All categories

3 years ago

What do you call the warming effect some gases have on the atmosphere by absorbing and radiating heat?

1 answer:

hjlf3 years ago

3 0

The greenhouse effect is the process by which radiation from a planet's atmosphere warms the planet's surface to a temperature above what it would be without its atmosphere.

You might be interested in

The higher the amplitude of the wave, the greater its intensity and the greater its loudness.

lyudmila [28]

True,when you turn the volume up on your television , you're actually turning up the amplitude<span>!

</span>

4 0

3 years ago

Suppose we take a 1 m long uniform bar and support it at the 33 cm mark. Hanging a 0.15 kg mass on the short end of the beam res

MakcuM [25]

Answer:

The mass of the beam is 0.074 kg

Explanation:

Given;

length of the uniform bar, = 1m = 100 cm

Set up this system with the given mass and support;

0-----------------33cm-----------------------------------100cm

↓ Δ ↓

0.15kg m

Where;

m is mass of the uniform bar

Apply the principle of moment to determine the value of "m"

sum of anticlockwise moment = sum of clockwise moment

0.15kg(33 - 0) = m(100 - 33)

0.15(33) = m(67)

$m = \frac{0.15kg(33 \ cm)}{67 \ cm}\\\\m = 0.074 \ kg$

Therefore, the mass of the beam is 0.074 kg

8 0

3 years ago

1. Na jaką wysokość wzniesie się piłka kopnięta pionowo w górę z prędkością 20 m/s ? 2. Piłka o masie 20 dag spada z wysokości 1

Tresset [83]

Answer:

its gonn abe 23

Explanation:

6 0

3 years ago

Read 2 more answers

The greater the pull of gravity on an object, the greater the what of that object

Nikitich [7]

Answer:

weight

Explanation:

" the greater the pull of gravity on an object, the greater the weight of that object." In physics, weight is measured in newtons (N), the common unit for measuring force.

3 0

3 years ago

To practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.7 kg and length 3.0 m

Zanzabum

Answer:

$T_{1}$ = 14.88 N

Explanation:

Let's begin by listing out the given variables:

M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,

g = 9.8 m/s²

At equilibrium, the sum of all external torque acting on an object equals zero

τ(net) = 0

Taking moment about $T_{1}$ we have:

(M + m) g * 0.5L - $T_{2}$ (L - d) = 0

⇒ $T_{2}$ = [(M + m) g * 0.5L] ÷ (L - d)

$T_{2}$ = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)

$T_{2}$ = 59.535 ÷ 2.4

$T_{2}$ = 24.80625 N ≈ 24.81 N

Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N

Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N

Using sum of equilibrium in the vertical direction, we have:

$T_{1}$ + $T_{2}$ = W + w ------- Eqn 1

Substituting T2, W & w into the Eqn 1

$T_{1}$ + 24.81 = 26.46 + 13.23

$T_{1}$ = <u>14.88</u> N

8 0

3 years ago