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kkurt [141]
4 years ago
9

You throw a ball straight up, it peaks out, and then cones back down to you. During this motion, the velocity and acceleration

Physics
1 answer:
madreJ [45]4 years ago
8 0

Answer:

The answer is C sometimes point in the same direction, and other times point in opposite to each other.

Explanation:

When you throw a ball straight up velocity direction head up to up side but the acceleration points opposite direction due to gravitation of earth. Gr aviation slows down the ball when it goes up, when it reaches the summit and starts to fall down both velocity and acceleration points the same way. The ball speeds up and drops down.

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A 2 kg block is pushed by an external force against a spring with spring constant 131 N/m until the spring is compressed by 2.1
Andrej [43]

Answer:

   d = 19.92 m

Explanation:

As in this exercise there is friction we must use the relationship between work and energy

          W = ΔEm

Look for energy in two points

Initial. Fully compressed spring

          Em₀ = K_{e} = ½ k x²

Final. When the block stopped

      Em_{f} = 0

Let's look for the work of the rubbing force

       W = fr d cos θ

Since rubbing is always contrary to movement, θ = 180

      W = - fr d

Let's use Newton's second Law, to find the force of friction

Y Axis

          N- w = 0

          N = mg

The equation for the force of friction is

         fr = μ N

         fr = μ mg

We substitute in the work equation

          W = - μ m g d

We write the relationship of work and energy

     -μ m g d = 0 - ½ k x²

     d = ½ k x² / μ m g

Let's calculate

     d = ½ 131 2.1 2 / (0.74 2 9.8)

     d = 19.92 m

7 0
4 years ago
WILL GIVE BRAINLIEST! THANK YOU :)
andrezito [222]

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6 0
3 years ago
Radio waves are a type of______
Verizon [17]

Answer:

I already know the first one is electromagnetic but I don't know the second one sorry

Explanation:

5 0
3 years ago
A) Calculate the pressure of water at the bottom of a well if the depth
Molodets [167]

Answer: 98000pa

Explanation:Given,

Depth(h)=10m

gravity(g)=9.8m/s

density(δ)=1000kg/m^3

we know,

P=hdg

P=10*1000*9.8

P=98000pa

7 0
3 years ago
A small sphere with mass mcarries a positive chargeqand is attached to one end of a silk fiber of lengthL. The other end of the
Aleksandr-060686 [28]

Answer:

(a):  The magnitude of the electric force on the small sphere = \dfrac{q\sigma}{2\epsilon_o}.

(b): Shown below.

Explanation:

<u>Given:</u>

  • m = mass of the small sphere.
  • q = charge on the small sphere.
  • L = length of the silk fiber.
  • \sigma = surface charge density of the large vertical insulating sheet.

<h2>(a):</h2>

When the dimensions of the sheet is much larger than the distance between the charge and the sheet, then, according to Gauss' law of electrostatics, the electric field experienced by the particle due to the sheet is given as:

\rm E = \dfrac{\sigma}{2\epsilon_o}.

<em>where,</em>

\epsilon_o is the electrical permittivity of the free space.

The electric field at a point is defined as the amount of electric force experienced by a unit positive test charge, placed at that point. The magnitude electric field at a point and the magnitude of the electric force on a charge q placed at that point are related as:

\rm F_e=qE.

Thus, the magnitude of the electric force on the small sphere is given by

\rm F_e = q\times \dfrac{\sigma }{2\epsilon_o}=\dfrac{q\sigma}{2\epsilon_o}.

The sheet and the small sphere both are positively charged, therefore, the electric force between these two is repulsive, which means, the direction of the electric force on the sphere is away from the sheet along the line which is perepndicular to the sheet and joining the sphere.

<h2>(b):</h2>

When the sphere is in equilibrium, the tension in the fiber is given by the resultant of the weight of the sphere and the electric force experienced by it as shown in the figure attached below.

According to the fig.,

\rm \tan \theta = \dfrac{F_e}{W}.

<em>where,</em>

  • \rm F_e = electric force on the sphere, acting along left.
  • \rm W = weight of the sphere, acting vertically downwards.

<em />

\rm F_e = \dfrac{q\sigma}{2\epsilon_o}\\\\W=mg\\\\Therefore,\\\\\tan\theta = \dfrac{\dfrac{q\sigma}{2\epsilon_o}}{mg}=\dfrac{q\sigma}{2mg\epsilon_o}.\\\Rightarrow \theta=\tan^{-1}\left ( \dfrac{q\sigma}{2mg\epsilon_o}\right ) .

g is the acceleration due to gravity.

6 0
4 years ago
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