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Brut [27]
3 years ago
13

What climate is characterized by constant high temperatures at sea level with an average temperature of 18 °C or higher, and is

found near the equator? A) arid B) polar C) tropical D) temperate
Physics
2 answers:
NeTakaya3 years ago
4 0

Answer:

c. tropical

Explanation:

Tropical Climate is characterized by constant high temperature at sea level with an average temperature of 18°C or higher and is found near the equator.

Areas with tropical climate have high humidity and temperature. Countries in tropical region have high precipitation rate. Tropical climate is also known as equatorial climate. Tropical regions are found near equator. Due to high convection rate, there is abundant rainfall.

kvasek [131]3 years ago
3 0

Correct answer id option C that is tropical.

Tropical Climate is characterized by constant high temperature at sea level with an average temperature of 18°C or higher and is found near the equator.

Areas with tropical climate have high humidity and temperature. Countries in tropical region have high precipitation rate. Tropical climate is also known as equatorial climate. Tropical regions are found near equator. Due to high convection rate, there is abundant rainfall.

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a satellite with a mass of 1800 kg is placed at an altitude of 594 km above the surface of Europa. What is the strength of the g
saul85 [17]
The formula for gravitational potential energy is Ep= mgh. Lets convert 594 km into m: 594 x 1,000 = 594,000 m. So we do: Ep= (1,800kg)(9.8N/kg)(594,000m) = (17,640N)(594,000) = 10,478,160,000 J.
4 0
3 years ago
A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+88 determines the height of the rock above
babymother [125]

1) 88 ft

2) 4.09 s

3) 1.38 s

4) 118.2 m

Explanation:

1)

For an object thrown upward and subjected to free fall, the height of the object at any time t is given by the suvat equation:

h(t) = h_0 + ut - \frac{1}{2}gt^2 (1)

where

h_0 is the height at time t = 0

u is the initial vertical velocity

g=32 ft/s^2 is the acceleration due to gravity

The function that describes the height of the rock above the surface at a time t in this problem is

f(t)=-16t^2+44t+88 (2)

By comparing the terms with same degree of eq(1) and eq(2), we observe that

h_0 = 88 ft

which means that the rock is at height h = 88 ft when t = 0: therefore, this means that the height of the bridge above the water is 88 feet.

2)

The rock will hit the water when its height becomes zero, so when

f(t)=0

which means when

0=-16t^2+44t+88

First of all, we can simplify the equation by dividing each term by 4:

0=-4t^2+11t+22

This is a second-order equation, so we solve it using the usual formula and we find:

t_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-11\pm \sqrt{(11)^2-4(-4)(22)}}{2(-4)}=\frac{-11\pm \sqrt{121+352}}{-8}=\frac{-11\pm 21.75}{-8}

Which gives only one positive solution (we neglect the negative solution since it has no physical meaning):

t = 4.09 s

So, the rock hits the water after 4.09 seconds.

3)

Here we want to find how many seconds after being thrown does the rock reach its maximum height above the water.

For an object in free fall motion, the vertical velocity is given by the expression

v=u-gt

where

u is the initial velocity

g is the acceleration due to gravity

t is the time

The object reaches its maximum height when its velocity changes direction, so when the vertical velocity is zero:

v=0

which means

0=u-gt

Here we have

u=+44 ft/s (initial velocity)

g=32 ft/s^2 (acceleration due to gravity)

Solving for t, we find the time at which this occurs:

t=\frac{u}{g}=\frac{44}{32}=1.38 s

4)

The maximum height of the rock can be calculated by evaluating f(t) at the time the rock reaches the maximum height, so when

t = 1.38 s

The expression that gives the height of the rock at time t is

f(t)=-16t^2+44t+88

Substituting t = 1.38 s, we find:

f(1.38)=-16(1.38)^2 + 44(1.38)+88=118.2 m

So, the maximum height reached by the rock during its motion is

h_{max}=118.2 m

Which means 118.2 m above the water.

3 0
4 years ago
Light travels at a speed of 3.00 x 10^11 cm/s. What is the speed of light in kilometers/hour?
rewona [7]

As 1 km = 1000 m = 1000,00 cm,

So, 1 cm = (1/1000,00) km

1 hour = 60 × 60 s = 3600 s

So,  1 s = (1 / 3600) hour

The light travels at a speed of 3.00 \times 10^{11}  \ cm/s.

In kilometer/hour,

3.00 \times 10^{11}  \ cm/s = 3.00 \times 10^{11} \frac{(1/1000,00) km}{ (1 / 3600) hour} = 108 \times 10^8 \ km/hour


4 0
4 years ago
Find the magnitude of the emf induced in the rod when it is moving toward the right with a speed 7.50 m/s.
Flauer [41]

Answer:

3 volts

Explanation:

It is given that,

Magnetic field, B = 0.8 T

Length of a conducting rod, l = 50 cm = 0.5 m

Velocity of the conducting rod, v = 7.5 m/s

We need to find the magnitude of the emf induced in the rod when it is moving toward the right. When a rod is moved in a magnetic field, an emf is induced in it and it is given by :

\epsilon=Blv

Putting all the values,

\epsilon=0.8\times 0.5\times 7.5\\\\\epsilon=3\ V

So, the magnitude of the emf induced in the rod is 3 volts.

6 0
3 years ago
A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.7 1010 m (inside the orbit
Lubov Fominskaja [6]

Answer:

58515.9 m/s

Explanation:

We are given that

d_1=4.7\times 10^{10} m

v_i=9.5\times 10^4 m/s

d_2=6\times 10^{12} m

We have to find the speed (vf).

Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.

K_i+U_i=K_f+U_f

\frac{1}{2}mv^2_i-\frac{GmM}{d_1}=\frac{1}{2}mv^2_f-\frac{GmM}{d_2}

\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2}=\frac{1}{2}v^2_f

v^2_f=2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})

v_f=\sqrt{2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})}

Using the formula

v_f=\sqrt{v^2_i+2GM(\frac{1}{d_2}-\frac{1}{d_1})}

v_f=\sqrt{(9.5\times 10^4)^2+2\times 6.7\times 10^{-11}\times 1.98\times 10^{30}(\frac{1}{6\times 10^{12}}-\frac{1}{4.7\times 10^{10})}

Where mass of sun=M=1.98\times 10^{30} kg

G=6.7\times 10^{-11}

v_f=58515.9 m/s

4 0
3 years ago
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