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
Masja [62]
3 years ago
11

It is generally a good idea to gain an understanding of the "size" of units. Consider the objects listed and calculate the magni

tude of their momentum in SI units. A ladybug weighing 31.80 mg flies by your head at 2.53 mi/h.
Physics
1 answer:
Alex3 years ago
5 0

Answer:

Explanation:

The standard unit of momentum Is kgm/s

And momentum is the product of mass and velocity

Then,

Momentum = mass × velocity

P = mv

Where

m is mass in kg

v is velocity in m/s

Now,

Given that,

M = 31.8mg

1mg = 10^-3g

So, m = 31.8 × 10^-3g

Also, 1000g = 1kg → 1g = 10^-3 kg

So,

m = 31.8 × 10^-3 × 10^-3kg

m = 31.8 × 10^-6 kg

v = 2.53mi/hr

1mile = 1609.34 m

1 hour = 3600seconds

Then,

v = 2.53 × 1609.34m/ 3600s

v = 1.13 m/s

So, momentum is given as

P = mv

P = 31.8 × 10^-6 × 1.13

P = 3.597 × 10^-5 kgm/s

You might be interested in
A 20 liter cylinder of helium at a pressure of 150 atm and a temperature of 27ÁC is used to fill a balloon at 1.00 atm and 37ÁC.
djyliett [7]

<u>Answer</u>

D) 3100 Liters


<u>Explanation</u>

To get the volume if the balloon you need to use the combined equation of the low of gases.

P₁V₁/T₁ = P₂V₂/T₂

(20×150)/(27+273) = (1×V₂)/(37+273)

3000/300 = V₂/310

10 = V₂/310

V₂ = 10 × 310

    = 3100 Liters


7 0
3 years ago
Read 2 more answers
Because of interstellar dust, astronomers can see at most about 5 kpc into the disk of the galaxy at visual wavelengths. What pe
NNADVOKAT [17]

Answer:

96%

Explanation

Let A the total area of the galaxy, is modeled as a disc:

A = πR^2 = π (25 kpc)^2

And let a be the area that astronomers are able to see:

a = πr^2 = π(5 kpc)^2

The percentage that can be seen is equal to 100 times the ratio of the areas, of the galaxy and the "visible" part:

P = 100 a/A = (5/25)^2 = 100/25 = 4%

Therefore, the percentage of the galaxy not included, i.e. not seen is:

(100-4)% = 96%

5 0
3 years ago
Jose gets up from his seat on the bus to move closer to the front. Just as he begins to walk forward, the bus stops at a light.
aleksklad [387]

Answer:

She falls forward

Explanation:

Dunno, just factss

5 0
3 years ago
Read 2 more answers
Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at
sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0
3 years ago
A bat moving at 3.7 m/s is chasing a ying insect. The bat emits a 36 kHz chirp and receives back an echo at 36.79 kHz. At what s
3241004551 [841]

Answer:

The speed the bat is gaining on its prey is 0.03m/s

Explanation:

Given;

speed of the bat, v₀ = 3.7 m/s

frequency of the bat, F₀ = 36 kHz

frequency of the source, Fs = 36.79

This is relative motion between a source of the sound and the observer.  The phenomenon is known as Doppler effect.

Apply the following equation to determine the speed of the insect which is the source;

F_0 = F_s[\frac{v+v_0}{v-v_s} ]\\\\\frac{F_0}{F_s} = [\frac{v+v_0}{v-v_s} ]\\\\\frac{36.79}{36} = \frac{340+3.7}{340-v_s}\\\\1.0219 = \frac{343.7}{340-v_s}\\\\  340-v_s = \frac{343.7}{1.0219}\\\\340-v_s = 336.33\\\\v_s = 340-336.33\\\\v_s = 3.67 \ m/s

The speed the bat is gaining on its prey = 3.7m/s - 3.67m/s = 0.03 m/s

Therefore, the speed the bat is gaining on its prey is 0.03m/s

8 0
3 years ago
Other questions:
  • Calculate the power developed in a 6.0o resistor with a potential drop of 12 volts.
    6·2 answers
  • Hurricanes are formed due to
    14·2 answers
  • 3. An atom has 11 protons in the nucleus. What is the atomic number? What is the name of this
    11·1 answer
  • What are weather balloons?
    7·1 answer
  • A proton (mass m 1.67 x 1027 kg) is being accelerated along a straight line at 1.4 x 1015 m/s2 in a machine. The proton has an i
    5·1 answer
  • If a pitcher throws a baseball 10 m in 30 seconds - what would be the average speed of the baseball
    12·1 answer
  • William drew a diagram of a box containing a gas for his science project. His drawing is shown.
    12·1 answer
  • (a) As you ride on a Ferris wheel, your apparent weight is different at the top and at the bottom. Explain. (b) Calculate your a
    11·1 answer
  • Continuous signals are characterized as . signals that are broken up into binary code are characterized as . an am/fm radio is a
    15·1 answer
  • You redo the primitive yo yo experiment (Figure 1), but instead of holding the free end of the string stationary, you move your
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!