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ExtremeBDS [4]
2 years ago
9

Suppose that at room temperature, a certain aluminum bar is 1.0000 m long. The bar gets longer when its temperature is raised. T

he length l of the bar obeys the following relation: l=1.0000+2.4×10−5T, where T is the number of degrees Celsius above room temperature. What is the change of the bar's length if the temperature is raised to 16.1 ∘C above room temperature?
Physics
1 answer:
Alja [10]2 years ago
7 0

Answer:

The change of the bar's length is 3.9\times10^{-4} m

Explanation:

The bar length is a function of temperature T above room temperature:

L(T)=1.0000+2.4\times10^{-5} T

So, if we evaluate at T= 16.1 C above room temperature

L(16.1)=1.0000+2.4\times10^{-5} (16.1)

L=1.00039 m

Now we can find the change of the bar length with the difference of L and Lo (the length at room temperature)

L- L_0=1.00039-1.0000 = 3.9\times10^{-4} m

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(HELP!!! 30 pts if answered right. )What formula gives the strength of an electric field, E, at a distance from a known source c
umka2103 [35]

Answer:

E=\frac{k\,Q}{d^2}

Explanation:

The strength of an electric field E produced by a single charge Q at a distance d from it is given by the formula: E=\frac{k\,Q}{d^2}, where K represents the Coulomb constant.

Since the electric field E is derived from the Coulomb Force per unit charge using a positive test charge, the field's units will be in units of Newtons/Coulomb, and be the formula for the Coulomb electric force between to charges (Q1 and Q2),

F_C=k\frac{Q_1\,Q_2}{d^2}

but modified with only one charge showing in the numerator of the expression.

8 0
2 years ago
I pull the throttle in my racing plane at a = 12.0 m/s2. I was originally flying at v = 100. m/s. Where am I when t = 2.0s, t =
Helen [10]
Summary:
a= 12.0 m/(s^2)
v= 100m/s
t1= 2.0s => s1=?
t2=5.0s => s2=?
t3=10.0s => s3=?
——————
Solution:
• when t1=2.0 s, I have gone:
S1= v*t1 + 1/2*a*(t1^2)
=100.0 *2 + 1/2*12.0*(2.0^2)
=224 (m)

• when t2=5.0s, I have gone
S2=v*t2+ 1/2*a*(t2^2)
= 100*5.0+ 1/2*12.0*(5.0^2)
=650 (m)

•when t3= 10.0s, I have gone:
S3=v*t3+ 1/2*a*(t3^2)
=100*10.0+ 1/2*12*(10.0^2)
=1600 (m)
7 0
2 years ago
A 0.242 g sample of potassium is heated in oxygen. The result is 0.292 g of a crystalline compound. What is the formula of this
masha68 [24]

Answer:

Hello there Dude answer is B :D hope it helped mark me brainliest.

8 0
2 years ago
A fly has a mass of 1 gram at rest. how fast would it have to be traveling to have the mass of a large suv, which is about 3000
Zigmanuir [339]

We solve this using special relativity. Special relativity actually places the relativistic mass to be the rest mass factored by a constant "gamma". The gamma is equal to 1/sqrt (1 - (v/c)^2). <span>

We want a ratio of 3000000 to 1, or 3 million to 1. 

</span>

<span>Therefore:
3E6 = 1/sqrt (1 - (v/c)^2) 
1 - (v/c)^2 = (0.000000333)^2 
0.99999999999999 = (v/c)^2 
0.99999999999999 = v/c 
<span>v= 99.999999999999% of the speed of light ~ speed of light
<span>v = 3 x 10^8 m/s</span></span></span>

8 0
2 years ago
A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) th
zheka24 [161]

Answer:

(a) A = 0.650 m

(b) f = 1.3368 Hz

(c) E = 17.1416 J

(d)  K = 11.8835 J

     U = 5.2581 J

Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40    ⇒   f = 8.40 / (2π)

⇒   f = 1.3368 Hz

(c) the total energy,

we use the formula

E = m*ω²*A² / 2

⇒  E = (1.15)(8.40)²(0.650)² / 2

⇒  E = 17.1416 J

(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²)       (the kinetic energy)

and

U = (1/2)*m*ω²*x²              (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒  K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

⇒  U = 5.2581 J

4 0
2 years ago
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