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2 years ago

What is the temperature increase of 4.0 kg of water when heated by an 800-W immersion heater for 10 min? (cw = 4 186 J/kg⋅°C)

Physics

1 answer:

Llana [10]2 years ago

3 0

Answer:

option C

Explanation:

given,

mass of water = 4 Kg

Water is heated to = 800 W

time of immersion = 10 min

= 10 x 60 = 600 s

using equation of specific heat

Q = m S ΔT

S is the specific heat capacity of water which is equal to 4182 J/kg°C.

and another formula of heat

Q = Pt

now,

P t = m S ΔT

800 x 600 = 4 x 4182 x ΔT

ΔT = 29° C

temperature increased is equal to ΔT = 29° C

Hence, the correct answer is option C

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A factory has 1200 workers of which 720 are male and the rast are female what percent of workers are female

umka21 [38]

Answer:

Percent of Female Workers = 40%

Explanation:

The percentage of the female workers in the given group of workers can be easily found by the following formula:

$Percent\ of\ Female\ Workers = \frac{No.\ of\ Female\ Workers}{Total\ No.\ of\ Workers}\ x\ 100\%$

where,

Total No. of Workers = 1200

No. of Female Workers = Total Workers - No. of Male Workers

No. of Female Workers = 1200 - 720 = 480

Therefore,

$Percent\ of\ Female\ Workers = \frac{480}{1200}\ x\ 100\%$

<u>Percent of Female Workers = 40%</u>

7 0

2 years ago

Please help me with questions 1, 2 and 3. <br> i need a step by step explanation

kifflom [539]

Answer:

1) d

2) 5 m/s

3) 100

Explanation:

The equation of position x for a constant acceleration a and an initial velocity v₀, initial position x₀, time t is:

(i) $x=\frac{1}{2}at^2+v_0t+x_0$

The equation for velocity v and a constant acceleration a is:

(ii) $v=at+v_0$

1) Solve equation (ii) for acceleration a and plug the result in equation (i)

(iii) $a = \frac{v -v_0}{t}$

(iv) $x = \frac{v-v_0}{2t}t^2+v_0t + x_0$

Simplify equation (iv) and use the given values v = 0, x₀ = 0:

(v) $x=-\frac{v_0}{2}t + v_0t= \frac{v_0}{2}t$

2) Given v₀= 3m/s, a=0.2m/s², t=10 s. Using equation (ii) to get the final velocity v: $v=at+v_0=0.2\frac{m}{s^2} * 10s+3\frac{m}{s}=2\frac{m}{s}+3\frac{m}{s}=5\frac{m}{s}$

3) Given v₀=0m/s, t₁=10s, t₂=1s and x₀=0. Looking for factor f = x(t₁)/x(t₂) using equation(i) to calculate x(t₁) and x(t₂):

$f=\frac{x(t_1)}{x(t_2)}=\frac{\frac{1}{2}at_1^2 }{\frac{1}{2}at_2^2}=\frac{t_1^2}{t_2^2}=\frac{10^2}{1^2}=\frac{100}{1}$

5 0

3 years ago

The correct sequence of layers of the atmosphere from innermost to outermost is

serious [3.7K]

Troposphere, stratosphere, mesosphere, thermosphere, exosphere

4 0

2 years ago

When sugar is poured from the box into the sugar bowl, the rubbing of sugar grains creates a static electric charge that repels

Afina-wow [57]

Answer:

2.6×10⁻³ N

Explanation:

From coulomb's law,

F = kq'q/r²................ Equation 1

Where F = Repulsive force, q' = charge on the first sugar grain, q = charge on the second sugar grain, r = distance of separation between the sugar grain, k = proportionality constant.

From the question,

since q' = q

Then,

F = kq²/r²..................... Equation 2

Given: q = 1.79×10⁻¹¹ C, r = 3.45×10⁻⁵ m,

Constant: k = 9×10⁹ Nm²/kg².

Substitute into equation 2

F = 9×10⁹(1.79×10⁻¹¹)²/(3.45×10⁻⁵ )²

F = 9×10⁹(3.2041×10⁻²²)/(11.9025×10⁻¹⁰)

F = (28.8369×10⁻¹³)/(11.9025×10⁻¹⁰)

F = 2.6×10⁻³ N.

3 0

2 years ago

What phase difference between two otherwise identical harmonic waves, moving in the same direction along a stretched string, wil

lora16 [44]

Answer:

$\theta=145$

Explanation:

The amplitude of he combined wave is:

$B=2Acos(\theta/2)\\$

A, is the amplitude from the identical harmonic waves

B, is the amplitude of the resultant wave

θ, is the phase, between the waves

The amplitude of the combined wave must be 0.6A:

$0.6A=2Acos(\theta/2)\\ cos(\theta/2)=0.3\\\theta/2=72.5\\\theta=145$

5 0

2 years ago