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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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If a rock is thrown upward on the planet mars with a velocity of 11 m/s, its height (in meters) after t seconds is given by h =
Butoxors [25]
(a) 3.56 m/s 
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(c) t = 5.9 s 
(d) -11 m/s  
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule. 
y = 11t - 1.86t^2 
y' = 11 - 3.72t 
Now that you have the first derivative, it will give you the velocity as a function of t. 
(a) Velocity after 2 seconds. 
y' = 11 - 3.72t 
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(b) Velocity after a seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72a  
So the answer is 11 - 3.72a  
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.  
(d) Plug in the value of t calculated for (c) into the velocity function, so: 
y' = 11 - 3.72a
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3 0
2 years ago
What is the volume of an irregularly shaped object that has a mass 3.0 grams and a density of 6.0 g/mL
pogonyaev

Answer: The volume of an irregularly shaped object is 0.50 ml

Explanation:

To calculate the volume, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of object = 6.0g/ml

mass of object = 3.0 g

Volume of object = ?

Putting in the values we get:

6.0g/ml=\frac{3.0g}{\text{Volume of substance}}

{\text{Volume of substance}}=0.50ml

Thus the volume of an irregularly shaped object is 0.50 ml

4 0
2 years ago
If the mass of a material is 45 grams and the volume of the material is 8 cm^3, what would the density of the material be?
Svet_ta [14]
The density of the substance is the ratio of its mass over the space it occupies. In mathematical equation, this can be expressed as,

        ρ = m / v

where ρ is density, m is mass, and v is volume. 

Substituting the known values from the given,
 
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<em>ANSWER: 5.625 g/cm³</em>
3 0
2 years ago
A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of 1.055\times 10^{- 7} in its size

Solution:

As per the question:

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Distance covered by photon, d = 1 km = 1000 m

Mass of proton, m_{p} = 1.67\times 10^{- 27} kg

The initial size of the wave packet, \Delta t_{o} = 1 mm = 1\times 10^{- 3} m

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

E = m_{p}c^{2}

E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J

This energy of proton is \simeq 250 GeV

Thus the speed of the proton, v\simeq c

Now, the time taken to cover 1 km = 1000 m of the distance:

T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

Now, in accordance to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}

Thus the increase in wave packet's width is relatively quite small.

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\Delta t_{o} = \Delta t

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3 0
2 years ago
A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm
Airida [17]

Answer:

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Explanation:

50 g at 90 cm

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x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

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The mass of the meter stick is 126.99115 g

6 0
2 years ago
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