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2 years ago

Energy from the Sun is transferred from the Earth’s surface to the atmosphere, resulting in

Physics

2 answers:

natali 33 [55]2 years ago

4 0

Answer:

C make a the most sense hopefully this helps

Masteriza [31]2 years ago

4 0

Energy from the sun is important so the correct answer is c
Explanation

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Order the speed of sound through these materials from the slowest to the fastest.

Sholpan [36]

Speed of sound in cold air < Speed of sound in Warm air < Speed of sound in hot molten lead < Speed of sound in water

Explanation:

Step 1:

Speed of sound in water varies from 1450 to 1498 meters per second

Speed of sound in Hot Molten lead is approximately 1210 meters per second

Speed of sound in warm air is approximately 338.89 meters per second

Speed of sound in cold air is approximately 293.33 meters per second

Step 2:

In warm air sound travels faster than that of sound travelling nature in cold air.

∴ Speed of sound in cold air < Speed of sound in Warm air < Speed of sound in hot molten lead < Speed of sound in water

Speed of sound in cold air the slowest while Speed of sound in water is the fastest mean.

8 0

2 years ago

How long would it take an object to reach the ground from the top of a building that is 470 feet tall? Round to the nearest tent

Zinaida [17]

Answer:

It would take the object 5.4 s to reach the ground.

Explanation:

Hi there!

The equation of the height of a free-falling object at any given time, neglecting air resistance, is the following:

h = h0 + v0 · t + 1/2 · g · t²

Where:

h = height of the object at time t.

h0 = initial height.

v0 = initial velocity.

g = acceleration due to gravity (-32.2 ft/s² considering the upward direction as positive).

t = time

Let´s supose that the object is dropped and not thrown so that v0 = 0. Then:

h = h0 + 1/2 · g · t²

We have to find the time at which h = 0:

0 = 470 ft - 1/2 · 32.2 ft/s² · t²

Solving for t:

-470 ft = -16.1 ft/s² · t²

-470 ft / -16.1 ft/s² = t²

t = 5.4 s

3 0

3 years ago

A 0.15 kg baseball is hit by a baseball bat. Right before it is hit, the ball’s speed is 30 m/s, and right after it is hit, the

Roman55 [17]

Answer:

Impulse will be 12 kgm/sec

So option (b) will be correct option

Explanation:

We have given mass of the baseball m = 0.15 kg

Ball speed before hit $v_1=30m/sec$

Ball speed after hitting $v_2=-50m/sec$ ( negative direction due to opposite direction )

We have to find the impulse

We know that impulse is equal; to the change in momentum

So change in momentum = $m(v_1-v_2)=0.15(30-(-50))=0.15\times 80=12kgm/sec$

So option (b) will be correct option

5 0

2 years ago

A 5000-lb wrecking ball hangs from a 50-ft cable of density 10 lb/ft attached to a crane. Calculate the work done if the crane l

aniked [119]

Answer:

total work is = 52450 J

Explanation:

given data

mass = 5000-lb

density = 10 lb/ft

height = 50 ft

solution

as we will treat here cable and ball are separate

and

here work need to lift cable is

w = (10Δy )(9.8 y ) j

and

now summing all segment of cable

so passing limit Δy to 0

so total work need

= $\int\limits^{10}_0 {98y} \, dy$

= $[49 y^2]^{50}_0$

= 2450J

so lifting 5000 lb wrcking 50 m required additional 5000 + 2450

so total work is = 52450 J

3 0

2 years ago

If the radius of the equipotential surface of the point charge is 14.3 m at a potential of 2.20 kV, what will be the magnitude o

PSYCHO15rus [73]

Answer:

The magnitude of the point charge is 3.496 x 10⁻⁶ C

Explanation:

Given;

radius of the surface, r = 14.3 m

magnitude of the potential, V = 2.2 kV = 2,200 V

The magnitude of the point charge is calculated as follows;

$V = (\frac{1}{4\pi \epsilon _0} )(\frac{Q}{r} )\\\\V = \frac{KQ}{r} \\\\Q = \frac{Vr}{K} \\\\Q = \frac{2,200 \times 14.3}{9\times 10^9} \\\\Q = 3.496 \times 10^{-6} \ C\\\\Q = 3.496 \ \mu C$

Therefore, the magnitude of the point charge is 3.496 μC

6 0

2 years ago