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
sukhopar [10]
3 years ago
8

The equation that is used to solve second law problems is # F= ma.

Physics
1 answer:
maw [93]3 years ago
8 0
F= Force 
M=Mass
A= acceleration 
F=N 
Mass= in grams or kilo grams (mostly kg)
A= m/s 

You might be interested in
If your weight is 120 pounds and your mass is 54 kilograms how would those values change if you were on the moon
BlackZzzverrR [31]
The gravitation acceleration on the moon is different than on Earth. It is 1.6 m/s^2. If you weigh 120 lbs, then you would multiply 120 pounds by the gravitational acceleration on the moon and then divide by the acceleration on Earth.

(120 lbs * 1.6) / 9.8 = 20 pounds.

The mass will always be the same no matter what planet you’re on, so it’s still 54 kg.
4 0
3 years ago
Determine the slope of end a of the cantilevered beam. E = 200 gpa and i = 65. 0(106) mm4
DENIUS [597]

For E = 200 gpa and i = 65. 0(106) mm4,  the slope of end a of the cantilevered beam  is mathematically given as

A=0.0048rads

<h3>What is the slope of end a of the cantilevered beam?</h3>

Generally, the equation for the   is mathematically given as

A=\frac{PL^2}{2EI}+\frac{ML}{EI}

Therefore

A=\frac{10+10^2+3^2}{2*240*10^9*65*10^6}+\frac{10+10^3*3}{240*10^9*65*10^{-6}}

A=0.00288+0.00192=0.0048rads

A=0.0048rads

In conclusion,  the slope is

A=0.0048rads

Read more about Graph

brainly.com/question/14375099

5 0
2 years ago
Please Answer the question in the picture ASAP PLEASE
attashe74 [19]

Answer:

HERE IS YOUR ANSWER

Explanation:

PLEASE MARK MY ANSWER AS BRAINLIEST IF THE ANSWERS ARE CORRECT .

Beacuse of the loose connection of the wire .

Straight

5 0
3 years ago
Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time–when they\'re not sleep
Anit [1.1K]

Answer:

mice total momentum (-0.000250, 0.00639) Kg m

Explanation:

To calculate the moment of the mice we must multiply their mass by their velocities, remember that the moment is a vector quantity, so we use the components of velocity

mouse 1

  m1 = 0.0225 Kg

  V1 = (0.869, -0.283) m / s

 

  Px = m Vx

  Px1 = 0.0225 0.869

  Px1 = 0.01955 Kg m

  Py = m Vy

  Py1 = 0.0225 (-0.283)

  Py1 = -0.006368 Kg m

  P1 = (0.0196, -0.00637) Kg m

Mouse 2

 m2 = 0.0223 Kg

 Px2 = 0.0223 (-0.883) = -0.0196 Kg m

 Py2 = 0.0223 (-0.253) = -0.00564 Kg m

 P2 = (-0.0196, -0.00564) Kg m

Mouse 3

 m3 = 0.0197

 Px3 = 0.0197 0.345 = 0.00680 Kg m

 Py3 = 0.0197 0.803 = 0.0158 Kg m

 P3 = (0.00680, 0.0158) Kg m

Mouse4

  m4 = 0.0127 Kg

  Px4 = 0.0127 (-0.555) = -0.00705 Kg m

  Py4 = 0.0127 0.205 = 0.00260 Kg m

  P4 = (-0.00705, 0.00260) Kg m

To find the total momentum we must add each component of the individual moments

   Px = Px1 + Px2 + Px3 + Px4  

   Py = py1 + Py2 + Py3 + Py4

   Px = 0.0196 -0.0196 +0.00680 -0.00705

   Px = -0,000250 Kg m

   Py = -0.00637 -0.00564 +0.0158 +0.00260

   Py = 0.00639 Kg m

   P = (-0.000250, 0.00639) Kg m

7 0
3 years ago
A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v
atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

y(x, t) = Acos(kx -wt)

The relationship between velocity and propagation constant is

v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\

v = 15.87m/s

Time taken, t = \frac{\lambda}{v}

= \frac{1.3}{15.87}\\\\=0.0819 sec

t = 0.0819s

2)

The velocity of transverse wave is given by

v = \sqrt{\frac{T}{\mu}}

v = \sqrt{\frac{W}{\frac{m}{\lambda}}}

mass of string is calculated thus

mg = 0.0125N

m = \frac{0.0125N}{9.8N/s}

m = 0.00128kg

\omega = \frac{v^2m}{\lambda}

\omega = \frac{(15.87^2)(0.00128)}{1.30}

\omega = 0.25N

3)

The propagation constant k is

k=\frac{2\pi}{\lambda}

hence

\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}

\lambda = 0.036 m

No of wavelengths, n is

n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\

n = 36

4)

The equation of wave travelling down the string is

y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]

without, unit\\\\y(x , t)= Acos[172x + 2730t]

7 0
3 years ago
Other questions:
  • Kevin is a black high school senior. While walking home from a sporting event at school, he sees a police car and decides to tak
    14·1 answer
  • In the living room of your house you would like to run a 360 W stereo, a 30 W computer, a 720 W laser light projector, and four
    11·1 answer
  • Define momentum in terms of football.
    9·1 answer
  • An object is placed 50.0 cm in front of a convex mirror. where can be the image located if the focal length is 40 cm from the mi
    14·1 answer
  • A truck moving at 7.0 meters per second accelerates to a speed of 15.0 meters per second within a time of 1.5 seconds. Calculate
    5·1 answer
  • A bus accelerates to 60 m/s to the east in 10 s. What is the buses acceleration? 2. A car traveling at 10.0 m/s to the west acce
    12·1 answer
  • What is the velocity of a plane that traveled 3,000 miles from new york to california in 5.0
    6·1 answer
  • Help me please pleasee
    9·1 answer
  • A conveyor belt at a recycling plant launches bottles and bottle caps into the air, so that an automatic image recognition devic
    14·1 answer
  • Calculate the speed of a satellite moving in a stable circular orbit about the Earth at a height of 4930 km .
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!