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What amount of heat is removed to lower the temperature of 80 grams of water from 75ºC to 45ºC? The specific heat of liquid wate

amm1812

amount of heat is removed to lower the temperature of 80 grams of water from 75ºC to 45ºC with specific heat of liquid water is 4.18 J/gºC

= 4.18 J/gºC * 80g * (75-45)ºC

= 10032J

3 0

3 years ago

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A pendulum of length 130.0 cm has a period of oscillation, T₁. The bob is pulled and released to move in a horizontal circle of

Anastasy [175]

The ratio of T1 : T2 is 1.6

<h3>What is Simple Harmonic Motion ?</h3>

SHM is the periodic motion of a particle along a line such that the acceleration of the body is directed toward a fixed point and proportional to its displacement.

Given that a pendulum of length 130.0 cm has a period of oscillation, T₁.

Then, The period T₁ = 2 $\pi$ $\sqrt{\frac{l}{g} }$

T₁ = 2 x 3.142 $\sqrt{\frac{1.3}{10} }$

T₁ = 6.284 x 0.36

T₁ = 2.266

But if T = 0

then, m $v^{2}$ /r = mg

the m will cancel out

Then v = $\sqrt{rg}$

where r = 50/100 = 0.5m

Substitute into the formula

v = $\sqrt{ 0.5 * 10}$

v = 2.24 m/s

Also, given that the bob is pulled and released to move in a horizontal circle of radius 50.0 cm and the period of rotation is T2,

w = 2 $\pi$ f

w = 2 $\pi$ /T

but v = wr

w = v/r

Therefore,

v/r = 2 $\pi$ /T

Substitute all the parameters into the formula

2.24/0.5 = (2 x 3.1423) /T

T = 6.284/4.472

T = 1.405 = $T_{2}$

The ratio T₁: T2 will be 2.266/1.405

The ratio is 1.6

Learn more about Simple Harmonic Motion here: brainly.com/question/24646514

#SPJ1

4 0

2 years ago

Why are antennas needed for radio, television, and cell phone transmission>

erastova [34]

Answer:

A.without a long antenna, 're signal will not travel far

5 0

3 years ago

A 0.25 kg mass is placed on a vertically oriented spring that is stretched 0.56 meters from its equilibrium position. If the spr

Vitek1552 [10]

Answer:

The speed of the ball when it reaches equilibrium position is 3.31 m/s

Explanation:

Given;

mass of the object, m = 0.25 kg

initial displacement of the object, h₁ = 0.56 m

spring constant, k = 105 N/m

displacement at equilibrium position, h₂ = 0

initial velocity of the object, v₁ = 0

velocity of the object at equilibrium position = v₂

The change in gravitational potential energy at the equilibrium position is given as;

ΔP.E = mg(h₂ - h₁)

The change in kinetic energy of the object at the equilibrium position is given as;

ΔK.E = ¹/₂m(v₂² - v₁²)

Apply the principle of conservation of mechanical energy;

ΔK.E + ΔP.E = 0

¹/₂m(v₂² - v₁²) + mg(h₂ - h₁) = 0

¹/₂m(v₂² - 0) + mg(0 - h₁) = 0

¹/₂mv₂² - mgh₁ = 0

¹/₂mv₂² = mgh

¹/₂v₂² = gh

v₂² = 2gh

v₂ = √2gh

v₂ = √(2 x 9.8 x 0.56)

v₂ = 3.31 m/s

Therefore, the speed of the ball when it reaches equilibrium position is 3.31 m/s

8 0

3 years ago

A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i

sukhopar [10]

Answer:

The velocity of the truck after the collision is 20.93 m/s

Explanation:

It is given that,

Mass of car, m₁ = 1200 kg

Initial velocity of the car, $v_{Ci}=25\ m/s$

Mass of truck, m₂ = 9000 kg

Initial velocity of the truck, $v_{Ti}=20\ m/s$

After the collision, velocity of the car, $v_{Cf}=18\ m/s$

Let $v$ is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.

$initial\ momentum=final\ momentum$

$1200\ kg\times 25\ m/s+9000\ kg\times 20\ m/s=1200\ kg\times 18+9000\ kg\times v$

$210000-21600=9000\ kg\times v$

$v=20.93\ m/s$

So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.

8 0

3 years ago