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3 years ago

8

An unknown material has a mass of 0.447 kg, and its temperature increases by 2.87°C when 943 J of heat are added. What is the sp

ecific heat of the material? (Unit=J(kg*C))

1 answer:

Sergeeva-Olga [200]3 years ago

4 0

Answer:

735 J/kg/C

Explanation:

Q = mcT

943 = (0.447)( c )(2.87)

1.28289c = 943

c = <u>7</u><u>3</u><u>5</u><u> </u><u>J</u><u>/</u><u>k</u><u>g</u><u>/</u><u>C</u><u> </u><u>(</u><u>3</u><u> </u><u>s</u><u>f</u><u>)</u>

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I need help with 23 please.

Ne4ueva [31]

The answer is c.

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3 years ago

5. The wire in consists of two segments of different diameters but made from the same metal. The current in segment 1 is I1. a.

Volgvan

Answer:

hello your question is incomplete attached below is the complete question

answer :

a) I1 = I2

b) J1 > J2

c) E 1 > E2

d) ( vd1 ) > ( vd2 )

Explanation:

a) The currents in the two segments are the same i.e. I1 = I2 and this is because the segments are connected in series

b) Comparing the current densities J1 and J2 in the two segments

note : current density ∝ 1 / area

The area of the second segment is > the area of first segment therefore

J1 > J2

J1 ( current density of first segment )

J2 ( current density of second segment )

c) Comparing the electric field strengths E1 and E2

note : electric field strength ∝ current density

since current density of first segment is > current density of second segment and conductivity of the materials are the same hence

E 1 > E2

d) Comparing the drift speeds Vd1 and Vd2

( vd1 ) > ( vd2 )

this because ; vd ∝ current density

7 0

3 years ago

if the current in a wire is 2.0 amperes and the potential difference across the wire is 10 volts what is the resistance of the w

Pavlova-9 [17]

Answer:

R = 2Ω

Explanation:

Potential difference (V) = current (I) * Resistance (R)

V = IR

I = 2.0A

V = 10v

R = ?

V = IR

R = V / I

R = 10 / 2

R = 2Ω

The resistance across the wire is 2Ω

3 0

3 years ago

Read 2 more answers

The displacement of a 500 g mass, undergoing simple harmonic motion, is defined by the function :

Delicious77 [7]

The maximum kinetic energy, maximum potential energy and the maximum mechanical energy are equal to 7.56J.

<h3>What is simple harmonic motion?</h3>

Simple harmonic motion, in physics, repetitive movement back and forth through an equilibrium, or central, position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side.

Simple Harmonic Motion

The given equation of the simple harmonic motion is

$x=3.5 sin (\frac{\pi }{2t} + \frac{5\pi }{4} )$

Data;

ω = π/2

k = 1.254N/m

Solving this

$\frac{dx}{dt} = -3.5 X \frac{\pi }{2} cos (\frac{x\pi t}{2}+\frac{5\pi }{4} )$

Let's calculate the maximum velocity.

$V_{m} =\frac{3.5\pi }{2}$

This is only possible when cos θ = -1

The maximum kinetic energy is

$K_m =\frac{1}{2} mv^2 = \frac{1}{2} X \frac{500}{1000} X \frac{7^2\pi ^2}^{4} ^2$

$w^2 = \frac{k}{m} \\k = w^2m\\k = \frac{\pi ^2}{4} X \frac{500}{1000} \\k =1.254 N/m$

Using the value of spring constant, we can find the maximum potential energy.

$P.E =\frac{1}{2} k x^2\\P.E =\frac{1}{2} X 1.234 X 3.5^2 \\P.E = 7.56 J$

The maximum potential energy is 7.56J

The maximum mechanical energy is equal to the sum of maximum potential energy and the maximum kinetic energy.

ME = K.E + P.E

ME = 7.56J

From the calculations above, the maximum kinetic energy, maximum potential energy and the maximum mechanical energy are equal to 7.56J.

Learn more on simple harmonic motion here;

brainly.com/question/15556430

#SPJ1

8 0

2 years ago

2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate

adoni [48]

Answer:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

v is image distance
u is object distance, u is 10 cm
f is focal length, f is 5 cm

${ \tt{ \frac{1}{v} + \frac{1}{10} = \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{v} = \frac{1}{10} }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}$

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

» Multiply throughout by fv

${ \tt{fv( \frac{1}{v} + \frac{1}{u} ) = fv( \frac{1}{f} )}} \\ \\ { \tt{ \frac{fv}{v} + \frac{fv}{u} = \frac{fv}{f} }} \\ \\ { \tt{f + f( \frac{v}{u} ) = v}}$

• But we know that, v/u is M

${ \tt{f + fM = v}} \\ { \tt{f(1 +M) = v }} \\ { \tt{1 +M = \frac{v}{f} }} \\ \\ { \boxed{ \mathfrak{formular : } \: { \tt{ M = \frac{v}{f} - 1 }}}}$

v is image distance, v is 10 cm
f is focal length, f is 5 cm
M is magnification.

${ \tt{M = \frac{10}{5} - 1 }} \\ \\ { \tt{M = 5 - 1}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}$

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

Image is magnified
Image is erect or upright
Image is inverted
Image distance is identical to object distance.

4 0

2 years ago