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natima [27]
3 years ago
9

An insulator________the loss of heat energy.

Physics
2 answers:
Lostsunrise [7]3 years ago
8 0

Answer:

A) speed up

is the correct option

wariber [46]3 years ago
4 0

Answer:

slows down

Explanation:

As the temperature difference decreases the rate of transfer slows (hope this helps)

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A swimming pool is 4.0 m in depth; a swimmer at this depth feels discomfort in the ear. Calculate the net force on a 0.50-cm-dia
Mashcka [7]

The net force on a 0.50-cm-diameter eardrum is mathematically given as

F= 0.76969 N

<h3>What is the net force on a 0.50-cm-diameter eardrum?</h3>

Generally, the equation for Pressure is  mathematically given as

P = ρgh

Therefore

P= 1000*9.8*4

P= 39200 Pa

Where

A= pi*(0.005/2)^2

Generally, the equation for Net force is  mathematically given as

F = PA

F= 39200 *( pi*(0.005/2)^2)

F= 0.76969 N

In conclusion, The net force is

F= 0.76969 N

Read more about Pressure

#SPJ1

5 0
2 years ago
When measuring espresso for a drink, which instrument would give the<br> greatest precision?
Gemiola [76]

How many mL is an espresso?

One shot of espresso is generally about 30–50 ml (1–1.75 oz), and contains about 63 mg of caffeine (3). Important point: The “golden ratio” for espresso is this: a single shot is 30 to 44 mL (1 to 1.5 ounces) of water and 7 grams of coffee

5 0
3 years ago
1)Determine, in terms of unit vectors, the resultant of the five forces illustrated in the figure, Consider F1=20 N, F2= 12 N, F
LiRa [457]

Explanation:

1) F₁ lies in a plane perpendicular to the xy plane, 60° from the x axis.  The angle between F₁ and the +z axis is 30°.  Therefore, the vector is:

<F₁> = 20 (sin 30° cos 60° i + sin 30° sin 60° j + cos 30° k)

<F₁> = 20 (¼ i + ¼√3 j + ½√3 k)

<F₁> = 5 i + 5√3 j + 10√3 k

F₂ is in the xy plane.  Its slope is -24/7.  The vector is:

<F₂> = 12 (-⁷/₂₅ i + ²⁴/₂₅ j + 0 k)

<F₂> = -3.36 i + 11.52 j

F₃ is parallel to the +x axis.  The vector is:

<F₃> = 17 (i + 0 j + 0 k)

<F₃> = 17 i

F₄ is parallel to the -z axis.  The vector is:

<F₄> = 15 (0 i + 0 j − k)

<F₄> = -15 k

F₅ is in the xy plane.  It forms a 15° angle with the -y axis.  The vector is:

<F₅> = 9 (-sin 15° i − cos 15° j + 0 k)

<F₅> = -9 sin 15° i − 9 cos 15° j

The resultant vector is therefore:

<F> = (5 − 3.36 + 17 − 9 sin 15°) i + (5√3 + 11.52 − 9 cos 15°) j + (10√3 − 15) k

<F> = 16.31 i + 11.49 j + 2.32 k

2) Sum of forces at point B in the x direction:

∑F = ma

Tbc cos 40° − ¹⁵/₁₇ Tab = 0

Tbc cos 40° = ¹⁵/₁₇ Tab

Tbc = 1.15 Tab

Sum of forces at point B in the y direction:

∑F = ma

Tbc sin 40° + ⁸/₁₇ Tab − mAg = 0

Tbc sin 40° + ⁸/₁₇ Tab = (2 kg) (10 m/s²)

(1.15 Tab) sin 40° + ⁸/₁₇ Tab = 20 N

1.21 Tab = 20 N

Tab = 16.52 N

Tbc = 19.02 N

Sum of forces at point C in the x direction:

∑F = ma

Tcd sin 25° − Tbc cos 40° = 0

Tcd sin 25° = Tbc cos 40°

Tcd = 1.81 Tbc

Tcd = 34.48 N

3(a) When the crane is on the verge of tipping, the center of gravity is directly over point F.  Relative to point A:

3.7 m = [ (390 kg) (0.9 m) + (90 kg) (9 m cos θ + 1.7 m) + (80 kg) (9 m cos θ + 2.9 m) ] / (390 kg + 90 kg + 80 kg)

2072 kgm = 351 kgm + 810 kgm cos θ + 153 kgm + 720 kgm cos θ + 232 kgm

1336 kgm = 1530 kgm cos θ

θ = 29.17°

3(b) 3.7 m = [ (390 kg) (0.9 m) + (90 kg) (x + 1.7 m) + (80 kg) (x + 2.9 m) ] / (390 kg + 90 kg + 80 kg)

2072 kgm = 351 kgm + (90 kg) x + 153 kgm + (80 kg) x + 232 kgm

1336 kgm = (170 kg) x

x = 7.86 m

4) Find the lengths of the cables.

Lab = √((2 m)² + (3 m)² + (5 m)²)

Lab = √38 m

Lac = √((2 m)² + (3 m)² + (5 m)²)

Lac = √38 m

Lde = √((2 m)² + (3 m)²)

Lde = √13 m

Sum of forces in the x direction:

∑F = ma

-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0

Sum of forces in the y direction:

∑F = ma

2/√38 Fab − 2/√38 Fac = 0

Fab = Fac

Sum of forces in the z direction:

∑F = ma

3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0

Sum of moments about the y-axis:

∑τ = Iα

(3/√38 Fab) (5 m) + (3/√38 Fac) (5 m) + (3/√13 Fde) (2 m) − (mg) (2 m) = 0

Substitute Fab = Fac and simplify:

6/√38 Fab + 3/√13 Fde − mg = 0

30/√38 Fab + 6/√13 Fde − 2mg = 0

Double first equation:

12/√38 Fab + 6/√13 Fde − 2mg = 0

Subtract from the second equation:

28/√38 Fab = 0

Fab = 0

Fac = 0

Solve for Fde:

3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0

3/√13 Fde = mg

3/√13 Fde = (1.7 kg) (10 m/s²)

Fde = 20.43 N

Solve for Rx:

-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0

Rx = 2/√13 Fde

Rx = 11.33 N

8 0
3 years ago
(b) Fig. 1.1 shows two airports A and C. north SE с sea land क WE not to scale Fig. 1.1 An aircraft flies due north from A for a
GalinKa [24]

The addition of vectors and the uniform motion allows to find the answers for the questions about distance and time are:

  • The distance to go between airports A and C  is 373.6 10³ m
  • The time to go from airport A to B is 2117 s

Vectors are quantities that have modulus and direction, so their addition must be done using vector algebra.

In this case the plane flies towards the North a distance of y = 360 10³ m at an average speed of v = 170 m / s, when arriving at airport B it turns towards the East and travels from x = 100 10³ m, until' it the distance reaches the airport C

Let's use the Pythagoras theorem to find the distance traveled

               R = Ra x² + y²

               R =   10³

               R = 373.6 10³ m

They indicate the average speed for which we can use the uniform motion ratio

               v = \frac{\Delta y }{t}

                t = \frac{\Delta y}{v}

They ask for the time in in from airport A to B, we calculate

                t = 360 10 ^ 3/170

                t = 2.117 10³ s

In conclusion we use the addition of vectors the uniform motion we can find the answer for the question of distance and time are:

  • The distance to go between airports A and C B is 373.6 10³ m
  • The time to go from airport A to B is 2117 s

Learn more here: brainly.com/question/15074838

5 0
3 years ago
A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it aroun
VARVARA [1.3K]

Answer:

(a) f= 622.79 Hz

(b) f= 578.82 Hz

Explanation:

Given Data

Frequency= 600 Hz

Distance=1.0 m

n=120 rpm

Temperature =20 degree

Before solve this problem we need to find The sound generator moves on a circular with tangential velocity

So

Speed of sound is given by

c = √(γ·R·T/M) ............in an ideal gas

where γ heat capacity ratio

R universal gas constant

T absolute temperature

M molar mass

The speed of sound at 20°C is

c = √(1.40 ×8.314472J/molK ×293.15K  / 0.0289645kg/mol)

c= 343.24m/s

The sound moves on a circular with tangential velocity

vt = ω·r.................where   ω=2·π·n

vt= 2·π·n·r

vt= 2·π · 120min⁻¹ · 1m

vt= 753.6 m/min  

convert m/min to m/sec

vt= 12.56 m/s

Part A

For maximum frequency is observed  

v = vt

f = f₀/(1 - vt/c )

f= 600Hz / (1 - (12.56m/s / 343.24m/s) )  

f= 622.789 Hz

Part B

For minimum frequency is observed

v = -vt

f = f₀/(1 + vt/c )

f= 600Hz / (1 + (12.56m/s / 343.24m/s) )

f= 578.82 Hz

3 0
3 years ago
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