Observe the given figure and find the the gravitational force between m1 and m2.​

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

41.23

v(B,E)=40 GOING EAST

V(R,E)=10 COMING DOWN

then 41.23

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

Answer:

ballance

i just took the quiz

Answer:

55

Explanation:95-40=55

i hope i did the math right if i didnt please tell me

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

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  the most important safety feature on a car is B. Safety Belt

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

https://brainly.com/question/25820562?referrer=searchResults

#SPJ3

Answer:

The answer should be: 1.20 A

Explanation:

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

[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

The 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

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

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

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.

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

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]

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

[tex]Option \: (b)[/tex]

Explanation:

Second STATEMENT SAYS REGARDING TO THE CHEMICAL poison

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]

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.

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

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.

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

Answer:

A is the answer!

Explanation:

Edge 2021

Answer:

A

Explanation:

Edge

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.