Observe the given figure and find the the gravitational force between m1 and m2.
Answers
Answer:
The gravitational force between m₁ and m₂, is approximately 1.06789 × 10⁻⁶ N
Explanation:
The details of the given masses having gravitational attractive force between them are;
m₁ = 20 kg, r₁ = 10 cm = 0.1 m, m₂ = 50 kg, and r₂ = 15 cm = 0.15 m
The gravitational force between m₁ and m₂ is given by Newton's Law of gravitation as follows;
[tex]F =G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}[/tex]
Where;
F = The gravitational force between m₁ and m₂
G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
r₂ = 0.1 m + 0.15 m = 0.25 m
Therefore, we have;
[tex]F = 6.67430 \times 10^{-11} \ N \cdot m^2/kg \times \dfrac{20 \ kg\times 50 \ kg}{(0.1 \ m+ 0.15 \ m)^{2}} \approx 1.06789 \times 10^{-6} \ N[/tex]
The gravitational force between m₁ and m₂, F ≈ 1.06789 × 10⁻⁶ N
Related Questions
A thunderclap sends a sound wave through the air and the ocean below. The
thunderclap sound wave has a constant frequency of 100 Hz. What is the
wavelength of the sound wave in water? (The equation for the speed of a
wave is v= f x 1.)
Water
Diamond
Glass
Air
1,493
12,000
5,640
346
Speed of
sound
(m/s)
A. 11.00 m
B. 12.00 m
C. 14.93 m
D. 3.46 m
Answers
Answer:
C. 14.93 m
Explanation:
The given frequency of the wave, f = 100 Hz
The given equation for the wave speed, v, is presented as follows;
v = f × λ
The speed of sound in water, v = 1,493 m/s
Therefore, we get;
The wavelength, λ = v/f
∴ λ = 1,493 m/s/(100 Hz) = 14.93 m
The wavelength, λ = 14.93 m.
A toy airplane is flying at a speed of 3 m/s with an acceleration of 1.1 m/s squared how fast is it flying after five seconds
Answers
Answer:
[tex]8.5\frac{m}{s}[/tex]
Explanation:
Use Kinematics:
[tex]v = v_0 + at[/tex]
[tex]v = 3 + 1.1 * 5[/tex]
[tex]v = 8.5 \frac{m}{s}[/tex]
A pendulum of mass 18 kg is released from rest at some height, as shown by
point A in the image below. At the bottom of its arc at point B, it is traveling at
a speed of 17 m/s. What is the approximate amount of energy that has been
lost due to friction and air resistance? (Recall that a=98 m/s2
Answers
By the work-energy theorem, the total work done on the mass as it swings is
W = ∆K = 1/2 (18 kg) (17 m/s)² = 153 J
No work is done by the tension in the string, since it's directed perpendicular to the mass at every point in the arc. Similarly, the component of the mass's weight mg pointing perpendicular to the arc also performs no work.
If we ignore friction/drag for the moment, the only remaining force is the parallel component of weight, which performs mgh = (176.4 N) h of work, where h is the vertical distance between points A and B.
Now, if w is the amount of work done by friction/air resistance, then
(176.4 N) h - w = 153 J
If you know the starting height h, then you can solve for w.
A force of 825N is required to push
a car across a lot.Tow students push the car 35m.How much work is done?
Answers
Answer:
the work done by the two students is 28,875 J
Explanation:
Given;
applied force, f = 825 N
distance through which the car is pushed, d = 35 m
The work done by the two students who pushed the 825 N across 35 m lot is calculated as follows;
Work done = force x distance
Work done = 825 N x 35 m
Work done = 28,875 Nm = 28,875 J
Therefore, the work done by the two students is 28,875 J
Pls help asap!!
A bucket contains hot water at 95°c. A man wants to bath with water at 40°c. What is the ratio of the mass of hot water to the mass of cold water that he needs.
Answers
Answer:
55
Explanation:95-40=55
i hope i did the math right if i didnt please tell me
Draw a closed circuit diagram of the battery of 2 cells arranged in series, connecting wire, switch and bulb; mark the direction of the current.
pls ill give brainly
Answers
how far should an object be from a converging mirror of radius 36cm to form a real image one - ninth of its size
Answers
Answer:
focal length :
[tex]{ \tt{f = \frac{r}{2} }} \\ focal \: length = \frac{36}{2} = 18 \: cm[/tex]
From linear magnification:
[tex]{ \tt{ \frac{1}{m} + 1 = \frac{u}{f} }} \\ \frac{1}{9} + 1 = \frac{u}{18} \\ \\ u = \frac{18 \times 10}{9} \\ u = 20 \: cm[/tex]
The object must be at 20 cm
A parallel combination of a 1.47 μF capacitor and a 2.95 μF capacitor is connected in series to a 4.89 μF capacitor. This three‑capacitor combination is connected to a 15.5 V battery. Determine the charge on each capacitor.
Answers
Answer:
a. i. 35.96 μC b. i. 11.98 μC ii. 24.04 μC
Explanation:
We need to find the total capacitance of the system C.
The total capacitance of the parallel combination of a 1.47 μF capacitor and a 2.95 μF capacitor is C' = 1.47 μF + 2.95 μF = 4.42 μF.
C' = 4.42 μF is in series with the 4.89 μF capacitor and for a series combination of capacitors, we have the total capacitance, C from
1/C = 1/4.42 μF + 1/4.89 μF
1/C = (4.42 μF + 4.89 μF)/(4.42 μF × 4.89 μF)
1/C = 9.31 μF/21.6138 μF²
C = 21.6138/9.31 μF
C = 2.32 μF
So, the total charge in the circuit Q = CV where C = total capacitance = 2.32 μF and v = voltage = 15.5 V
So, Q = CV
Q = 2.32 μF × 15.5 V
Q = 35.96 μC
i. The charge on the 4.89 μF capacitor
Since the 4.89 μF is in series with C', the total charge flowing i the circuit is the total charge in the 4.89 μF capacitor. So, its charge Q = 35.96 μC
b. The charge in the 1.47 μF and 2.95 μF capacitors.
To find the charge in the 4.89 μF and 2.95 μF capacitors, we need to find the voltage across the combined parallel combination of a 1.47 μF capacitor and a 2.95 μF capacitor. The voltage, V' across the 4.89 μF capacitor, since Q = CV', V' = Q/C = 35.96 μC/4.89 μF = 7.35 V
So, the voltage V" across the combined parallel combination of a 1.47 μF capacitor and a 2.95 μF capacitor, C' is V" = 15.5 V - V' (since V' + V" = 15.5 V).
So, V" = 15.5 V - V'
V" = 15.5 V - 7.35 V
V" = 8.15 V
i. The charge on the 1.47 μF capacitor
Using Q' = CV" where Q' = charge across capacitor, C = 1.47 μF and V" = 8.15 V.
So, Q' = CV"
Q' = 1.47 μF × 8.15 V
Q' = 11.98 μC
ii. The charge on the 2.95 μF capacitor
Using Q" = CV" where Q' = charge across capacitor, C = 2.95 μF and V" = 8.15 V.
So, Q" = CV"
Q" = 2.95 μF × 8.15 V
Q" = 24.04 μC
Photosynthesis is a process in which plants prepare food using carbon dioxide, chlorophyll, and water in the presence of sunlight.
Answers
HELP ASAP
A. 1.09 A
B. 1.20 A
C.0.910 A
D. 0.830 A
Answers
Answer:
The answer should be: 1.20 A
Explanation:
What symbol is a concentration be
Answers
[tex]Option \: (b)[/tex]
Explanation:
Second STATEMENT SAYS REGARDING TO THE CHEMICAL poison
What is the relationship between electric field lines and equipotential lines that you observed in doing the lab
Answers
Answer:
Explained below
Explanation:
Generally speaking, we know in physics that Electric field lines are lines which usually start at positive charges and deflect away from them to terminate at the negative charges. Meanwhile Equipotential lines are lines that are used to connect points located on the same electric potential.
Finally, in conclusion, electric field lines are usually lines that go through in a perpendicular manner across every equipotential lines.
which is part of the convection cycle in earths atmosphere?
a. hot, denser air rises
b. cold, denser air rises
c. hot, less-dense air rises
d. cold, less-dense air falls
Answers
I hope this helps !!!!!!!!
một vật m=1kg có động lượng bằng 10kg.m/s. động năng của vật là
Answers
A rectangular coil of wire, 22.0 cm by 35.0 cm and carrying a current of 1.40 A, is oriented with the plane of its loop perpendicular to a uniform 1.50-T magnetic field pointing into the plane of the loop. Let the loop be in x-y Cartesian plane so that the long and short sides of the loop are parallel to x- and y-axis, respectively. The loop center is at the origin of x-y Cartesian plane. Note that the magnetic field is in the direction of the negative z-axis.a. Calculate: (i) the net force that the magnetic field exerts on the coil; (ii) the torque about the z-axis that the magnetic field exerts on the coil.b. The plane of the coil is now rotated through +30º from its initial orientation (the x-y plane of the Cartesian coordinate system that remains the same). Calculate: (i) the net force that the magnetic field exerts on the coil; (ii) the torque about the rotation axis that the magnetic field exerts on the coil.
Answers
Answer:
a) [tex]F_{net}=0[/tex]
b) [tex]T=0[/tex]
Explanation:
From the question we are told that:
Dimensions:
[tex]L*B=22.0*35.0cm[/tex]
Current [tex]I=1.40A[/tex]
Magnetic field [tex]B=1.40[/tex]
Therefore
[tex]Area=L*B[/tex]
[tex]A=22.0*35.0cm[/tex]
[tex]A=770cm=>770*0^{-4}[/tex]
a)
Generally Force on Looping gives
[tex]F_1-F_2[/tex]
[tex]F_3=F_4[/tex]
Therefore
[tex]F_{net}=0[/tex]
b)
Generally the equation for Torque is mathematically given by
[tex]T=i*Asin \theta[/tex]
Since A and B are on opposite direction
[tex]\theta=180[/tex]
Therefore
[tex]T=1.40*770*10^{-4}sin 180[/tex]
[tex]T=0[/tex]
Which statement describes why energy is released in a nuclear fission reaction based on mass-energy equivalence?
A. For large nuclei, the mass of the original nucleus is greater than the mass of the products.
B. For large nuclei, the mass of the original nucleus is less than the mass of the products.
C. For small nuclei, the binding energy of the lighter nuclei is greater than the binding energy of the heavier nucleus.
D. For small nuclei, the binding energy of the lighter nuclei is less than the binding energy of the heavier nucleus.
Answers
Answer:
A is the answer!
Explanation:
Edge 2021
Answer:
A
Explanation:
Edge
what is projectile motion
Answers
[tex]\boxed{\large{\bold{\textbf{\textsf{{\color{blue}{Answer}}}}}}:)}[/tex]
[tex]\sf{\qquad{\qquad{\underline{\underline{ Projectile~motion }}}}}[/tex]
If an object is given an initial velocity in any direction and then allowed to travel freely under gravity only, it is called a projectile motion.
It is basically 3 types
horizontally projectile motion oblique projectile motion included plane projectile motionThe path followed by a projectile is called its trajectory.
Projectile motion is when an object moves in a bilaterally symmetrical, parabolic path.
The path that the object follows is called its trajectory.
Projectile motion only occurs when there is one force applied at the beginning, after which the only influence on the trajectory is that of gravity
Which activity is best described as a scientific endeavor?
A. Designing devices astronauts can use to help them stay healthy
while in space
B. Developing a procedure for testing how extended stays in space
affect the human body
C. Performing tests on devices used by astronauts to ensure they
can be used safely
D. Creating a process for improving the design of equipment used to
monitor astronauts' health
Answers
Answer:
B
Explanation:
BECAUSE TO DO THE TESTS YOU NEED TO DO THE SCIENTIFIC METHOD.
FOR EXAMPLE: OBSERVATIONS AND EXPERIMENTS TO OBTAIN RESULTS.
ANYWAY I LEAVE YOU THE LINK:
https://gscourses.thinkific.com
Use the information below the answer the following 3 questions.
A 50 kg crate is being dragged across a floor by a force of 225 N at an angle of 40o from the horizontal. The crate is dragged a distance of 5.0 m and the frictional force is 60 N.
Question 2 (2 points)
Question 2 options:
The work done on the crate by the applied force is ___x102 Nm. (Give your answer with the correct number of sign digs and do not include units).
Question 3 (2 points)
Question 3 options:
The work done on the crate by the frictional force is -___x102 Nm. (Give your answer with the correct number of sign digs and do not include units).
Question 4 (2 points)
Question 4 options:
The net work done on the crate is ___x102 Nm. (Give your answer with the correct number of sign digs and do not include units).
Hint: Do not use rounded answers in subsequent calculations
Answers
Answer:
2. 8.62×10² Nm
3. 2.30×10² Nm
4. 6.32×10² Nm
Explanation:
2. Determination of the work done by the applied force.
Force (F) = 225 N
Distance (d) = 5 m
Angle (θ) = 40°
Workdone (Wd) =?
Wd = Fd × Cos θ
Wd = 225 × 5 × Cos 40
Wd = 8.62×10² Nm
3. Determination of the work done by the frictional force.
Frictional Force (Fբ) = 60 N
Distance (d) = 5 m
Angle (θ) = 40°
Workdone (Wd) =?
Wd = Fբd × Cos θ
Wd = 60 × 5 × Cos 40
Wd = 2.30×10² Nm
4. Determination of the net work done.
We'll begin by calculating the net force acting on the crate
Force applied (F) = 225 N
Frictional Force (Fբ) = 60 N
Net force (Fₙ) =?
Fₙ = F – Fբ
Fₙ = 225 – 60
Fₙ = 165 N
Finally, we shall determine the net Workdone. This can be obtained as follow:
Net force (Fₙ) = 165 N
Distance (d) = 5 m
Angle (θ) = 40°
Workdone (Wd) =?
Wd = Fₙd × Cos θ
Wd = 165 × 5 × Cos 40
Wd = 6.32×10² Nm
A cyclist is travelling eastwards at a velocity of 40 ms and rain is falling vertically at a speed of 10 ms. Find the velocity of the rain relative to the cyclist.using vectors to solve it
Answers
41.23
v(B,E)=40 GOING EAST
V(R,E)=10 COMING DOWN
then 41.23
8. A copper container of 84g mass contains 84g of water at 20°C. 46g of water at 200°C is mixed with water in the copper ontainer. What is the final temperature of the water? Specific heat capacity of water = 4200 J kg-1 °C-1, Specific heat capacity of copper = 400 J kg-1 °C-1
Answers
Answer:
80 °C
Explanation:
The heat transfer parameters for the water and copper container are;
Mass of the copper container, m₁ = 84 g
Mass of the water in the container, m₂ = 84 g
Initial temperature of the water in the container, T₂ = 20°C
Mass of the hot water added, m₃ = 46 g
Initial temperature of the hot water, T₃ = 200°C
Specific heat capacity of water, c₂ = 4,200 J·kg⁻¹·°C⁻¹
Specific heat capacity of copper, c₁ = 400 J·kg⁻¹·C⁻¹
The formula for the specific heat, ΔQ = m·c·ΔT
The heat lost by the hot water = The heat gained by the container the and the cold water
The formula for the specific heat of the mixture is presented as follows;
m₃ × c₃ × (T₃ - T) = m₁ × c₁ × (T - T₁) + m₂ × c₂ × (T - T₂)
Where T represents the final temperature of the water
Therefore, by plugging in the values, we get;
46 × 4200 × (200 - T) = 84 × 400 × (T - 20) + 84 × 4200 × (T - 20)
38640000 - 193200·T = 386400·T - 7728000
38640000 + 7728000 = 46368000 = 386400·T + 193200·T = 579,600·T
∴ T = 46368000/579,600 = 80
The final temperature of the water, T = 80°C
What is the final step in the fourth stage of technological design, after a product has been improved abs approved
Answers
Answer:
Evaluate the solution.
Explanation:
A technological design is designed as the design and study of a solution that can be provided from the solution by identifying the root cause or problem and trying to solve by various means.
A good technological design requires the minimum effort and resources while meeting the requirement of the problem.
The steps involved in the technological design are :
1. search and identify the problem or need.
2. design a solution
3. Implement a solution.
4. Evaluate the solution.
Therefore, the final step or the fourth step in the process of a technological design is " evaluating or communicating the final design solution".
For saving lives, what is the most important safety feature on a car? A. Air bag B. Safety Belt C. Anti-lock brakes
Answers
For saving lives the most important safety feature on a car is B. Safety Belt
What are safety features of a car ?Safety features of a car is a feature of a product designed to ensure or increase safety.
Air bag and Anti-lock brakes are the supplemental protection and designed to work best with combination with seat bells.
Air bag reduce the chance that upper body or head will strike the vehicle's interior during a crash alongside with belt that will also hold your upper body
so, the primary safety feature is seat belt and Air bag and Anti-lock brakes comes in secondary safety feature as they increases the safety and risk of getting an injury during any accident
correct answer is B. Safety Belt
learn more about Safety features
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A girl is standing 150m in front of a tall building, fires a shot with starting pistol. A boy standing 350m behind her, hears two bangs 1s apart. From this information what is speed of sound in air?
Answers
Answer:
300 m/s
Explanation:
The difference in time between the two bangs is 1 s.
Thus;
t2 - t1 = 1
We know that distance/time = speed.
Thus;
d2/v - d1/v = 1
Multiply through by v to get;
d2 - d1 = v
Where v is speed of sound in air.
d1 = 350 m
d2 = (150 × 2) + 350 = 650 m
Thus;
v = d2 - d1 = 650 - 350 = 300 m/s
how can mass be measured?
spring scale
balance
scale
bathroom scale
Answers
Answer:
ballance
i just took the quiz
Which of the following is evidence for continental drift?
Answers
Solid to gas and gas to solid on the basis of kinetic model.
Answers
Explanation:
The kinetic molecular theory of matter states that: ... The average amount of empty space between molecules gets progressively larger as a sample of matter moves from the solid to the liquid and gas phases. There are attractive forces between atoms/molecules, and these become stronger as the particles move closer together.
an object 5 cm in size is placed at 30 cm in front of a concave mirror of focal length 45 CM at what distance from the mirror should a screen be placed in order to obtain a shop image find the magnification size and nature of the image
Answers
Answer: Given: h₀=5cm
p=30cm
hi
f=45cm
require: q=?
hi=?
formulas:
1/f=1/p+1/q
hi/h₀=q/p
calculations:1/q=1/f-1/p
1/q=1/45-1/30
1/q=0.022=0.033
1/q= = -0.011
q=1/-0.011
q= -90.91cm
hi=p × h₀/q
hi=30ₓ5/-90.91
hi=150/-90.91
hi= -1.65cm
image is virtual , erect and diminished
please mark as brainliest plzzzzzzzzzz
if the pelican in item 3 was traveling at the same speed but was only 2.7m above the water, how far would the fish travel horizontally before hitting the water?
Answers
Answer:
5.66 m
Explanation:
From online sources, the speed in item 3 being referred to was discovered to be 7.62 m/s
Now, let's get the time of flight from one of Newton's equation of motion;
S = ut + ½gt²
Considering the vertical component, we have u = 0 m/s.
Thus;.
S = ½gt²
Plugging in the relevant values;
2.7 = ½ × 9.8 × t²
t² = 2.7/4.9
t = √(2.7/4.9)
t = 0.7423 s
Now, when we consider the horizontal component of the motion, we have;
S = vt
Where;
S is the distance the fish will travel horizontally before hitting the water.
v = 7.62 m/s
t = 0.7423
Thus
s = 7.62 × 0.7423
s ≈ 5.66 m
The net charge difference across the membrane, just like the charge difference across the plates of a capacitor, is what leads to the voltage across the membrane. How much excess charge (in picocoulombs, where 1 pc = 1x10-12 C) must lie on either side of the membrane of an axon of length 2 cm to provide this potential difference (0.07 V) across the membrane? You may consider that a net positive charge with this value lies just outside the axon cell wall, and a negative charge with this value lies just inside cell wall, like the equal and opposite charges on capacitor plates. Again, make sure to include LaTeX: \kappa=7κ = 7 for the lipid bilayer.
Answers
Answer:
a) Q = 1.24 10⁻² pC, b) Q = 8.68 10⁻² pC
Explanation:
a) the capacitance is defined
C = [tex]\frac{Q}{\Delta V} = \epsilon_o \frac{A}{d}[/tex]
Q = ε₀ [tex]\frac{A}{d} \ \Delta V[/tex]
let's calculate
Q = 8.85 10⁻¹² 0.07 [tex]\frac{A}{d}[/tex]
Q = 0.6195 10⁻¹² [tex]\frac{A}{d}[/tex]
where a is the area of the membrane
A = d L
Q = 0.6195 10⁻¹² Ll
Q = 0.6195 10⁻¹² 0.02
Q = 1.24 10⁻¹⁰ C
Q = 1.24 10⁻² pC
B) the membrane is full of fat with k = 7
C = [tex]\frac{Q}{\Delta V} = k \epsilon_o \ \frac{A}{d}[/tex]
Q = k ε₀ [tex]\frac{A}{d} \ \Delta V[/tex]
Q = k Q₀
Q = 7 1.24 10⁻²
Q = 8.68 10⁻² pC
