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kumpel [21]
3 years ago
15

A volleyball player's hand applies a 39 N force while in contact with a volleyball for 2 seconds. What is the impulse on the bal

l? 20 N • s 37 N • s 41 N • s 78 N • s
Physics
1 answer:
antiseptic1488 [7]3 years ago
8 0

Answer:78 N

Explanation:Impulse=Force*time

=39*2 =78 Ns

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A wall has a negative charge distribution producing a uniformhorizontal electric field. A small plastic ball of mass .01kg carry
Sauron [17]

Answer:

a)  E = -4 10² N / C , b) x = 0.093 m, c)     a = 10.31 m / s², θ=-71.9⁰

Explanation:

For that exercise we use Newton's second Law, in the attached we can see a free body diagram of the ball

X axis

             F_{e} - T_{x} = m a

Axis y

            T_{y} - W = 0

Initially the system is in equilibrium, so zero acceleration

            Fe = T_{x}  

            T_{y} = W

Let us search with trigonometry the components of the tendency

            cos θ = T_{y} / T

            sin θ = T_{x}  / T

           T_{y} = cos θ

           T_{x}  = T sin θ

We replace

            q E = T sin θ

            mg = T cosθ

             

a) the electric force is

                F_{e} = q E

                E = F_{e} / q

                E = -0.032 / 80 10⁻⁶

                E = -4 10² N / C

b) the distance to this point can be found by dividing the two equations

                q E / mg = tan θ

                θ = tan⁻¹ qE / mg

Let's calculate

              θ = tan⁻¹ (80 10⁻⁶ 4 10² / 0.01 9.8)

              θ = tan⁻¹ 0.3265

               θ = 18 ⁰

               sin 18 = x/0.30

               x =0.30 sin 18

               x = 0.093 m

c) The rope is cut, two forces remain acting on the ball, on the x-axis the electric force and on the axis and the force gravitations

X axis

           F_{e} = m aₓ

            aₓ = q E / m

           aₓ = 80 10⁻⁶ 4 10² / 0.01

           aₓ = 3.2 m / s²

Axis y

           W = m a_{y}

           a_{y} = g

           a_{y} = 9.8 m/s²

The total acceleration is can be found using Pythagoras' theorem

             a = √ aₓ² + a_{y}²

             a = √ 3.2² + 9.8²

             a = 10.31 m / s²

The Angle meet him with trigonometry

               tan θ = a_{y} / aₓ

               θ = tan⁻¹ a_{y} / aₓ

               θ = tan⁻¹ (-9.8) / 3.2

               θ = -71.9⁰

Movement is two-dimensional type with acceleration in both axes

8 0
3 years ago
a skydiver jumps out of a plane. describe how gravitational potential energy changes as the skydiver falls. describe how the sky
Oliga [24]

gravitational potential is directly proportional to the height of the object relative to a reference line and is given as

PE = mgh                        

where m = mass of object , g = acceleration due to                    

gravity   and  h = height of the object above the reference line .


as the skydiver falls , its height above the ground decrease and hence the gravitational potential energy of the skydiver decrease.

as per conservation of energy , total energy of the skydiver must remain constant all the time . hence the decrease in potential energy appears as increase in kinetic energy by same amount to keep the total energy constant

KE + PE = Total energy

so as the skydiver falls , it gains speed and hence the kinetic energy of skydiver increase since kinetic energy is directly proportional to the square of the speed.


when the parachute opens, the skydiver experience force in upward which tries to balance the weight of the skydiver. hence the speed of the skydiver decrease until upward force becomes equal to the downward force. hence the kinetic energy decrease just after the parachute opens

4 0
3 years ago
What is the magnitude of the momentum of a 3400 kg airplane traveling at a speed of 450 miles per hour?
Vitek1552 [10]

Answer:

The magnitude of momentum of the airplane is 6.83\times 10^5\ kg-m/s.

Explanation:

Given that,

Mass of the airplane, m = 3400 kg

Speed of the airplane, v = 450 miles per hour

Since, 1 mile per hour = 0.44704 m/s

v = 201.16 m/s

We need to find the magnitude of momentum of the airplane. It is given by the product of mas and velocity such that,

p=mv

p=3400\ kg\times 201.16 \ m/s

p=683944\ kg-m/s

or

p=6.83\times 10^5\ kg-m/s

So, the magnitude of momentum of the airplane is 6.83\times 10^5\ kg-m/s. Hence, this is the required solution.

7 0
3 years ago
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Ierofanga [76]
Don’t trust those links they usually pull up your IP
4 0
3 years ago
What is principal focus and focus length with examples​
Shkiper50 [21]

Explanation:

Principle focus is the point on the axis of a convex lens, where the parallel rays of light from one side of the lens. meet on other side after refraction. Distance between optical centre to principle focus point is the focal length.

7 0
2 years ago
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