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2 years ago

Describe the effects of heat

Physics

2 answers:

astraxan [27]2 years ago

8 0

1. Heat raises the temperature.

2. It increases volume.

3. It changes state.

4. Brings about chemical action.

5. Changes physical properties.

<h3>Hope this helps :)</h3>

In-s [12.5K]2 years ago

3 0

Answer:

Prolonged exposure to extreme heat can cause heat exhaustion, heat cramps, heat stroke, and death, as well as exacerbate preexisting chronic conditions, such as various respiratory, cerebral, and cardiovascular diseases.

hope this helps! :)
-goodluck!!

also, please mark me as brainliest, it would really help me out! thanks :)

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What term would you use to describe the movement of thermal energy between a person’s body and the frigid water she jumped into

inysia [295]

It's called heat transfer

3 0

4 years ago

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Two wires carry current I1 = 73 A and I2 = 31 A in the opposite directions parallel to the x-axis at y1 = 3 cm and y2 = 13 cm. W

bogdanovich [222]

Answer:

The position on the y-axis where the magnetic field is zero is at y = 10 cm

Explanation:

The magnetic field B due to a long straight wire carrying a current, i at a distance R from the wire is given by

B = μ₀i/2πR

Now, let y be the point where the magnetic fields of both wires are equal.

So, the magnetic field due to wire 1 carrying current i₁ = 73 A is

B₁ = μ₀i₁/2π(y - 3) and

the magnetic field due to wire 2 carrying current i₂ = 31 A is

B₂ = μ₀i₂/2π(13 - y)

At the point where the magnetic field is zero, B₁ = B₂. So,

μ₀i₁/2π(y - 3) = μ₀i₂/2π(13 - y)

cancelling out μ₀ and 2π, we have

i₁/(x - y) = i₂/(13 - y)

cross-multiplying, we have

(13 - y)i₁ = (y - 3)i₂

Substituting the values of i₁ and i₂, we have

(13 - y)73 = (y - 3)31

949 - 73y = 31y - 93

Collecting like terms, we have

949 + 93 = 73y + 31y

1042 = 104y

dividing through by 104, we have

y = 1042/104

y = 10.02 cm

y ≅ 10 cm

So, the position on the y-axis where the magnetic field is zero is at y = 10 cm

6 0

3 years ago

PLS HELP. WILL GIVE BRAINLIEST. PLS THIS IS WORTH 35 POINTS ON MY TEST

Nata [24]

Answer:

18m/s^2

Explanation:

Vf = Vi + at

t = distance/ average velocity

(120 + 0)/2 = 60 (average velocity)

400m/60m/s = 20/3 s

insert into first equation:

120 = 0 + a(20/3)

360 = 20a

18 = a

HOPE THIS HELPS!!!

5 0

2 years ago

Read 2 more answers

An 80.0-kg man jumps from a height of 2.50 m onto a platform mounted on springs. As the springs compress, he pushes the platform

frosja888 [35]

Answer:

6.16 m/s

0.0105 m

Explanation:

Let the ground 0 for potential reference be at where the spring is compress 0.24 m. The the man would jump from a height h = 2.5 + 0.24 = 2.74 m from it. We can apply the law of energy conservation knowing that as the man jumps, his potential energy converts to kinetic energy, then finally to elastic energy:

$E_p = E_e$

$mgh = kx^2/2$

where m = 80 kg is the man mass, g = 9.81 m/s2 is the gravitational acceleration, h = 2.74 m is the potential distance he travels, k N/m is the spring constant and x = 0.24 is the distance it compresses

$80*9.81*2.74 = k0.24^2/2$

$2150.352 = 0.0288k$

$k = 74665 N/m$

Similarly at the position where it compresses by 0.12 m, it's 0.24 - 0.12 = 0.12 m far from ground 0.

$E_p = E_{e2} + E_k + E_{p2}$

$mgh = kx_2^2 + mv^2/2 + mgh_2$

$2150.352 = 74665*0.12^2/2 + 80v^2/2 + 80*9.81*0.12$

$2150.352 = 537.588 + 40v^2 + 94.176$

$40v^2 = 1518.588$

$v^2 = 37.9647$

$v = \sqrt{37.9647} = 6.16 m/s$

When he steps gently, then his gravity force would equal to his spring force

$mg = kx_3$

$x_3 = mg/k = 80*9.81/74665 = 0.0105m$

7 0

3 years ago

Determine the centroid of the shaded area shown in figure 2. Determine the moment of inertia about y-axis of the shaded area sho

Nady [450]

Answer:

centroid: (x, y) = (81.25 mm, 137.5 mm)
I = 8719.31 mm^2 for unit mass

Explanation:

Finding the desired measures requires we know a differential of area. That, in turn, requires we have a way to describe a differential of area. Here, we choose to use a vertical slice, which requires we know the area boundaries as a function of x.

The upper boundary is a line with a slope of 125/156.25 = 0.8, and a y-intercept of 125. That is, ...

y1 = 0.8x +125

The lower boundary is given in terms of y, but we can solve for y to find ...

100x = y^2

y2 = 10√x

Then our differential of area is ...

dA = (y1 -y2)dx

The centroid is found by computing the first moment about the x- and y-axes, and dividing those values by the area of the figure.

The area will be ...

$\displaystyle A=\int_0^{156.25}{dA}=\int_0^{156.25}{(y_1-y_2)}\,dx$

The y-coordinate of the centroid is ...

$\displaystyle \overline{y}=\dfrac{S_x}{A}=\dfrac{1}{A}\int_0^{156.25}{\dfrac{y_1+y_2}{2}}\,dA=\dfrac{1}{A}\int_0^{156.25}{\dfrac{y_1+y_2}{2}(y_1-y_2)}\,dx=137.5$

Similarly, the x-coordinate is ...

$\displaystyle \overline{x}=\dfrac{S_y}{A}=\dfrac{1}{A}\int_0^{156.25}{x}\,dA=\dfrac{1}{A}\int_0^{156.25}{x(y_1-y_2)}\,dx=81.25$

That is, centroid coordinates are (x, y) = (81.25, 137.5) mm.

The moment of inertia is the second moment of the area. If we normalize by the "mass" (area), then the integral looks a lot like the one for $\overline{x}$ , but multiplies dA by x^2 instead of x.

The attachment shows that value to be ...

I ≈ 8719.31 mm^2 (normalized by area)

The area is 16276.0416667 mm^2, if you want to "un-normalize" the moment of inertia.

7 0

3 years ago

Describe the effects of heat​

Describe the effects of heat