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2 years ago

6

Kathmandu lies at high altitude than biratnagar from sea level.Where does an object has more weight between two places?Give reas

on

1 answer:

Alla [95]2 years ago

8 0

Answer:

Kathmandu

Explanation:

As the altitude get higher, the gravitational pull of the earth on the object increases, therefore, the mass is higher up above.

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The length of a 60 W, 240 Ω light bulb filament is 60 cm Remembering that the current in the filament is proportional to the ele

faust18 [17]

Answer:

Finally current will be

i = 0.35 A

Explanation:

As we know that power of the bulb is given by the formula

$P = \frac{V^2}{R}$

now we have

$P = 60 W$

R = 240 ohm

so we have

$60 = \frac{V^2}{240}$

$V = 120 Volts$

now the current in the bulb is given as

$i = \frac{V}{R}$

$i = \frac{120}{240} = 0.5 A$

now when length of the filament is double

so the resistance of the wire also gets double

so we have

$P = \frac{V^2}{R}$

$60 = \frac{V^2}{480}$

$V = 169.7 volts$

now the current in the bulb is given as

$V = i R$

$169.7 = i(480)$

$i = 0.35 A$

8 0

2 years ago

When explaining chemical reactions to a friend, Brianna models a reaction by taking several colors of modeling clay and making a

Drupady [299]

Answer: synthesis

Explanation:

5 0

2 years ago

Read 2 more answers

A force of 250 N is applied to a hydraulic jack piston that is 0.02 m in diameter. If the piston that supports the load has a di

Vikentia [17]

Answer:

option B

Explanation:

given,

Force exerted by the hydraulic jack piston = F₁ = 250 N

diameter of piston, d₁ = 0.02 m

r₁ = 0.01 m

diameter of second piston, d₂ = 0.15 m

r₂ = 0.075 m

mass of the jack to lift = ?

now,

$\dfrac{F_1}{A_1} =\dfrac{F_2}{A_2}$

$\dfrac{250}{\pi r_1^2} =\dfrac{F_2}{\pi r_2^2}$

$\dfrac{250}{0.01^2} =\dfrac{F_2}{0.075^2}$

$F_2= \dfrac{250}{0.01^2}\times {0.075^2}$

F₂ = 14062.5 N

F = m g

$m = \dfrac{F}{g}$

$m = \dfrac{14062.5}{9.8}$

m = 1435 Kg

hence, the correct answer is option B

5 0

2 years ago

When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa

MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

$\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2$

In parallel

$\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}$

$\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0$

Solving the above quadratic equation

$\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}$

$\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega$

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0

2 years ago

A metal has a strength of 414 MPa at its elastic limit and the strain at that point is 0.002. Assume the test specimen is 12.8-m

ser-zykov [4K]

To solve this problem, we will start by defining each of the variables given and proceed to find the modulus of elasticity of the object. We will calculate the deformation per unit of elastic volume and finally we will calculate the net energy of the system. Let's start defining the variables

Yield Strength of the metal specimen

$S_{el} = 414Mpa$

Yield Strain of the Specimen

$\epsilon_{el} = 0.002$

Diameter of the test-specimen

$d_0 = 12.8mm$

Gage length of the Specimen

$L_0 = 50mm$

Modulus of elasticity

$E = \frac{S_{el}}{\epsilon_{el}}$

$E = \frac{414Mpa}{0.002}$

$E = 207Gpa$

Strain energy per unit volume at the elastic limit is

$U'_{el} = \frac{1}{2} S_{el} \cdot \epsilon_{el}$

$U'_{el} = \frac{1}{2} (414)(0.002)$

$U'_{el} = 414kN\cdot m/m^3$

Considering that the net strain energy of the sample is

$U_{el} = U_{el}' \cdot (\text{Volume of sample})$

$U_{el} = U_{el}'(\frac{\pi d_0^2}{4})(L_0)$

$U_{el} = (414)(\frac{\pi*0.0128^2}{4}) (50*10^{-3})$

$U_{el} = 2.663N\cdot m$

Therefore the net strain energy of the sample is $2.663N\codt m$

6 0

1 year ago