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
zzz [600]
3 years ago
9

How would i solve these problems, nobody in my class understands and there is a substitute

Physics
1 answer:
drek231 [11]3 years ago
6 0

find acceleration force divided by Mass

a=f/m

You might be interested in
Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are
Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

d_{12} = \sqrt{3^{2}+3^{2}  }  = \sqrt{18} m : distance from q₁ to q₂

(d₁₂)² = 18 m²

d_{13} =\sqrt{1^{2}+3^{2}  } = \sqrt{10} m  : distance from q₁ to q₃

(d₁₃)² = 10 m²

d_{14} =\sqrt{3^{2}+2^{2}  } = \sqrt{13} m  : distance from q₁ to q₄

(d₁₄)² = 13 m²

K=  8.98755 × 10⁹ N *m²/C²

q₁=  7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45°  ,  β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N  

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }

F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }

Fn₁= 23.95*10⁻⁹ N

8 0
3 years ago
Which of the following best describes a property of water?
NNADVOKAT [17]

Answer:

C. weak cohesive forces exist between its molecules.

Explanation:

This is because water has less intermolecular forces than solids, but more than gases. Also their cohesive forces is low.

4 0
3 years ago
You kick a ball with a speed of 14 m/s at an angle of 51°. How far away does the ball land?
In-s [12.5K]
-- The vertical component of the ball's velocity is 14 sin(<span>51°) = 10.88 m/s

-- The acceleration of gravity is 9.8 m/s².

-- The ball rises for 10.88/9.8 seconds, then stops rising, and drops for the
same amount of time before it hits the ground.

-- Altogether, the ball is in the air for (2 x 10.88)/(9.8) = 2.22 seconds
==================================

-- The horizontal component of the ball's velocity is  14 cos(</span><span>51°) = 8.81 m/s

-- At this speed, it covers a horizontal distance of (8.81) x (2.22) = <em><u>19.56 meters</u></em>
before it hits the ground.


As usual when we're discussing this stuff, we completely ignore air resistance.
</span>
4 0
3 years ago
Read 2 more answers
Which percentage best describes the visibility of a waxing crescent moon
Rus_ich [418]

Answer:

It be 0,1% - 49.9%

This is because the new moon is 0.0% and the first quarter is 50%.

3 0
2 years ago
The potential energy for a certain mass moving in one dimension is given by U(x)=(2.0J/m3)x3−(15J/m2)x2+(36J/m)x−23JU(x)=(2.0J/m
Angelina_Jolie [31]

Answer:x=2 and x=3

Explanation:

Given

Potential Energy for a certain mass is

U(x)=2x^3-15x^2+36x-23

and we know force is given by

F=-\frac{\mathrm{d} U}{\mathrm{d} x}

F=-(2\times 3x^2-15\times 2x+36)

For Force to be zero F=0

\Rightarrow 6x^2-30x+36=0

\Rightarrow x^2-5x+6=0

\Rightarrow x^2-2x-3x+6=0

\Rightarrow (x-2)(x-3)=0

Therefore at x=2 and x=3 Force on particle is zero.

8 0
3 years ago
Other questions:
  • A non reactive metal can be identified from other metals because it's surface is?
    13·1 answer
  • sitting on the dock of the bay wasting time with my sister. i get bored and push her off the 2 m dock. How fast is she moving wh
    9·1 answer
  • The number of protons in an atom is that element’s __________________ number.<br> PLEASE HELP
    14·1 answer
  • A child with mass 40 kg sits on the edge of a merry-go-round at a distance of 3.0 m from its axis of rotation. The merry-go-roun
    9·1 answer
  • How will a positive and a negative charge react to each other?
    14·1 answer
  • Emma is working in a shoe test lab measuring the coefficient of friction for tennis shoes on a variety of surfaces. The shoes ar
    10·1 answer
  • When discussing Newton’s laws of motion, which terms do people most likely use when talking about Newton’s third law of motion?
    11·2 answers
  • Two in-phase loudspeakers are 3.0 m apart. they emit sound with a frequency of 490 hz. a microphone is placed half-way between t
    12·1 answer
  • An engine creat a 2230N force that pushes a car forward 58.3m in 9.4s
    6·1 answer
  • For an object on a flat surface, the force of gravity is 10 newtons downward and the normal force is 10 newtons upward. The appl
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!