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2 years ago

7

The specific heat capacity of copper is 385 J/kg/°C. Calculate the energy needed to increase the temperature of a 4 kg block of

copper from 15°C to 33°C.

1 answer:

Vesnalui [34]2 years ago

6 0

Answer:

heat energy =mc∆t

Q = 4×385×[33-15]

Q = 4×385×18

Q = 27720joule

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A cannonball is launched with initial velocity of magnitude v0 over a horizontal surface. At what minimum angle

melisa1 [442]

Answer:

A) θmin= 76°.

Explanation:

Given that

Speed of cannonball = Vo

Lets take θ is the angle measure from horizontal

We know that

Range R

$R=\dfrac{V_o^2sin2\theta }{g}$

Height h

$h=\dfrac{V_o^2sin^2\theta }{2g}$

Given that h should be larger than R

h > R

$\dfrac{V_o^2sin^2\theta }{2g}>\dfrac{V_o^2sin2\theta }{g}$

${sin^2\theta }>2{sin2\theta }$

${sin^2\theta }>4{sin\theta\ cos\theta }$

${tan\theta }>4$

θ >75.96°

So minimum angle should be 75.96° or 76°.

4 0

2 years ago

Ask Your Teacher An electric utility company supplies a customer's house from the main power lines (120 V) with two copper wires

natta225 [31]

Answer:

The potential difference at the customer's house is 117.1 V.

Explanation:

a) The potential difference at the customer's house can be calculated as follows:

$\Delta V_{h} = \Delta V_{p} - \Delta V_{l}$

<u>Where</u>:

$V_{h}$ : is the potential difference at the customer's house

$V_{p}$ : is the potential difference from the main power lines = 120 V

$V_{l}$ : is the potential difference from the lines

$\Delta V_{h} = \Delta V_{p} - IR$

The resistance, R, is:

$\frac{0.109 \Omega}{300 m}*2*34.0 m = 0.025 \Omega$

Now, the potential difference at the customer's house is:

$\Delta V_{h} = 120 V - 116A*0.025 \Omega = 117.1 V$

Therefore, the potential difference at the customer's house is 117.1 V.

I hope it helps you!

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3 years ago

8.92 A 45.0 kg woman stands up in a 60.0 kg canoe 5.00 m long. She walks from a point 1.00 m from one end to a point I .00 m fro

omeli [17]

The distance that the canoe moves in this process is 1.29 meters.

We first have to find the center of mass

$X = \frac{MsXs+McXc}{Mw+Mc} \\\\$

Where

Ms = Woman's mass = 45

Mc = Canoe's mass = 60kg

Xs = position from left= 1 cm

Xc = position from left end of canoe's mass = 2.5cm

When we put these values into the equation we have:

$X=\frac{45*1+60*2.5}{45+60} \\\\= 1.857\\$

The center of gravity lies at the center of this boat. Therefore,

$Xc = \frac{L}{2} \\\\L = 5 m long\\\\\frac{5}{2} =2.5$

5.00 - 1. 00 = 4 meters

$\frac{45*4+60*2.5}{45+60} = 3.14meters\\\\$

To get the distance that is moved by this canoe

distance = 3.143-1.857

= 1.286

≈ 1.29 meters

The distance that the canoe moves in this process is 1.29 meters.

Read more on brainly.com/question/13198009?referrer=searchResults

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1 year ago

Answer the question correctly. Look at the 2 pictures.

Alexxandr [17]

Answer:

The answer is 3.15

Explanation:

I added lol

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2 years ago

Read 2 more answers

A bicycle traveled 150 meters west from point A to point B. Then it took the same route and came back to point A. It took a tota

Katen [24]

B i really
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3 years ago