1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
MaRussiya [10]
2 years ago
15

Lucia kicks a ball on a level playing field with an initial velocity of 11.3 m/s at an angle of 35° above the horizontal. Find:

Time of flight? Horizontal distance traveled? Maximum height?
Physics
1 answer:
grandymaker [24]2 years ago
6 0

Explanation:

Given that,

Initial velocity, u = 11.3 m/s

Angle above the horizontal, \theta=35^{\circ}

Time of flight :

t=\dfrac{2u\sin\theta}{g}\\\\t=\dfrac{2\times 11.3\times \sin(35)}{9.8}\\\\t=1.32\ s

Horizontal distance traveled  is given by :

x = ut

x = 11.3 m/s × 1.32 s

x = 14.916 m

Maximum height is given by :

H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(11.3)^2\times \sin^2(35)}{2\times 9.8}\\\\H=2.14\ m

Hence, time of flight is 1.32 s, horizontal distance is 14.916 m and maximum height is 2.14 m.

You might be interested in
A missile is moving 1350 m/s at a 25.0° angle. It needs to hit a target 23,500 m away in a 55.0° direction in 10.20 s. What is d
QveST [7]

Answer:

  The target's velocity is about 1320 m/s in the direction 265.7°.

Explanation:

In order for there to be a collision between missile and target, we must have ...

  (target starting position) + (target movement) = (missile movement)

assuming the missile starts from the origin of all measurements. The missile moves 10.2 seconds before impact, so moves a distance of ...

  (10.2 s)(1350 m/s) = 13,770 m

__

We are interested in the target movement, so we can solve for that:

  (target movement) = (missile movement) - (target starting position)

In terms of meters, this is ...

  (target movement) = 13770∠25° - 23500∠55° ≈ 13467.74∠-94.3°

The target covers this distance in the same 10.2 seconds before collision, so its speed is (13467.74 m)/(10.2 s) ≈ 1320.4 m/s.

As a positive angle, the target's direction is ...

  -94.3° +360° = 265.7°

The direction of the target's velocity is 265.7°.

_____

If you're calculating this by hand, there are a couple of ways you can do it. You can convert to rectangular coordinates and back (perhaps least confusing), or you can use the law of cosines to solve the triangle, then translate angles back to the x-y coordinate plane.

Using rectangular coordinates, we have ...

  13770∠25° = 13770(cos(25°), sin(25°)) ≈ (12479.9, 5819.45)

  23500∠55° = 23500(cos(55°), sin(55°)) ≈ (13479.0, 19250.1)

Then the difference is ...

  (12479.9, 5819.45) -(13479.0, 19250.1) ≈ (-999.188, -13430.6)

and the (3rd-quadrant) angle is ...

  target direction = arctan(-13430.6/-999.188) ≈ -94.3° = 265.7°

__

The target's speed is found by dividing the distance it covers by the time it takes.

  √(13430.6² +999.188²)/10.2 ≈ 1320.36 . . . m/s

3 0
3 years ago
A mass on a spring with k=88.7 N/m oscillates 15 times in 9.24s. what is the objects mass? unit=kg?
sweet [91]

The mass on the spring is 0.86 kg

Explanation:

The period of a mass-spring system is given by the equation

T=2\pi \sqrt{\frac{m}{k}}

where

m is the mass

k is the spring constant

In this problem, we have:

k = 88.7 N/m is the spring constant

The system makes 15 oscillations in 9.24 s: therefore, the period of the system is

T=\frac{9.24}{15}=0.62 s

Now we can re-arrange the first equation  to solve for the mass:

m=k(\frac{T}{2\pi})^2=(88.7)(\frac{0.62}{2\pi})^2=0.86 kg

Learn more about period:

brainly.com/question/5438962

#LearnwithBrainly

3 0
3 years ago
A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne
Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0
3 years ago
A student shakes a rope such that 20 complete vibrations are made in 4.00 seconds. Determine the vibrational frequency of the ro
fgiga [73]

Answer:

The vibrational frequency of the rope is 5 Hz.

Explanation:

Given;

number of complete oscillation of the rope, n = 20

time taken to make the oscillations, t = 4.00 s

The vibrational frequency of the rope is calculated as follows;

Frequency = \frac{number \ of \ complete \ vibrations}{time \ taken} \\\\Frequency = \frac{20 }{4 \ s} \\\\Frequency = 5 \ Hz

Therefore, the vibrational frequency of the rope is 5 Hz.

8 0
2 years ago
A 54.0 kg ice skater is gliding along the ice, heading due north at 4.10 m/s . The ice has a small coefficient of static frictio
lesya [120]

Answer:

a. 2.668 m/s

b. 0.00494

Explanation:

The computation is shown below:

a. As we know that

W = F\times d

KE = 0.5\times m\times v^2

As the wind does not move the skater to the east little work is performed in this direction. All the work goes in the direction of the N-S. And located in that direction the component of the Force.

F = 3.70 cos 45 = 2.62 N

W = F \times d = 2.62 N \times 100 m

W = 261.6 N\times m

We know that

KE1 = Initial kinetic energy

KE2 = kinetic energy following 100 m

The energy following 100 meters equivalent to the initial kinetic energy less the energy lost to the work performed by the wind on the skater.

So, the equation is

KE2 = KE1 - W

0.5 m\times v2^2 = 0.5 m\ v1^2 - W

Now solve for v2

v2 = \sqrt{v1^2 - {\frac{2W}{M}}}

= \sqrt{4.1 m/s)^2 - \frac{2 \times 261.6 N\times m}{54.0 kg}}

= 2.668 m/s

b. Now the minimum value of Ug is

As we know that

Ff = force of friction

Us = coefficient of static friction

N = Normal force = weight of skater

So,

Ff = Us\times N

Now solve for Us

= \frac{Ff}{N}

= \frac{3.70 N \times cos 45 }{54.0 kg \times 9.81 m/s^2}

= 0.00494

4 0
3 years ago
Other questions:
  • To hoist himself into a tree, a 72.0-kg man ties one end of a nylon rope around his waist and throws the other end over a branch
    5·1 answer
  • Suppose that the electric field in the Earth's atmosphere is E = 1.16 102 N/C, pointing downward. Determine the electric charge
    15·1 answer
  • What is the best description of the chromosomes by the end of metaphase of mitosis?
    10·1 answer
  • 1) The "Ring of Fire" is an area along the edge of the Pacific Ocean with many volcanoes. Most of
    15·1 answer
  • You ride your bike for a distance of 30 km. You travel at a speed of 0.75 km/ minute. How many minutes does this take?
    10·2 answers
  • A woman lifts her 100-newton child up one meter and carries her for a distance of 50 meters to the child's bedroom. How much wor
    15·2 answers
  • A boat is headed due north on a river that's flowing directly east. What is the general direction of travel for this boat with r
    5·2 answers
  • When the flashlight is in the air and the refracted ray enters the water, how does the angle of refraction compare with the angl
    8·1 answer
  • A 35.0-g object connected to a spring with a force constant of 40.0 N/m oscillates with an amplitude of 4.00 cm on a frictionles
    11·1 answer
  • Cold water flows to the solar panels at 15°C. During the day, the panels supply 3.8 kg of hot water
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!