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qaws [65]
2 years ago
12

Water has a specific heat of 4.184 J/g°C and requires 23,000 J of heat to raise the

Mathematics
1 answer:
nasty-shy [4]2 years ago
3 0

Answer:

m=80.840 kg

Step-by-step explanation:

Given Data,

Q=23000 J

c=4.184 J/g °c

Δt= 68 °c

m=?

By using this equation,

Q=mcΔt

23000=m\times 4.184\times 68\\\\m=\frac{23000}{4.184\times68} \\\\\\m=80.840 kg

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Katena32 [7]
3y+21+8y
11y+21 is the answer
7 0
2 years ago
Read 2 more answers
Part A
nydimaria [60]

Answer:

1) D

2) 10

Step-by-step explanation:

sin(30) = opposite/hypotenuse

sin(30) = 5/x

½ = 5/x

x = 5/0.5

x = 10

5 0
3 years ago
What is the length of the curve with parametric equations x = t - cos(t), y = 1 - sin(t) from t = 0 to t = π? (5 points)
zzz [600]

Answer:

B) 4√2

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Parametric Differentiation

Integration

  • Integrals
  • Definite Integrals
  • Integration Constant C

Arc Length Formula [Parametric]:                                                                         \displaystyle AL = \int\limits^b_a {\sqrt{[x'(t)]^2 + [y(t)]^2}} \, dx

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle \left \{ {{x = t - cos(t)} \atop {y = 1 - sin(t)}} \right.

Interval [0, π]

<u>Step 2: Find Arc Length</u>

  1. [Parametrics] Differentiate [Basic Power Rule, Trig Differentiation]:         \displaystyle \left \{ {{x' = 1 + sin(t)} \atop {y' = -cos(t)}} \right.
  2. Substitute in variables [Arc Length Formula - Parametric]:                       \displaystyle AL = \int\limits^{\pi}_0 {\sqrt{[1 + sin(t)]^2 + [-cos(t)]^2}} \, dx
  3. [Integrand] Simplify:                                                                                       \displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx
  4. [Integral] Evaluate:                                                                                         \displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx = 4\sqrt{2}

Topic: AP Calculus BC (Calculus I + II)

Unit: Parametric Integration

Book: College Calculus 10e

4 0
3 years ago
The LATERAL Surface Area for the shape below is ____________. (assume the bases are the larger faces on the top and bottom.)
Zolol [24]

Answer:

LA=210\ cm^2

Step-by-step explanation:

we know that

The lateral area of the figure is given by

LA=Ph

where

P is the perimeter of the base (larger faces)

h is the height

Find the perimeter of the base P

P=2(11+10)=42\ cm

we have

h=5\ cm

substitute in the formula of lateral area

LA=42(5)=210\ cm^2

3 0
3 years ago
PLEASE ITS DUE 5 MINS
kolezko [41]

Answer:

I'm sorry pls retake the pic really quickly.. I can't see the entire problem!

Then I'll add my answer if I can! Thx!

5 0
2 years ago
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