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3 years ago

How much heat is absorbed by a 34g iron skillet when its temperature rises from 11oC to 26oC?

1 answer:

4 0

Answer is: 229,5 J of heat is absorbed.
m(mass of iron - Fe) = 34 g.
ΔT( change in temperature) = 26°C - 11°C = 15°C.
cp( specific heat capacity for iron) = 0,450 J/g·°C.
Q( heat absorbed) = m(Fe) · ΔT · cp.
Q = 34 g · 15°C · 0,450 J/g·°C.
Q = 229,5 J = 0,2295 kJ.

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Nesterboy [21]

The position of the object at time t =2.0 s is 6.4 m.

Velocity vₓ of a body is the rate at which the position x of the object changes with time.

Therefore,

$v_x= \frac{dx}{dt}$

Write an equation for x.

$dx=v_xdt\\ x=\int v_xdt$

Substitute the equation for vₓ =2t² in the integral.

$x=\int v_xdt\\ =\int2t^2dt\\ =\frac{2t^3}{3} +C$

Here, the constant of integration is C and it is determined by applying initial conditions.

When t =0, x = 1. 1m

$x= \frac{2t^3}{3} +C\\ x_0=1.1\\ x= (\frac{2t^3}{3} +1.1)m$

Substitute 2.0s for t.

$x= (\frac{2t^3}{3} +1.1)m\\ =\frac{2(2.0)^3}{3} +1.1\\ =6.43 m$

The position of the particle at t =2.0 s is 6.4m

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2 years ago

What are parts of a pulley

PtichkaEL [24]

Wheel, axel and rope

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2 years ago

During heavy rain, a section of a mountainside measuring 2.5 km horizontally, 0.80 km up along the slope, and 2.0 m deep slips i

Crank

Answer:

The mass of the mud is 3040000 kg.

Explanation:

Given that,

length = 2.5 km

Width = 0.80 km

Height = 2.0 m

Length of valley = 0.40 km

Width of valley = 0.40 km

Density = 1900 Kg/m³

Area = 4.0 m²

We need to calculate the mass of the mud

Using formula of density

$\rho=\dfrac{m}{V}$

$m=\rho\times V$

Where, V = volume of mud

$\rho$ = density of mud

Put the value into the formula

$m=1900\times4.0\times0.40\times10^{3}$

$m =3040000\ kg$

Hence, The mass of the mud is 3040000 kg.

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3 years ago

Hooke's law describes a certain light spring of unstretched length 31.8 cm. when one end is attached to the top of a doorframe a

lubasha [3.4K]

Answer:

Explanation:

extension in the spring = 40.4 - 31.8 = 8.6 cm = 8.6 x 10⁻² m .

kx = mg

k is spring constant , x is extension , m is mass

k x 8.6 x 10⁻² = 7.52 x 9.8

k = 856.93 N/m

= 857 x 10⁻³ KN /m

b ) Both side is pulled by force of 188 N .

Tension in spring = 188N

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856.93 x = 188

x = .219.38 m

= 21.938 cm

= 21.9 cm .

length of spring = 31.8 + 21.9

= 53.7 cm .

6 0

2 years ago

A 15.0-μF capacitor is charged by a 130.0-V power supply, then disconnected from the power and connected in series with a 0.280-

SVETLANKA909090 [29]

The resonant frequency of a circuit is the frequency $\omega_0$ at which the equivalent impedance of a circuit is purely real (the imaginary part is null).

Mathematically this frequency is described as

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Where

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$C = 15*10^{-6}\mu F$

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Replacing we have,

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$f= 2455.81Hz$

From this relationship we can also appreciate that the resonance frequency infers the maximum related transfer in the system and that therefore given an input a maximum output is obtained.

For this particular case, the smaller the capacitance and inductance values, the higher the frequency obtained is likely to be.

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3 years ago