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
frez [133]
3 years ago
10

Consider a projectile of mass 20 kg launched with a speed 9 m/s at an elevation angle of 45 degrees. Taking the launch point as

the origin of your coordinate system---
a. What is the magnitude of the angular momentum at launch? kg m/s
b. What is the magnitude of the angular momentum at the instant it reaches maximum height? kg m/s
c.What is the magnitude of the angular momentum at impact?
Physics
1 answer:
viva [34]3 years ago
5 0

Answer:

a) L=0. b) L = 262 k ^   Kg m²/s and c)  L = 1020.7 k^   kg m²/s

Explanation:

It is angular momentum given by

      L = r x p

Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum

One of the easiest ways to make this vector product is with the use of determinants

{array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]

Let's apply this relationship to our case

Let's start by breaking down the speed

      v₀ₓ = v₀ cosn 45

      voy =v₀ sin 45

      v₀ₓ = 9 cos 45

      voy = 9 without 45

      v₀ₓ = 6.36 m / s

      voy = 6.36 m / s

a) at launch point r = 0 whereby L = 0

. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero

   vfy² = voy²- 2 g y

   y = voy² / 2g

   y = (6.36)²/2 9.8

   y = 2.06 m

Let's calculate the angular momentum

L= \left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]

L = -px y k ^

L = - (m vox) (2.06) k ^

L = - 20 6.36 2.06 k ^

L = 262 k ^   Kg m² / s

The angular momentum is on the z axis

c) At the point of impact, at this point the height is zero and the position on the x-axis is the range

     R = vo² sin 2θ / g

     R = 9² sin (2 45) /9.8

     R = 8.26 m

L = \left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]

L = - x py k ^

L = - x m voy

L = - 8.26 20 6.36 k ^

L = 1020.7 k^   kg m² /s

You might be interested in
A 1350 kg uniform boom is supported by a cable. The length of the boom is l. The cable is connected 1/4 the
olchik [2.2K]

Answer:

Tension= 21,900N

Components of Normal force

Fnx= 17900N

Fny= 22700N

FN= 28900N

Explanation:

Tension in the cable is calculated by:

Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium

FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)

Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)

Ftorque= 2/3FBcostheta+ 4/3FWcostheta

Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°

Ftorque= 21900N

b) components of Normal force

Efx=FNx-FTcos(90-theta)=0 static equilibrium

Fnx=21900cos(90-55)=17900N

Fy=FNy+ FTsin(90-theta)-FB-FW=0

FNy= -FTsin(90-55)+FB+FW

FNy= -21900sin(35)+(1350+2250)×9.81=22700N

The Normal force

FN=sqrt(17900^2+22700^2)

FN= 28.900N

4 0
3 years ago
A vector has an x component of 25.0 units and a y
sammy [17]
Here is the answer to your question

8 0
3 years ago
A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w
frosja888 [35]

Answer:

(a) n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) n_{2} = 1.349

(c) v_{1} = 2.04\times 10^{8}\ m/s

(d) v_{2} = 2.22\times 10^{8}\ m/s

Solution:

As per the question:

Refractive index of medium 1, n_{1} = 1.47

Angle of refraction for medium 1, \theta_{1} = 59^{\circ}

Angle of refraction for medium 2, \theta_{1} = 69^{\circ}

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

n_{1}sin\theta_{1} = n_{2}sin\theta_{2}

where

n_{2} = Refractive Index of medium 2

Now,

n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}

n_{2} = 1.349

(c) To calculate the velocity of light in medium 1:

We know that:

Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}

Thus for medium 1

n_{1} = \frac{c}{v_{1}

v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s

(d) To calculate the velocity of light in medium 2:

For medium 2:

n_{2} = \frac{c}{v_{2}

v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s

5 0
2 years ago
Read 2 more answers
Calculate the average speed between seconds 1 and 4 PLS HELP MEEEEEEEEE
sesenic [268]

Answer:

8.7

Explanation:

3 0
3 years ago
What is linear motion?
zalisa [80]

Answer:

Linear motion is a one-dimensional motion along a straight line, and can therefore be described mathematically using only one spatial dimension.

Explanation:

5 0
3 years ago
Read 2 more answers
Other questions:
  • Can objects in a system have momentum while the momentum of the system is zero? explain your answer.
    12·1 answer
  • Emmy is standing on a moving sidewalk that moves at +2 m/s. Suddenly, she realizes she might miss her flight, so begins to speed
    11·1 answer
  • What type of plant tissue is NOT part of the vascular bundle?
    8·2 answers
  • Given two vectors A⃗ = 4.20 i^+ 7.00 j^ and B⃗ = 5.70 i^− 2.60 j^ , find the scalar product of the two vectors A⃗ and B⃗ .
    5·1 answer
  • Cho q1= 8.10 mũ -8C ; q2 = -8.10 mũ -8C . Đặt tại 2 điểm A và B cách nhau 6cm trong không khí . Xác định lực điện dung dịch lên
    11·1 answer
  • Pressure, V is volume , T is temperature of a gas and R is gas constant.
    9·1 answer
  • Lydia is 9kg underweight. Caroline has an ideal weight for her height. Viv is 8kg overweight. Who is most likely to have the hig
    12·2 answers
  • Through what potential difference should electrons be accelerated so that their speed is 1.0 % of the speed of light when they h
    8·1 answer
  • WILL GIVE BRAINLIEST!!!
    12·1 answer
  • The idea that the Earth sits motionless in the Universe at the center of a revolving globe of stars, with the Moon and planets i
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!