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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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A vertical cylinder is divided into two parts by a movable piston of mass m. The piston and cylinder system is well insulated (t
Mekhanik [1.2K]

Answer:

Final temperature will be 438.076 K

Explanation:

We have given temperature T_1=323K

Volume V_1=V\ and\ V_2=\frac{V}{2}

As there is no heat transfer so this is an adiabatic process

For and adiabatic process TV^{\gamma -1}=constant

Here \gamma =1.4

So T_1V_1^{\gamma -1}=T_2V_2^{\gamma -1}

T_2=\left ( \frac{V_1}{V_2} \right )^{\gamma -1}\times T_1

T_2=\left ( \frac{V}{\frac{V}{2}} \right )^{1.4 -1}\times 332=2^{0.4}\times 332=438.076K

4 0
2 years ago
A particle at 9 AM is moving towards the east at 4 ms At 12 noon, it changes its velocity and starts moving towards the north un
nexus9112 [7]

Answer:

Explanation:

Acceleration is the time rate of change of velocity.

Acceleration and velocity are vectors

If east and north are the positive directions, the east moving vector is reduced to zero and the north moving vector increases from zero to 4 m/s.

There are 3 hours or 10800 seconds between 10 AM and 1 PM

a1 = √((-4)² + 4²) / 10800 = (√32) / 10800 m/s² ≈ 4.2 x 10⁻⁴ m/s²

There are 14400 seconds between 10 AM and 2 PM

The velocity changes are still the same

a2 = √((-4)² + 4²) / 10800 = (√32) / 14400 m/s² ≈ 3.9 x 10⁻⁴ m/s²

7 0
1 year ago
Describe how private citizens have a voice in which projects the federal government will fund
erastovalidia [21]
<span>Two main ways, free elections, casting your vote for the candidate supporting your goals. Second, Lobbying the sitting gov.</span>
3 0
2 years ago
A ball is thrown upward from the top of a 25.0 m tall building. The ball’s initial speed is 12.0 m/sec. At the same instant, a p
zimovet [89]
<h2>Person must have 8.18 m/s to catch the ball</h2>

Explanation:

Consider the vertical motion of ball

We have equation of motion s = ut + 0.5at²

Initial velocity, u = 12 m/s

Acceleration, a = -9.81 m/s²

Displacement, s = -25 m

Substituting

             -25 = 12 x t + 0.5 x -9.81 x t²

               4.905 t² -12t - 25 = 0

              t = 3.79 sec

Ball hits ground after 3.79 seconds.

So person need to cover 31 m in 3.79 seconds

Consider the horizontal motion of person

We have equation of motion s = ut + 0.5at²

Initial velocity, u = ?

Acceleration, a = 0 m/s²

Displacement, s = 31 m

Time, t = 3.79 seconds

Substituting

             31 = u x 3.79 + 0.5 x 0 x 3.71²

               u = 8.18 m/s

Person must have 8.18 m/s to catch the ball

6 0
1 year ago
A vertical spring of constant k = 400 N/m hangs at rest. When a 2 kg mass is attached to it, and it is released, the spring exte
Viefleur [7K]

Answer:

4.9 cm

Explanation:

From Hook's Law,

F = ke......................... Equation 1

Where F= force, e = extension, k = spring constant.

Note: the Force acting on the the spring is the weight of the mass.

W = mg.

F = mg.................... Equation 2

Where m = mass, g = acceleration due to gravity

Substitute equation 2 into equation 1

mg = ke

make e the subject of the equation

e = mg/k............... Equation 3.

Given: m = 2 kg, g = 9.8 m/s², k = 400 N/m

e = (2×9.8)/400

e = 19.6/400

e = 0.049 m

e = 4.9 cm

3 0
2 years ago
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