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sladkih [1.3K]
2 years ago
5

PLS HELP! I’ll give Brainly!

Physics
2 answers:
mojhsa [17]2 years ago
8 0
<h2><em><u>Answer:</u></em></h2><h2><em><u>Answer:The sun is what makes the water cycle work. The sun provides what almost everything on </u></em><em><u>Earth</u></em></h2>

<em><u> needs to go—energy, or heat.</u></em>

<em><u>please</u></em><em><u> </u></em><em><u>follow </u></em><em><u>me </u></em><em><u>mark </u></em><em><u>me </u></em><em><u>as </u></em><em><u>brainlist </u></em>

Zanzabum2 years ago
5 0

Answer:

4

Explanation:

water cycle is driven primarily by the energy from the sun

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A snowball is thrown with an initial x velocity of 7.5 m/s and an initial y velocity of 8.4 m/s . Part A How much time is requir
elena55 [62]

In order to calculate the time taken by the snowball to reach the highest point in its journey, we need to consider the variables along the y-direction.

Let us list out what we know from the question so that we can decide on the equation to be used.

We know that Initial Y VelocityV_{iy} = 8.4 m/s

Acceleration in the Y direction a_{y} = -9.8 m/s^{2}, since the acceleration due to gravity points in the downward direction.

Final Y Velocity V_{fy} = 0 because at the highest point in its path, an object comes to rest momentarily before falling down.

Time taken t = ?

From the list above, it is easy to see that the equation that best suits our purpose here is V_{fy} = V_{iy} + a_{y}t

Plugging in the numbers, we get 0 = 8.4 - (9.8)t

Solving for t, we get t = 0.857 s

Therefore, the snowball takes 0.86 seconds to reach its highest point.

8 0
3 years ago
A 500 kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 30 N/m. The blo
m_a_m_a [10]

Answer:

x = 0.396 m

Explanation:

The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is   spring

Data the putty has a mass m1 and velocity vo1, the block has a mass m2 .  t's start using the moment to find the system speed.

Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash

    p₀ = m1 v₀₁

Moment after shock

    p_{f} = (m1 + m2) v_{f}

   p₀ = p_{f}

   m1 v₀₁ = (m1 + m2) v_{f}

  v_{f} = v₀₁ m1 / (m1 + m2)

   v_{f}= 4.4 600 / (600 + 500)

  v_{f} = 2.4 m / s

With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring

Before compressing the spring

   Em₀ = K = ½ (m1 + m2) v_{f}²

After compressing the spring

   E_{mf} = Ke = ½ k x²

As there is no rubbing the energy is conserved

   Em₀ = E_{mf}

   ½ (m1 + m2) v_{f}² = = ½ k x²

   x = v_{f} √ (k / (m1 + m2))

   x = 2.4 √ (11/3000)

   x = 0.396 m

7 0
3 years ago
What is the difference between a physical quantity and a unit​
Shalnov [3]

Answer: What is the difference of physical quanity and a unit?

Explanation: If you are working science or math problems, the answer to this question is that quantity is the amount or numerical value, while the unit is the measurement. For example, if a sample contains 453 grams, the quantity is 453 while the unit is grams.

4 0
3 years ago
An uniform electric field of magnitude E = 100 N/C is oriented along the positive y-axis. What is the magnitude of the flux of t
Ede4ka [16]

Answer:

The magnitude of the flux of electric field through a square of surface area is zero.

Explanation:

E=100 NC^{-1}\\\\A=2 m^2\\\\Electic\,\,flux\,\,flux\,\,is\,\,given\,\,as:\\\\\phi_E=E.A\,cos\,\theta

It is given that square box is parallel to yz-plane which has normal vector perpendicular to plane in x-direction. Angle between normal vector of area and electric field is 90°. Substituting in (1)

\phi_E=E.A\,cos\,(90^o)\\\\\phi_E=0

4 0
3 years ago
Cooling causes a material to
e-lub [12.9K]

Answer:

whats the question?

Explanation:

4 0
3 years ago
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