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
Norma-Jean [14]
3 years ago
11

From the earth, the moon subtends an angle of approximately 0.5°. If the distance to the moon is approximately 240,000 miles, fi

nd an approximation for the diameter of the moon accurate to the nearest hundred miles.
Physics
2 answers:
slega [8]3 years ago
8 0

The approximation for the diameter of the moon accurate to the nearest hundred miles is s =  +- 2100 miles

<h3>Further explanation </h3>

Arc Measure definition is,  in a circle, arc degree measure is equal to the measure of the central angle that intercepts the arc. Whereas the arc Length definition is, In a circle, the length of an arc is a portion of the circumference. Where the letter "s" is used to represent arc length.      

s = r * \theta (Radian Measure and Arc of a Circle).

There is a formula which relates the arc length of a circle of radius, r, to the central angle, θ in radians.

The radian measure θ of a central angle is defined as the ratio of the length of the arc  the angle subtends s, divided by the radius of the circle r.

where s = the arc length

\theta = angle  in radians

0.5 degrees *  \frac{\pi}{180}  = 0.008726646 radians \\ r = 240000 miles  \\ s  = 240000 * 0.008726646 \\

s = 2094.4 miles or

s =  +- 2100 miles

So the approximation for the diameter of the moon accurate to the nearest hundred miles is s =  +- 2100 miles

<h3>Learn more</h3>
  1. Learn more about the radian measure of the angle brainly.com/question/4675968

<h3>Answer details</h3>

Grade:  9

Subject:  physics

Chapter:  the radian measure of the angle

Keywords:  moon

laiz [17]3 years ago
3 0
The subtended angle and by an arc is given by:
S = r∅
Although the arc is circular, the huge amount of distance means that it is almost equivalent to the director.
D = 240,000 x (0.5 x π)/180
D = 2094 miles
You might be interested in
A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x (F
Marrrta [24]

Answer:

Part a)

x = 0.4 m

Part b)

v_i = 11.7 m/s

Part c)

Speed is more than the required speed so it will reach the top

Explanation:

Part a)

As we know that there is no frictional force while block is moving on horizontal plane

so we can use energy conservation on the block

\frac{1}{2}mv^2 = \frac{1}{2}kx^2

\frac{1}{2}0.500(12^2) = \frac{1}{2}(450)x^2

x = 0.4 m

Part b)

If the track has average frictional force of 7 N then work done by friction while block slides up is given as

W_f = -7( \pi R)

W_f = -7(\pi \times 1.00)

W_f = -22 J

work done against gravity is given as

W_g = - mg(2R)

W_g = -(0.500)(9.8)(2\times 1)

W_g = -9.8 J

Now by work energy equation we have

\frac{1}{2}mv_i^2 + W_f + W_g = \frac{1}{2}mv_f^2

\frac{1}{2}0.5(12^2) - 9.8 - 22 = \frac{1}{2}(0.5)v_f^2

v_f = 4.1 m/s

Part c)

now minimum speed required at the top is such that the normal force must be zero

mg = \frac{mv^2}{R}

v = \sqrt{Rg}

v = 3.13 m/s

so here we got speed more than the required speed so it will reach the top

5 0
3 years ago
A wheel is rotating at 30.0 rpm. The wheel then accelerates uniformly to 50.0 rpm in 10.0 seconds. Determine the – a) rate of an
Mademuasel [1]

Answer:

The angular acceleration is 0.209\ rad/s^2

Explanation:

Given that,

Angular velocity, \omega_{i} = 30.0\ rpm

Angular velocity, \omega_{f} = 50.0\ rpm

Time t = 10.0 sec

We need to calculate the angular acceleration

Using formula of angular acceleration

\alpha=\dfrac{\Delta \omega}{\Delta t}

\alpha=\dfrac{\omega_{f}-\omega_{i}}{\Delta t}

\alpha=\dfrac{50.0-30.0}{10.0}

Now, we change the angular velocity in rad/s.

\omega=20\times\dfrac{2\pi}{60}

\omega=2.09\ rad/s

\alpha=\dfrac{2.09}{10.0}

\alpha=0.209\ rad/s^2

Hence, The angular acceleration is 0.209\ rad/s^2

5 0
3 years ago
Read 2 more answers
3. Question
aalyn [17]
2. Newton's laws of motion
5 0
3 years ago
Read 2 more answers
Buhrs atomic model differed from ruthofords because it explained that
Scrat [10]

Buhrs atomic model differed from ruthofords because it explained that electrons exist in specified energy levels surrounding the nucleus. This means that, Ruthoford believed that electrons can't do very much. However, Buhrs' model showed that electrons are much more powerful than anyone else believes they can be.

6 0
3 years ago
Read 2 more answers
A field measuring 12 meters by 16 meters is to have a brick paver walkway installed all around it, increasing the total area to
kolezko [41]

Answer:

1.5 m

Explanation:

Length. L = 12 m

Width, W = 16 m

Area, A = 12 x 16 = 192 m^2

Let the width of pavement be d.

The new length, L' = 12 + 2d

the new width, W' = 16 + 2d  

New Area, A' = L' x W' = (12 + 2d)(16 + 2d) = 192 + 56 d + 4d^2

Difference in area = A' - A

285 =  192 + 56 d + 4d^2 - 192

93 =  56 d + 4d^2

4d^2 + 56 d - 93 = 0

d = \frac{-56\pm \sqrt{56^{2}+4\times 4\times 93}}{8}

d=\frac{-56\pm 87.72}{8}\

d = 1.5 m

Thus, the width of the pavement is 1.5 m.

6 0
4 years ago
Other questions:
  • The primary difference between a scientific theory and a hypothesis is that a theory is: A prediction of an experimental outcome
    7·1 answer
  • Explain why it is esaier to climb a mountain on a zigzag path rather than one straight up the side. is your increase in gravitat
    14·1 answer
  • A mass is connected to a spring on a horizontal frictionless surface. The potential energy of the system is zero when the mass i
    7·1 answer
  • The ratio of the lift force L to the drag force D for the simple airfoil is L/D = 10. If the lift force on a short section of th
    10·1 answer
  • two bowling balls each have a mass of 8kg. if they are 2 m apart, what is the gravitational force between them?
    15·1 answer
  • Why is the meteor shower is best observed after midnight?​
    12·1 answer
  • A .2.-kg soccer ball is rolling at 6.0 m/s toward a player. The player kicks the ball back and gives it a velocity of 14 m/s in
    5·1 answer
  • I need help with this problem on science 8th grade here is the picture of the assignment place help me
    14·2 answers
  • Anyone?????help with this questions!!!
    10·2 answers
  • A ball is thrown from a top of a building of height 20m. If the initial velocity of the ball is 15m/s at 370 above the horizonta
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!