I’m a god, a planet, and I measure heat. What am I?

Although the temperature gradient changes from region to region in the homosphere, there is one gradient that stays the same. it

bezimeni [28]

Answer: the answer is air gradient

Explanation:

3 0

2 years ago

A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift i

bija089 [108]

Answer

Mass m = 78 kg

Vertical height in each stage h = 11 m

(a).

Initial speed u = 0

Final speed v = 1.1 m / s

$v^2=u^2 + 2 as$

$1.1^2 = 2 a \times 11$

a = 0.055 m/s²

Work done

$W_a= m g h + \dfrac{1}{2}mv^2$

$W_a= 78\times 9.8 \times 11 + \dfrac{1}{2} 78 \times 1.1^2$

$W_a = 8408.4 + 47.19$

$W_a = 8455.59 J$

(b).Work done

$W_b= mgh$

W_b = 78× 9.8× 11

$W_b= 8408.4 J$

c)

Work done

$W= m g h + \dfrac{1}{2}m(v_f-v_i)^2$

Where V = final speed

= 0

v = 1.1 m / s

for deceleration a = -0.055 m/s²

now,

$F_L = 56 (-0.055+9.8) = 545.72 N$

W_c = 545.75 × 11

$W_c = 6003.25 J$

7 0

2 years ago

A rock hits a window and stops in 0.15 seconds. The net force on the rock is 58N during the collision. What is the magnitude of

nlexa [21]

Answer:

The change in momentum is $\Delta p = 0.7 \ kg\cdot m \cdot s^{-1}$

Explanation:

From the question we are told that

The time taken for the stone to stop is $\Delta t = 0.15 \ seconds$

The net force on the rock is $F = 58 \ N$

The impulse of the rock can be mathematically represented as

$I = F * \Delta t$

Substituting values

$I = 58 * 0.15$

$I = 0.7\ kg * m * s^{-1}$

Now impulse is defined as the rate at which momentum change

Hence the change in momentum $\Delta p$ of the rock is equal to the impulse of the rock

So

$\Delta p = I = 0.7 \ kg\cdot m \cdot s^{-1}$

7 0

3 years ago

Use the drop-down menus to complete each sentence about the layers of the atmosphere. If the did not exist, Earth might be destr

ozzi

Answer:

If the mesosphere did not exist, Earth might be destroyed by chunks of rock from space.

The stratosphere is located 12 to 50 kilometers from Earth's surface.

Both the trophosphere and mesosphere get colder as altitude increases.

The ozone in the stratosphere protects people from ultraviolet (UV) radiation.

The thermosphere has the highest temperature of any layer in Earth's atmosphere.

Explanation:

Different layers of the atmosphere are;

Troposphere

It is the lowest layer of atmosphere that begins at ground level and extends upward to about 10 km. In this layer, all organisms live, weather occurs and most clouds appear. As altitude increases, the temperature decreases i.e., the layer gets colder.

Stratosphere

This layer is located at the top of the troposphere and extends up to about 50 km from Earth's surface. The ozone layer present in this layer absorbs high-energy ultraviolet (UV) light from the Sun. Here, the temperature rises with altitude. The passenger jets are usually fly in this layer.

Mesosphere

Above the stratosphere, the mesosphere extends upward to about 85 km from Earth's surface. The chunks of rock from space such as meteors burn upon entering the mesosphere and protect the Earth. In this layer, the temperature gets colder as altitude increases and so the coldest temperatures (about -90° C) in Earth's atmosphere are found at the top of this layer.

Thermosphere

This layer is located above the mesosphere and the air here is really very thin. The thermosphere absorbs high-energy X-rays and UV radiation from the Sun and as a result, its temperature increases to hundreds or thousands of degrees i.e., 1,500° C or higher. This layer extends up to 500 to 1,000 km from the ground. Many satellites orbit Earth within this layer.

Ionosphere

The ionosphere is actually an extension of the thermosphere and so it is not a distinct layer like the others. It is filled with charged particles produced from the ionization of molecules by high temperatures in the thermosphere caused by the high-energy radiation from the Sun. The phenomenon called auroras (Northern Lights and Southern Lights) occurs in this layer. The different regions of the ionosphere reflecting the radio waves back to Earth and make the long-distance radio communication possible.

4 0

2 years ago

Read 2 more answers

A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c

saw5 [17]

Answer:

The final velocity of the thrower is $\bf{3.88~m/s}$ and the final velocity of the catcher is $\bf{0.029~m/s}$ .

Explanation:

Given:

The mass of the thrower, $m_{t} = 70~Kg$ .

The mass of the catcher, $m_{c} = 55~Kg$ .

The mass of the ball, $m_{b} = 0.0480~Kg$ .

Initial velocity of the thrower, $v_{it} = 3.90~m/s$

Final velocity of the ball, $v_{fb} = 33.5~m/s$

Initial velocity of the catcher, $v_{ic} = 0~m/s$

Consider that the final velocity of the thrower is $v_{ft}$ . From the conservation of momentum,

$&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s$

Consider that the final velocity of the catcher is $v_{fc}$ . From the conservation of momentum,

$&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s$

Thus, the final velocity of thrower is $3.88~m/s$ and that for the catcher is $0.029~m/s$ .

8 0

2 years ago