Answer:
F= 11.25*10⁵ N to the right.
Explanation:
Assuming that the three charges can be treated like point charges, they must obey Coulomb's Law.Due to the linearity of this Law, we can use superposition in order to find the resultant force on the 1.00 C charge due to the other two.First, we find the force that the -3.00 C charge (located 200 m to the left) exerts on the 1.00 C, as follows:[tex]F_{13} = \frac{K*q_{1}*q_{2}}{r_{13} ^{2} } = \frac{9e9*(-3.00C)(1.00C)}{(200m)^{2}} = -6.75e5 N (1)[/tex]
Then, in the same way, we can find the force that the 2.00 C exerts on the 1.00 C charge, located 100 m away to the left:[tex]F_{23} = \frac{K*q_{3}*q_{2}}{r_{23} ^{2} } = \frac{9e9*(2.00C)(1.00C)}{(100m)^{2}} = 18e5 N (2)[/tex]
Since both vectors are on the same line, their sum is directly the algebraic sum, as follows:F₃ = F₁₃ + F₂₃ = -6.75*10⁵ N + 18.00*10⁵ N = 11.25*10⁵ N to the right, assuming this direction as positive.Power is measured in unit of Joules per second or
a) seconds
b) hertz
c) joules
d) watts
e) newtons
An object is moving with an initial velocity of 3.3m/s it is subject to a constant acceleration of 3.7 m/s2 for 10 s. How far will it have traveled during the time of its acceleration
Answer:
218m
Explanation:
Given parameters:
Initial velocity = 3.3m/s
acceleration = 3.7m/s²
time = 10s
Unknown:
How far will it travel during the time of acceleration = ?
Solution:
We use of the kinematics equations to solve this problem;
S = ut + [tex]\frac{1}{2}[/tex] at²
S is the distance
u is the initial velocity
t is the time
a is the acceleration
So;
S = (3.3x10) + ([tex]\frac{1}{2}[/tex] x 3.7 x 10²) = 218m
Falling to Earth
Using this graphic organizer, create an argument that supports the assumption that gravity pulls all objects down toward the center of Earth. Provide evidence to support your claim answer
Answer:
gravity pulls objects down to earth.
Explanation
Because Isaac Newton stated that. In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.
Define joule conversion factor?
Explanation:
Joule, unit of work or energy in the International System of Units (SI); it is equal to the work done by a force of one newton acting through one metre. ... In electrical terms, the joule equals one watt-second—i.e., the energy released in one second by a current of one ampere through a resistance of one ohm.
How many pets do you have???!!!??
Answer:
Right now I have three.
Explanation: Thanks for the points luv ^-^.
Answer:
I have two
Explanation:
51. If two cylinders of dissimilar geometry are connected together to form a closed hydrostatic system, why would one cylinder piston travel a greater distance than the other
Answer:
see that for the same pressure the displaced height in each cylinder is different because its diameter is different. Pascal's principle
Explanation:
The pressure on a system is given by the relations
P = ρ g h
P = F / A
where ρ is the density of the liquid, h the height and A the area
The expressions above we see that if for the same height the pressure is the same regardless of the shape of the cylinder.
With the second expression we see that if the system has a different area, the pressure is
P = [tex]\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }[/tex]
where we use subscript 1 for one body and subscript 2 for the other body
F₁ = [tex]\frac{A_{1} }{A_{2} } F_{2}[/tex]
The cylinder displacement is
V = A h
where V is the volume and h the height, in general the liquids are incompressible therefore the displaced volume is constant in the two bodies
V = A₁ h₁ = A₂ h₂
[tex]\frac{A_{1} }{A_2} = \frac{h_2}{h_1}[/tex]
we substitute
F₁ = [tex]\frac{h_2}{h_1}[/tex] F2
From here we see that for the same pressure the displaced height in each cylinder is different because its diameter is different.
If the diameter is the same, the offset height is the same
With a wooden ruler, you measure the length of a rectangular piece of sheet metal to be 14 mm. With micrometer calipers, you measure the width of the rectangle to be 4.98 mm. Give your answers to the following questions using the correct number of significant figures.
What is:
a. the area of the rectangle.
b. the ratio of the rectangle’s width to its length.
c. the perimeter of the rectangle.
d. the difference between the length and the width.
e. the ratio of the length to the width.
Answer:
Follows are the solution to the given question:
Explanation:
The rectangular part has a length of [tex]14 \ mm[/tex] and its rectangular part has a width of [tex]4.98 \ mm[/tex].
In option A
Calculating the area of the rectangular throgh the given piece:
[tex]\to A_R = WL=(14 mm) (4.98 mm) =69.72 \ mm^2[/tex]
In option B
Calculating the ratio of rectangle's width which is rectangle's length:
[tex]\to R_{WL}=\frac{W}{L}= \frac{4.98 \ mm}{14 \ mm} = 0.3557[/tex]
So, the ratio of rectangle's width to rectangle's length is 0.3557 .
In option C
Calculating the Perimeter of the rectangle:
[tex]\to P_R=2(W+L)=2(14 \ mm+ 4.98 \ mm)= 2(18.98) = 37.96 \ mm[/tex]
In option D
Calculating the difference between length and width:
[tex]\to D_{LW} = L- W = 14\ mm -4.98 \ mm =9.02 \ mm[/tex]
In option E
Calculating the ratio of length to width:
[tex]\to R_{LW}=\frac{L}{W} =\frac{14\ mm}{4.98 \ mm} = 2.811[/tex]
When an electromagnetic wave such as light interacts with a solid and polarizes the electron clouds within the atomic structure of that solid, what are the (2) primary consequences of this interaction
Answer:
1. Absorption or Emission of the light
2. Light induced changes in the matter
Explanation:
When an electromagnetic wave such as light interacts with solid, two consequences are for sure:
1. Absorption or Emission of the light
2. Light induced changes in the matter
When light travels through the solid, the intensity of light decreases as a result of addition of light energy to the body to which it interacts. If the medium or body to which light interacts is low in absorbing due to its atomic structure inside then light passing through it will show it. On the contrary, if a material is high in absorbing, very less intensive light will travel out.
Moreover, there will ionization of the atoms inside the medium to which light interacts. As light carries energy and when it interacts with atoms of the body, atoms gets energy and excited or de-excited accordingly.
Hence, above are the two primary consequences of this interaction.
Explaining How Temperature
Does sound travel faster in a warm room or a cold room?
Explain your answer.
Intro
Do
Answer:
Explanation:
Sample Response: Sound travels faster in a warm room because temperature affects the speed of a wave. In a warm room, the particles of air move faster and have higher chances of bumping into each other, which then increases the instances of energy transfer.
Place the lunar phases in the correct order. !QUICK! (I WILL GIVE BRAINLEST!!!!!)
Answer:
Here:
Explanation:
These eight phases are, in order, new Moon, waxing crescent, first quarter, waxing gibbous, full Moon, waning gibbous, third quarter and waning crescent. The cycle repeats once a month (every 29.5 days).
I need an expert in physics for help
How long will it take a planet 150 million kilometers from a star
of mass 1.99 x 1030
kg to complete one full orbit?
Answer: 370 days.
Explanation:
It will take the planet 16.69 minutes to complete one full orbit.
Data Given;
r = 150*10^6kmmass of star = 1.99*10^30kgGravitational ForceApplying gravitational and centripetal force
[tex]F = \frac{Gm_1m_2}{r^2}\\ F = \frac{mv^2}{r}\\ \frac{mv^2}{r}=\frac{Gm_1M2}{r^2}\\ v = \frac{2\pi r}{T} \\ \frac{m(2\pi r/T)^2}{r} = \frac{GM_1M_2}{r^2}\\ \frac{4\pi ^2r^2}{T^2} = \frac{GM}{r} \\ T^2 = \frac{4\pi^2 r^3}{Gm}\\ [/tex]
Let's substitute the values into the equation
[tex]T^2 = \frac{4\pi ^2 * (150*10^6)^3}{6.67*10^-11 * 1.99 *10^30}\\ T^2 = 1002799.605 \\ T = 1001.398s\\ T = \frac{1001.398}{60} = 16.69min [/tex]
It will take the planet 16.69 minutes to complete one full orbit.
Learn more on planetary rotation here;
https://brainly.com/question/21222010
Question:Which kind of nuclear decay is shown below? Please answer!!!
Answer:
This is alpha decay
Explanation:
what are the SI units of thermal conductivity?
Answer:
The SI unit of thermal conductivity is watts per meter-kelvin (W/(m⋅K)).
Explanation:
hope this will help u
Why is Pluto considered less of an oddball planet now?
Answer:
The International Astronomical Union (IAU) downgraded the status of Pluto to that of a dwarf planet because it did not meet the three criteria the IAU uses to define a full-sized planet. Essentially Pluto meets all the criteria except one—it “has not cleared its neighboring region of other objects
Explanation:
A sports car accelerates from rest to 42 m/s in 6.2 s, what is its acceleration? (6.8 m/s) with proof please
Answer:
Explanation:
As the car Started from rest means that Initial Velocity "Vi = 0 m/s" and final Velocity is given "Vf = 42m/s". Time is given "t = 6.2s"
Acceleration is required a =?
Use Formula;; a = [tex]\frac{Vf-Vi}{t}[/tex]
a = [tex]\frac{42-0}{6.2}[/tex]
a = 6.7741 m/s² ≈ 6.8 m/s²
Mark me as brainliest if you got it...
HELPPPP if this object is accelerating to the right at 2.5 m/s2, arrow C
7.0 N
65 N
27 N
23 N
1.
A proton has a __________ charge.
positive
negative
neutral
2.
___________ and ___________ are two subatomic particles found in the nucleus of an atom
electrons; neutrons
protons; neutrons
protons; electrons
3.
A(n) ______________ is a subatomic particle that has a negative charge.
proton
electron
neutron
Answer:
positive
proton and neutron
electron
what happen to kinetic energy of a body if it speed is doubled
Answer:
The energy possessed by a body because of its motion, equal to one half the mass of the body times the square of its speed is called its kinetic energy. Hence, when velocity is doubled, kinetic energy becomes 4 times.
Explanation:
A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. After the wheel has undergone 14 rotations, what is the radial component of the acceleration (in m/s2) on the edge the wheel?
Answer:
[tex]a=368.97\ m/s^2[/tex]
Explanation:
Given that,
Initial angular velocity, [tex]\omega=0[/tex]
Acceleration of the wheel, [tex]\alpha =7\ rad/s^2[/tex]
Rotation, [tex]\theta=14\ rotation=14\times 2\pi =87.96\ rad[/tex]
Let t is the time. Using second equation of kinematics can be calculated using time.
[tex]\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s[/tex]
Let [tex]\omega_f[/tex] is the final angular velocity and a is the radial component of acceleration.
[tex]\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s[/tex]
Radial component of acceleration,
[tex]a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2[/tex]
So, the required acceleration on the edge of the wheel is [tex]368.97\ m/s^2[/tex].
The radial component of the acceleration (in m/s2) on the edge the wheel is 369.45 m/s²
Using ω² = ω₀² + 2αθ, we find the final angular speed after 14 revolutions, ω where
ω₀ = initial angular speed = 0 rad/s (since the wheel starts from rest), α = angular acceleration = 7 rad/s and θ = number of revolutions = 14 = 14 rev × 2π rad/rev = 87.965 rad.So, substituting the values of the variables into the equation, we have
ω² = ω₀² + 2αθ,
ω² = (0 rad/s)² + 2 × 7 rad/s² × 87.965 rad.
ω² = 0 rad²/s² + 1231.504 rad²/s²
ω² = 1231.504 rad²/s²
ω = √(1231.504 rad²/s²)
ω = 35.09 rad/s
We know that the radial acceleration a = rω² where
r = radius of wheel = diameter of wheel/2 = 60 cm/2 = 30 cm = 0.30 m and ω = final angular acceleration of wheel = 35.09 rad/s.So, substituting the values of the variables into the equation, we have
a = rω²
a = 0.30 m × (35.09 rad/s)²
a = 0.30 m × 1231.504 rad²/s²
a = 369.45 m/s²
So, the radial component of the acceleration (in m/s2) on the edge the wheel is 369.45 m/s².
Learn more about radial acceleration here:
https://brainly.com/question/25243603
If you can run at a speed of 8 miles per hour and you want to run to the store 16 miles away, how
much time will it take you to get there?
Equation needed
work
Answer:
16÷8=2
Explanation:
if you run 8 mi an hour than in 16 mi you would have run 2 hours
Needddd helppppppp!!!
Answer:
2/9 times as strong.
Explanation:
From the question given above, the following assumptions were made:
Initial mass of 1st planet (M₁ ) = M
Initial mass of 2nd planet (m₁ ) = m
Initial distance apart (r₁) = r
Initial Force of attraction (F₁) = F
Final mass of 1st planet (M₂) = 2M
Final mass of 1st planet (m₂) = constant = m
Final distance apart (r₂) = 3r
Final force of attraction (F₂) =?
Next, we shall obtain an expression to determine the new force. This can be obtained as follow:
F = GMm / r²
Cross multiply
Fr² = GMm
Divide both side by Mn
G = Fr² / Mm
Since G is constant, then we have
F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂
Finally, we shall determine the new force as follow:
Initial mass of 1st planet (M₁ ) = M
Initial mass of 2nd planet (m₁ ) = m
Initial distance apart (r₁) = r
Initial Force of attraction (F₁) = F
Final mass of 1st planet (M₂) = 2M
Final mass of 1st planet (m₂) = constant = m
Final distance apart (r₂) = 3r
Final force of attraction (F₂) =?
F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂
Fr² / Mm = F₂ × (3r)² / 2M × m
Fr² / Mm = F₂ × 9r² / 2Mm
Cross multiply
Fr² × 2Mm = F₂ × 9r² × Mm
Divide both side by 9r² × Mm
F₂ = Fr² × 2Mm / 9r² × Mm
F₂ = F × 2 / 9
F₂ = 2/9 F
Thus, the new force is 2/9 times the original force i.e 2/9 times as strong.
V (V)
1 (A)
0
135
9
3
4
1
30
2
2
2
2.
3
4
1
5
0
6
5
6
50
3
Answer:
6755727gvbu7euyeue77377365353663636
Explanation:
ghfjfkjfjfn has a long way is a good relationship is a different person than the difference is that the answer is 8AM is a good relationship with the other 4AM in a certain way is a different type and have different effects and feelings and the difference is the difference is that you can do it for a long period and the same way you know what to smome but you can get to the claire family a lot of times and the difference between the two is that the person you want and you know it would have different opinions on the subject and then use the difference between the two and a half hour and the difference is that the answer is yes but if you're not going anywhere in the past two and your life you can do that with your 6 or an 6th year old girl who is a good things go to
Determine the torque caused by the cable pulling horizontally on the inclined drawbridge. The force the cable exerts on the bridge is 5000N and the bridge length is 8 m.
Answer:
[tex]\tau=20000\ Nm[/tex]
Explanation:
Given that,
The force the cable exerts on the bridge is 5000N
The length of the bridge is 8 m
We need to find the torque caused by the cable pulling horizontally on the inclined draw bridge. The torque acting on it is given by :
[tex]\tau=Fr[/tex]
Putting all the values,
[tex]\tau=5000\ N\times 8\ m\\\\=20000\ Nm[/tex]
So, the required torque is 20000 Nm.
A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate
the following using the energy/work formulae only:
1.The kinetic energy at the top of the cliff
2.The total mechanical energy at the top of the cliff
3.The kinetic energy of the Rock half way down 4.the speed of the Rock half way down
5.The speed of the Rock as it hits the ground
Answer:
this is the answer......
On a horizontal surface is located
an object of mass 20 kg and it is observed that
To set it in motion, a
80 N force, whereas if the object is
in motion, only one
40 N force, to keep it at speed
constant. Determine the values of the coefficients
static and kinetic friction, respectively.
(g = 10 m / s2).
A) 0.32; 0.11
B) 0.36; 0.15
C) 0.40; 0.2
D) 0.45; 0.22
By Newton's second law, the net vertical force acting on the object is 0, so that
n - w = 0
where n = magnitude of the normal force of the surface pushing up on the object, and w = weight of the object. Hence n = w = mg = 196 N, where m = 20 kg and g = 9.80 m/s².
The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of static friction µ is such that
80 N = µ (196 N) → µ = (80 N)/(196 N) ≈ 0.408
Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of kinetic friction ν is
40 N = ν (196 N) → ν = (40 N)/(196 N) ≈ 0.204
And so the closest answer is C.
(Note: µ and ν are the Greek letters mu and nu)
Automobiles must be able to sustain a frontal impacl The automobile design must allow low speed impacts with little sustained damage, while allowing the vehicle front end structure to deform and absorb impact energy at higher speeds. Consider a frontal impact test of a 1000 kg mass vehicle. (a) For a low speed test at 2.5 rnls, compute the energy in the vehicle just prior to impacl If the bumper is a pure elastic element, what is the effective design stiffness required to limit the bumper maximum deflection during impact to 4 em
Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m
Explanation:
Given that;
mass of vehicle m = 1000 kg
for a low speed test; V = 2.5 m/s
bumper maximum deflection = 4 cm = 0.04 m
First we determine the energy of the vehicle just prior to impact;
W_v = 1/2mv²
we substitute
W_v = 1/2 × 1000 × (2.5)²
W_v = 3125 J
now, the the effective design stiffness k will be:
at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;
hence;
W_v = 1/2kx²
we substitute
3125 = 1/2 × k (0.04)²
3125 = 0.0008k
k = 3125 / 0.0008
k = 3906250 N/m
Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m
What force (in N) must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2100 kg car (a large car) resting on the slave cylinder
Complete Question:
What force (in N) must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2100 kg car (a large car) resting on the slave cylinder ? The master cylinder has a 2.00-cm diameter and the slave has a 24.0-cm diameter
Answer:
[tex]F_1 = 142.92N[/tex]
Explanation:
Given
[tex]m = 2100kg[/tex] --- mass
[tex]D_1 = 2.00\ cm[/tex] --- diameter of the large cylinder
[tex]D_2 = 24.0\ cm[/tex] --- diameter of the slave cylinder
To do this, we apply Archimedes' principle of buoyancy which implies that:
[tex]P = \frac{F_1}{A_1} = \frac{F_2}{A_2}[/tex]
Where
[tex]F_1 = Force\ on\ the\ master\ cylinder[/tex]
[tex]F_2 = Force\ on\ the\ slave\ cylinder[/tex]
[tex]A_1 = Area\ of\ the\ master\ cylinder[/tex]
[tex]F_2 = Area\ of\ the\ small\ cylinder[/tex]
Calculating the area of the master cylinder.
[tex]A_1 = \pi r_1^2[/tex]
[tex]r_1 = \frac{1}{2}D_1 = \frac{1}{2} * 2.00cm = 1.00cm[/tex]
[tex]A_1 = \pi* 1^2[/tex]
[tex]A_1 = \pi * 1[/tex]
[tex]A_1 = \pi[/tex]
Calculating the area of the slave cylinder.
[tex]A_2 = \pi r_2^2[/tex]
[tex]r_2 = \frac{1}{2}D_2 = \frac{1}{2} * 24.00cm = 12.00cm[/tex]
[tex]A_2 = \pi* 12^2[/tex]
[tex]A_2 = \pi* 144[/tex]
[tex]A_2 = 144\pi[/tex]
Substitute these values in:
[tex]P = \frac{F_1}{A_1} = \frac{F_2}{A_2}[/tex]
[tex]\frac{F_1}{\pi} = \frac{F_2}{144\pi}[/tex]
Multiply both sides by [tex]\pi[/tex]
[tex]\pi * \frac{F_1}{\pi} = \frac{F_2}{144\pi} * \pi[/tex]
[tex]F_1 = \frac{F_2}{144}[/tex]
The force exerted on the slave cylinder (F2) is calculated as:
[tex]F_2 = mg[/tex]
[tex]F_2 = 2100 * 9.8[/tex]
[tex]F_2 = 20580[/tex]
Substitute 20580 for F2 in [tex]F_1 = \frac{F_2}{144}[/tex]
[tex]F_1 = \frac{20580}{144}[/tex]
[tex]F_1 = 142.92N[/tex]
Hence, the force exerted on the master cylinder is approximately 142.92N
Collision practice
Running at 2.0 m/s, Bruce, the 45.0-kg quarterback, collides with Biff, the 90.0 kg tackle, who is traveling at 7.0 m/s in the other direction. Upon collision,
Biff wraps up Biff for the tackle. What is the resultant velocity of the players.
V=————— m/s
Answer:
[tex]v_f=4\:\mathrm{m/s\: in\:Biff's\:direction}[/tex]
Explanation:
Using the Law of Conservation of Momentum, we can set up the following equation:
[tex]m_2v_2-m_1v_1=m_fv_f[/tex], where [tex]m_1v_1[/tex] is Bruce's momentum and [tex]m_2v_2[/tex].
Plugging in given values, we get:
[tex]90\cdot7-45\cdot2=(90+45)v_f, \\v_f=\frac{540}{135}=\fbox{$4\:\mathrm{m/s}$}[/tex].
An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the way down and passes a point a distance 50.0 mm below its starting point at a time 5.00 ss after it leaves the thrower's hand. Air resistance may be ignored.
1. What is the initial speed of the egg?
v=?m/s
2. How high does it rise above its starting point
h=?m
3. What is the magnitude of its velocity at the highest point?
v=?m/s
4. What is the magnitude of its acceleration at the highest point?
a=?m/s^2
Answer:
Explanation:
Displacement ( downwards ) = 50 m , initial speed = - u (upwards ) ,
time = 5 s .
acceleration due to gravity( downwards) = 9.8 m /s² .
s = ut + 1/2 g t²
50 = - 5 u + .5 x 9.8 x 5²
50 = -5u + 122.5
5u = 122.5 - 50
u = 14.5 m /s
2 )
final velocity v = 0 , height upto which it rises = h
v² = u² - 2 g h
0 = 14.5² - 2 x 9.8 h
h = 10.72 m
3 )
At the highest point velocity = 0
4 )
At the highest point acceleration = 9.8 m /s² downwards .
