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Mademuasel [1]
2 years ago
6

A ball is fired at an angle of 45 degrees, the angle that yields the maximum range in the absence of air resistance. What is the

ratio of the ball's maximum height to its range?

Physics
2 answers:
Nesterboy [21]2 years ago
7 0
Look at the picture for the answer

OleMash [197]2 years ago
5 0

Answer: 0.361:1

Explanation:

Since Range is U^2sin2Φ/g

And maximum height = U^2 sin^2Φ/2g

For range sinceΦ= 45

Range = U^2sin90/g

Sin90=1

Range= U^2/g

For maximum height

U^2sin^2Φ/2g

U^2sin(45)^2/2g

0.723U^2/2g

Ratio of maximum height to range

0.723U^2/2g*g/U^2

0.723/2

0.361

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Mrs. Brown's class is studying magnets and electricity. At the end of the unit Claire states that magnets and electricity are bo
Gwar [14]

Answer:

Magnets can create electricity and electricity can create a magnetic force.

Both electric charges and magnets do not have to touch an object in order to exert a force on it.

Electromagnets use electricity to create a magnetic force.

Explanation:

7 0
2 years ago
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Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 20.0 m above water wit
melamori03 [73]

Answer:

Explanation:

Given that,

Height of the bridge is 20m

Initial before he throws the rock

The height is hi = 20 m

Then, final height hitting the water

hf = 0 m

Initial speed the rock is throw

Vi = 15m/s

The final speed at which the rock hits the water

Vf = 24.8 m/s

Using conservation of energy given by the question hint

Ki + Ui = Kf + Uf

Where

Ki is initial kinetic energy

Ui is initial potential energy

Kf is final kinetic energy

Uf is final potential energy

Then,

Ki + Ui = Kf + Uf

Where

Ei = Ki + Ui

Where Ei is initial energy

Ei = ½mVi² + m•g•hi

Ei = ½m × 15² + m × 9.8 × 20

Ei = 112.5m + 196m

Ei = 308.5m J

Now,

Ef = Kf + Uf

Ef = ½mVf² + m•g•hf

Ef = ½m × 24.8² + m × 9.8 × 0

Ef = 307.52m + 0

Ef = 307.52m J

Since Ef ≈ Ei, then the rock thrown from the tip of a bridge is independent of the direction of throw

7 0
2 years ago
Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg
horrorfan [7]

Answer:

The value is t_1 =  9 \  s

Explanation:

Generally the velocity attained by the sled after t = 3.10 s is mathematically evaluated using the kinematic equation as follows

v  =  u   +  at

Here u = 0 \ m/s

a = 13.5 m/s^2

So

v  =  0   +  13.5 *  3.10

=> v  =  41.85 \ m/s

The is distance it covers at this time is

s =  u *  t  +  \frac{1}{2} a *  t^2

=> s =  +  \frac{1}{2} * 13.5 *  3.10^2

=> s =64.87

Now when sled stops its the final velocity is v_f =  0 m/s while the initial velocity will be the velocity after its acceleration i.e v  =  41.85 \ m/s

So

v_f  =  v  +  a_1t_1

Here  a_1 =  - 4.65, the negative sign shows that it is deceleration

So

           0  =  41.85  - 4.65 *  t_1

=> t_1 =  9 \  s

3 0
2 years ago
A 10N force pulls to the right and friction opposes 2N. If the object is 20kg,find the acceleraton.
zmey [24]

Force = mass * acceleration

10 N - 2 N = 20 kg * acceleration

8 N = 20 kg * acceleration

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8 0
2 years ago
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A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an init
Darina [25.2K]

Answer:4.05 s

Explanation:

Given

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Equating 1 &2 we get

\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}

\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

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t=4.05 s

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