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levacccp [35]
3 years ago
13

A soccer player heads the ball and sends it flying vertically upwards at a speed of 18.0 m/s . How high above the players ' head

does the ball travel ?​
Physics
1 answer:
Irina-Kira [14]3 years ago
6 0

Answer:

16.53 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 18.0 m/s.

Final velocity (v) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (h) =.?

The maximum height reached by the ball can be obtained as follow:

v² = u² – 2gh (since the ball is going against gravity)

0² = 18² – (2 × 9.8 × h)

0 = 324 – 19.6h

Rearrange

19.6h = 324

Divide both side by 19.6

h = 324 / 19.6

h = 16.53 m

Therefore, the maximum height reached by the ball is 16.53 m

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A student places blocks on a 100cm long see-saw as shown/
Hunter-Best [27]

Answer:

Part 1)

\tau_1 = 5 \times (0.50) = 2.5 N m

Part 2)

\tau_2 = 14 \times (0.30) = 4.2 N m

Part 3)

\tau_3 = 1.4 N m

Part 4)

Since torque on right side is more so here it will turn and slip over it

Explanation:

As we know that the block A is placed at distance

d = 50 cm from the hinge at 70 cm mark

So torque due to weight of A is given as

\tau_1 = 5 \times (0.50) = 2.5 N m

the block B is placed at distance

d = 30 cm from the hinge at 70 cm mark

So torque due to weight of B is given as

\tau_2 = 14 \times (0.30) = 4.2 N m

Now torque due to weight of the scale is given as

\tau_3 = 7(0.20)

\tau_3 = 1.4 N m

now torque on left side of scale is given as

\tau_{left} = \tau_1 + \tau_3

\tau_{left} = 2.5 + 1.4 = 3.9 N m

Torque on right Side is given as

\tau_{right} = \tau_2 = 4.2 Nm

Since torque on right side is more so here it will turn and slip over it

8 0
3 years ago
An example of potential energy is a ball sitting _____ of the stairs.
expeople1 [14]

Answer:

at the top

Explanation:

Potential energy is the stored energy, mechanical energy,

or energy possessed by by virtue of the position of an object.an example of potential energy is the energy that a ball possesses by virtue of its sitting at the top of the stairs it being about to roll down the stairs.

3 0
3 years ago
How many lines per mm are there in the diffraction grating if the second order principal maximum for a light of wavelength 536 n
grandymaker [24]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

dSin\theta = m\lambda

Where

m= Any integer which represent the number of repetition of spectrum

\lambda= Wavelength

d = Distance between the slits.

\theta= Angle between the difraccion paterns and the source of light

Re-arrange to find the distance between the slits we have,

d = \frac{m\lambda}{sin\theta }

d = \frac{2*536*10^{-9}}{sin(24)}

d = 2.63*10^{-6}m

Therefore the number of lines per millimeter would be given as

\frac{1}{d} = \frac{1}{2.63*10^{-6} }

\frac{1}{d} = 379418.5\frac{lines}{m}(\frac{10^{-3}m}{1 mm})

\frac{1}{d} = 379.4 lines/mm

Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm

6 0
3 years ago
You're carrying a 3.2-m-long, 24kg pole to a construction site when you decide to stop for a rest. You place one end of the pole
Aleksandr [31]

Answer:

Tension of 132N

Explanation:

We need to apply Summatory of Force to find the tension in the hand.

We define te tensión in the hand as F_2 and the Tension in fence post as F_1, then

\sum F = 24(9.8)

F_1 + F_2= 24(9.8)

We apply summatory of moments then

F_2*1.25 = F_1*1.6

Where the Force 2 is 1.25m from the center of summatory,

We can note that,

1.6 m - 0.35m=1.25m

We have two equation and two incognites, then replacing (1) in (2)

1.6(235.2 -F_2) = 1.25F_2

376.32 = F2(1.6+1.25)

F_2= \frac{376.32}{2.85}

F_2 =132 N

5 0
3 years ago
A 65.0 kg diver is 4.90 m above the water, falling at speed of 6.40 m/s. Calculate her kinetic energy as she hits the water. (Ne
mojhsa [17]

Answer:

4452.5 J.

Explanation:

The diver have both kinetic and potential energy.

Ek = 1/2mv² ................. Equation 1

Where Ek = Kinetic Energy of the diver, m = mass of the diver, v = velocity of the diver.

Given: m = 65 kg, v = 6.4 m/s.

Substitute into equation 1

Ek = 1/2(65)(6.4²)

Ek = 1331.2 J.

Also,

Ep = mgh ............................ Equation 2

Where Ep =  Potential energy of the diver when its above the water, h = height of the diver above the water, g = acceleration due to gravity.

Given: m = 65 kg, h = 4.9 m, g = 9.8 m/s²

Substitute into equation 2.

Ep = 65(4.9)(9.8)

Ep = 3121.3 J.

Note: When she hits the water, the potential energy is converted to kinetic energy.

E = Ek+Ep

Where E = Kinetic energy of the diver when she hits the water.

E = 1331.2+3121.3

E = 4452.5 J.

3 0
3 years ago
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