 V_2

6 0
3 years ago
Which of the following statements about iron filings placed upon glass resting on top of a bar magnet is false?
leva [86]

The answer would be D. Because the iron fillings are attracted to the magnet underneath the glass.

5 0
3 years ago
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