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ArbitrLikvidat [17]
4 years ago
11

a ball is thrown inclined in to the air with a initial velocity of u. if it reaches the maximum height in 6 seconds , find the r

atio to the distance travelled in 1 second to the 7th second... plzz help me​
Physics
1 answer:
olga55 [171]4 years ago
8 0

Answer:

11:1

Explanation:

At constant acceleration, an object's position is:

y = y₀ + v₀ t + ½ at²

Given y₀ = 0, v₀ = u, and a = -g:

y = u t − ½g t²

After 6 seconds, the ball reaches the maximum height (v = 0).

v = at + v₀

0 = (-g)(6) + u

u = 6g

Substituting:

y = 6g t − ½g t²

The displacement between t=0 and t=1 is:

Δy = [ 6g (1) − ½g (1)² ] − [ 6g (0) − ½g (0)² ]

Δy = 6g − ½g

Δy = 5½g

The displacement between t=6 and t=7 is:

Δy = [ 6g (7) − ½g (7)² ] − [ 6g (6) − ½g (6)² ]

Δy = (42g − 24½g) − (36g − 18g)

Δy = 17½g − 18g

Δy = -½g

So the ratio of the distances traveled is:

(5½g) / (½g)

11 / 1

The ratio is 11:1.

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When you have two like charges in a line â where is the electric field the greatest? is there ever a point where the field will
Vesna [10]

The magnitude of the electric field will be the greatest at the point where it is closest,to its charges.

Yes ,there is a point where the field will be zero.

what is an electric field?

The region where an electrostatic force is experienced by a charged entity is known as the electric field at a point.

As per the principle of field lines and vectors,where the field lines are in a close manner together,the field will be strongest.However ,where the field lines are in a manner apart,the field will be the weakest.

As per the concept,the electric field will be the greatest at the point where it is closest to its charges.For like charges, the electric field will be zero closer to the smaller charge and will be along the line joining the two charges. For opposite charges of equal magnitude, there will not be any zero electric fields.

Thus,we can conclude that there will be a point where the electric field is zero

learn more about electric field from here: brainly.com/question/28197462

#SPJ4

4 0
2 years ago
child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a s
Strike441 [17]

Answer:

θ = 13.7º

Explanation:

  • According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
  • The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
  • As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:

       \Delta K + \Delta U = W_{nc} (1)

  • ΔK, is the change in kinetic energy, as follows:

       \Delta K = \frac{1}{2}* m* (v_{f} ^{2}  - v_{0} ^{2}) (2)

  • ΔU, is the change in the gravitational potential energy.
  • If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:

        \Delta U = 0 - m*g*h = -m*g*d* sin\theta (3)

  • Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

        W_{nc} = F_{f} * d * cos 180\º \\\\  = 0.2*m*g*d* cos 180\º = -0.2*m*g*d (4)

  • Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

      \frac{1}{2}* (v_{f} ^{2}  - v_{0} ^{2}) = g*d* sin\theta -0.2*g*d

  • Replacing by the givens and the knowns, we can solve for sin θ, as follows:              \frac{1}{2}*( (4.30 m/s) ^{2}  - (0.75 m/s)^{2}) = 9.8 m/s2*25.5m* sin\theta -0.2*9.8m/s2*25.5m\\ \\ 8.56 (m/s)2 = 250(m/s)2* sin \theta -50 (m/s)2\\ \\ sin \theta = \frac{58.6 (m/s)2}{250 (m/s)2}  = 0.236⇒ θ = sin⁻¹ (0.236) = 13.7º
8 0
3 years ago
A thin rod of length 0.64 m and mass 120 g is suspended freely from one end. It is pulled to one side and then allowed to swing
valina [46]

Answer:

1. Kinetic Energy = 0.0161 Joules

2. Height = 0.0137m

Explanation:

Given

Length of Rod, l = 0.64m

Mass, m = 120g = 0.12kg

Angular speed, w = 1.40 rad/s

a.

Calculating the Rod's kinetic energy

This is calculated by

Kinetic Energy = ½Iw²

Where I = rotational inertia of the rod about an axis.

This is calculated as follows;

I = Icm + mh²

I = ImL² + m(L/2)²

I = 1/12 * 0.12 * 0.64² + 0.12 * (0.64/2)²

I = 0.016384 kgm²

By substituton

KE = ½Iw² becomes

KE = ½ * 0.016384 * 1.40²

KE = 0.01605632J

KE = 0.0161 Joules

2. Using the total conservation of momentum;

K + U = Kf + V

Where K = Initial Kinetic Energy of the rod at lowest point.

U = Initial gravitational potential energy of the rod at lowest point

Kf = Final Kinetic Energy of the rod at maximum height = 0 J

V = Final gravitational potential energy of the rod at maximum height

So, K + U = Kf + V become

K + U = 0 + V

K + U = V

K = V - U = mgh

substitute 0.01605632J for K

0.01605632J = mgh

h = 0.01605632J/mg

h = 0.01605632J/(0.12 * 9.8)

h = 0.013653333333333

h = 0.0137m

4 0
3 years ago
Two air craft P and Q are flying at the same speed 300m/s. The direction along which P is flying is at right angles to the direc
Lerok [7]

Answer:

424.26 m/s

Explanation:

Given that Two air craft P and Q are flying at the same speed 300m/s. The direction along which P is flying is at right angles to the direction along which Q is flying. Find the magnitude of velocity of the air craft P relative to air craft Q

The relative speed will be calculated by using pythagorean theorem

Relative speed = sqrt(300^2 + 300^2)

Relative speed = sqrt( 180000 )

Relative speed = 424.26 m/s

Therefore, the magnitude of velocity of the air craft P relative to air craft Q is 424.26 m/s

7 0
3 years ago
A beam of light in air is incident at an angle of 30º to the surface of a rectangular block of clear plastic (n = 1.46). The lig
Aneli [31]

Answer:

θ = 30°

Explanation:

Firts, the angle when the beam of light passes through the block cam be calculated using Snell Law:

n_{1}sin(\theta_{1}) = n_{2}sin(\theeta_{2})

<u>Where</u>:

n₁: is the index of refraction of the incident medium (air) = 1

θ₁: is the incident angle = 30°

n₂: is the medium 2 (plastic) = 1.46

θ₂: is the transmission angle

Hence, θ₂ is:

sin(\theta_{2}) = \frac{n_{1}*sin(\theta_{1})}{n_{2}} = \frac{1*sin(30)}{1.46} = 0.34 \rightarrow \theta_{2} = 20.03 ^{\circ}

Now, when the beam of light re-emerges from the opposite side, we have:

n₁: is the index of refraction of the incident medium (plastic) = 1.46  

θ₁: is the incident angle = 20.03°                

n₂: is the medium 2 (air) = 1

θ₂: is the transmission angle

Hence, the angle to the normal to that surface (θ₂) is:

sin(\theta_{2}) = \frac{n_{1}*sin(\theta_{1})}{n_{2}} = \frac{1.46*sin(20.03)}{1} = 0.50 \rightarrow \theta_{2} = 30 ^{\circ}  

 

Therefore, we have that the beam of light will come out at the same angle of when it went in, since, it goes from air and enters to a plastic medium and then enters again in this medium to go out to air again. This was proved using the Snell Law.    

I hope it helps you!                

5 0
3 years ago
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