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3 years ago

The specific heat of water is 4.2 J/g • °C. How much heat is required to raise the temperature of 100 g of water by 5°C? *

Physics

1 answer:

Svetach [21]3 years ago

7 0

Answer:

2100 J

Explanation:

The heat required to increase the temperature of a substance is given by

$Q=mC\Delta T$

where

m is the mass of the substance

C is its specific heat capacity

$\Delta T$ is its change in temperature

For the water in this problem, we have:

m = 100 g is its mass

C = 4.2 J/g • °C is the specific heat capacity

$\Delta T=5^{\circ}C$ is the increase in temperature

So, the amount of heat needed is:

$Q=(100)(4.2)(5)=2100 J$

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Two cello strings, with the same tension and length, are played simultaneously. Their fundamental frequencies produce audible be

qwelly [4]

Explanation:

Let f₁ is the fundamental frequency, $f_1=8\ Hz$

Lower pitch frequency, $f_2=220\ Hz$

Fundamental frequency is, $f_1=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu_1}}$ .....(1)

Lower frequency is, $f_2=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu_2}}$ ..............(2)

Dividing equation (1) and (2) as :

$\dfrac{f_1}{f_2}=\sqrt{\dfrac{\mu_2}{\mu_1}}$

$\dfrac{\mu_2}{\mu_1}=(\dfrac{f_1}{f_2})^2$

$\dfrac{\mu_2}{\mu_1}=(\dfrac{8}{220})^2$

$\dfrac{\mu_2}{\mu_1}=0.00132$

So, the ratio of linear mass density μ of the string with the higher pitch to that of the string with the lower pitch is 0.00132. Hence, this is the required solution.

3 0

3 years ago

PLEASE PROVIDE AN EXPLANATION THANK YOU!

Rus_ich [418]

Answer:

(a) 0.993 s

(b) 14.0 N/m

(d) -6.01 m/s²

Explanation:

(a) The block's position can be modeled as a cosine wave:

x(t) = A cos(ωt)

where A is the amplitude (in this case, 50 cm) and ω is the angular frequency.

At t = 0.200 s, x(t) = 15.0 cm.

15.0 cm = 50.0 cm cos((0.200 s) ω)

0.3 = cos((0.2 s) ω)

1.266 rad = (0.2 s) ω

ω = 6.33 rad/s

The period is:

T = (2π rad) (1 s / 6.33 rad)

T = 0.993

(b) For a spring-mass system, ω = √(k/m). The mass of the block is 0.350 kg, so:

ω = √(k/m)

6.33 rad/s = √(k / 0.350 kg)

40.1 rad/s² = k / 0.350 kg

k = 14.0 N/m

EE₀ = EE + KE

½ kx₀² = ½ kx² + ½ mv²

kx₀² = kx² + mv²

(14.0 N/m) (0.50 m)² = (14.0 N/m) (0.15 m)² + (0.35 kg) v²

v = -3.02 m/s

Alternatively, we can take the derivative of our position equation:

v(t) = -Aω sin(ωt)

v = -(0.50 m) (6.33 rad/s) sin((6.33 rad/s) (0.2 s))

v = -3.02 m/s

(d) Sum of forces on the block:

∑F = ma

-kx = ma

a = -kx / m

a = -(14.0 N/m) (0.15 m) / (0.350 kg)

a = -6.01 m/s²

Alternatively, we can take the derivative of our velocity equation:

a(t) = -Aω² cos(ωt)

a = -(0.50 m) (6.33 rad/s)² cos((6.33 rad/s) (0.2 s))

a = -6.01 m/s²

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