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max2010maxim [7]
3 years ago
13

A 0.016-kg piece of iron absorbs 1086.75 joules of heat energy, and its

Physics
1 answer:
wel3 years ago
7 0

Answer:

c = 115.92 q/g C

Explanation:

0.016 kg = 16 grams

q = Joules

m = mass (g)

t = (final temp. - initial temp)

c = specific heat

c = (1086.75 joules) / (175 C - 25 C) (16 g)

c = 115.92 q/g C

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Two technicians are discussing torque wrenches. Technician A says that a torque wrench is capable of tightening a fastener with
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Chanice drives her scooter 7 kilometers north. she stops for lunch and then drives 5 kilometers and then 1 km east again. what d
lord [1]

Distance covered is given as follows

1). 7 km North

2). 5 km North

3). 1 km East

Now total distance covered will be given as

d = d_1 + d_2 + d_3

d = 7 km + 5 km + 1 km

d = 13 km

Now in order to find the displacement we will show all with their directions

\vec d_1 = 7 + 5 = 12 km towards North

\vec d_2 = 1 km towards East

So total displacement is

\vec d = \vec d_1 + \vec d_2

\vec d = 12 \hat j + 1 \hat i

so net displacement will be

d = \sqrt{12^2 + 1^2} = 12.04 km

so displacement is 12.04 km

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4 years ago
Swimmers at the beach are tanning on towels. Which method of heat transfer is responsible for their tan? *
Free_Kalibri [48]

Answer:

ratiation

Explanation:

3 0
3 years ago
Read 2 more answers
If the car speeds up at a steady 1.6 m/s2 , how long after starting is the magnitude of its centripetal acceleration equal to th
Rufina [12.5K]
Based on internet sources, <span>the basic formulas are: v^2/r = (at)^2/r = a ==> at^2 = r ==> t = sqrt(r/a). 
</span>
<span>Assuming the missing units are mutually compatible, as in the following example, they don't need to be known. </span>
<span>Acceleration = 1.6 cramwells/s^2 </span>
<span>Radius = 150 cramwells </span>
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8 0
3 years ago
Two light bulbs have resistances of 400 Ω and 800 ΩThe two light bulbs are connected in series across a 120- V line. Find the cu
Natasha2012 [34]

1) Current in each bulb: 0.1 A

The two light bulbs are connected in series, this means that their equivalent resistance is just the sum of the two resistances:

R_{eq}=R_1 + R_2 = 400 \Omega + 800 \Omega=1200 \Omega

And so, the current through the circuit is (using Ohm's law):

I=\frac{V}{R_{eq}}=\frac{120 V}{1200 \Omega}=0.1 A

And since the two bulbs are connected in series, the current through each bulb is the same.

2) 4 W and 8 W

The power dissipated by each bulb is given by the formula:

P=I^2 R

where I is the current and R is the resistance.

For the first bulb:

P_1 = (0.1 A)^2 (400 \Omega)=4 W

For the second bulb:

P_1 = (0.1 A)^2 (800 \Omega)=8 W

3) 12 W

The total power dissipated in both bulbs is simply the sum of the power dissipated by each bulb, so:

P_{tot} = P_1 + P_2 = 4 W + 8 W=12 W

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3 years ago
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