2. A lever has efficiency 60% and mechanical advantage 4, then
(i) Calculate effort required to lift 1000 N load. [250 N] (ii
(ii) Calculate velocity ratio of the lever. [VR = 6.66]
(iii) Find out output work to turn load up to 20 m. [20,000 Nm]
1:0​

Answers

Answer 1

Answer:

(i) Effort = 250 N

(ii) Velocity ratio = 6.667

(iii) work output = 20 000 Nm

Explanation:

Given: efficiency = 60%, mechanical advantage = 4.

(i) Mechanical advantage = [tex]\frac{Load}{effort}[/tex]

Load = 1000 N

⇒ Effort = [tex]\frac{Load}{Mechanical advantage}[/tex]

             = [tex]\frac{1000}{4}[/tex]

             = 250 N

Effort = 250 N

(ii) Efficiency = [tex]\frac{Mechanical advantage}{velocity ratio}[/tex] x 100%

Velocity ratio = [tex]\frac{Mechanical advantage}{Efficiency}[/tex] x 100%

                      = [tex]\frac{4}{60}[/tex] x 100

                      = 6.667

Velocity ratio = 6.667

(iii) Work output = Load x distance moved by load

                         = 1000 x 20

                         = 20 000

Work output = 20 000 Nm


Related Questions

An incidence that is caused by sleep and also occurs during sleep is known as a __________.

Answers

Answer:

parasomnia

Explanation:

Pretty sure it’s parasmonia (if I’m spelling that right

Chef Andy tosses an orange in the air, then catches it again at the same height. The orange is in the air for 0.75\,\text s0.75s0, point, 75, start text, s, end text. We can ignore air resistance.
What was the orange's velocity at the moment it was tossed into the air?

Answers

Answer:

90.82%digrash silsho

When the mallet hits the ball with an action force, the ball exerts a reaction 1 force on the mallet as explained by: 1) Newton's first law 2) Newton's second law 3) Newton's third law 4) all three of Newton's laws​

Answers

Answer:

It's Newton's third law. For every known reaction, there is a equal and opposite reaction.

Explanation:

The mallet which hits the ball with an action force, the ball exerts a reaction force on the mallet as explained by the Newton's third law. Thus, the correct option is 3.

What is the Newton's third law?

Newton's third law states that for every action (force) in nature there is an equal and opposite force or reaction. An example of third law of motion is that the mallet hits the ball with an action force, then the ball exerts a reaction.

If an object A exerts a force on the object B, then object B must also exert a force of equal magnitude and in the opposite direction back on object A. The law represents a certain symmetry in the nature that the forces always occur in pairs, and one body cannot exert a force on another without experiencing any kind of force itself.

Therefore, the correct option is 3.

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What is horizontal motion

Answers

Horizontal motion is defined as a projectile motion in a horizontal plane depending upon the force acting on it. For a short distance, the vertical and horizontal components of a projectile are perpendicular and independent of each other.

Answer:

A projectile moves along its path with a constant horizontal velocity

The most common elements in the Earth’s crust are rarely found on their own. They are usually found combined. Suggest why this is.

Answers

Answer:

Because there are other elements that get in touch with it. And over time with, evolution, erosion and temperature their particles tend to mix. So what once was very common became mixed over time.

The most common element throughout the Earth’s crust that are rarely found on their own will be "Astatine".

Astatine seems to be the very most uncommon or infrequent natural substance throughout Earth's crust that occurs just under 1 gram at any time.It's indeed exceedingly impossible to make Astatine-210, particularly in perhaps the most useful state, don't just in less and less astatine present in the fundamental natural world.

Thus the above answer is correct.

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12. A rolling ball has 18 J of kinetic energy and is rolling 3.0 m/s. Find its mass.

Answers

Answer:

4g

I hope it will be useful.

Explanation:

KE = (1/2)*m*v^2

Can you please help me with this physics question

Answers

Answer:

See the answers below

Explanation:

We can solve both problems using vector sum.

a)

Let's assume the forces that help the diver dive as positive downward, and the forces that oppose upward, as negative

[tex]F_{resultant}=100+30-85+900\\F_{resultant}=845[N][/tex]

The drag force is horizontal d this way in the horizontal direction we will only have the drag force that produces the water stream.

[tex]F_{drag}=50[N][/tex]

b)

Let's assume the forces that propel the rocket upwards as positive and forces like the weight of the rocket and other elements as negative forces.

[tex]F_{resultant}=960+7080-7700\\F_{resultant}=340 [kN][/tex]

3. A boat at sea is being moved by the ocean current at 5 metres/minute directly to the east. At the same time it is being moved directly to
the north at 3 metres/minute by a strong wind. Determine the magnitude and direction of the boat's overall velocity.

Answers

Explanation:

A boat at sea is being moved by the ocean current at 5 metres/minute directly to the east.

At the same time, it is being moved directly to  the north at 3 metres/minute by a strong wind.

We need to find the magnitude and direction of the boat's overall velocity.

As both velocities are moving perpendicular to each other. The resultant velocity is given by :

[tex]v=\sqrt{5^2+3^2} \\\\=\sqrt{25}\\\\=5\ \text{metres/minute}[/tex]

For direction,

[tex]\tan\theta=\dfrac{v_y}{v_x}\\\\=\dfrac{3}5{}\\\\\theta=\tan^{-1}(\dfrac{3}5{})\\\\=30.96^{\circ}[/tex]

Hence, this is the required solutin.

1. Which of the following is an example of projectile motion?
An elevator
Ajet taking off
O A football flying through the air
Dropping an aluminum can into the garbage

Answers

Answer:

A football flying through the air

Explanation:

when an object has projectile motion it is moving through the air

From the given options, the football flying through the air is the prime example of projectile motion. (option d)

Projectile motion refers to the motion of an object that is launched into the air and moves under the influence of gravity alone. It follows a curved path known as a projectile trajectory. Among the options provided, the example that best represents projectile motion is a football flying through the air.

When a football is kicked or thrown, it experiences an initial force that propels it forward. However, as soon as it leaves the kicker's or thrower's hand or foot, the only force acting on it is gravity. Gravity pulls the football downward, causing it to follow a curved path. The shape of the path depends on the angle at which the football is launched and its initial velocity.

During its flight, the football follows a parabolic trajectory, moving both horizontally and vertically. The horizontal component of its motion remains constant, while the vertical component experiences a downward acceleration due to gravity. This combination of horizontal and vertical motion creates the characteristic curved path of projectile motion.

Therefore, option (d) is correct.

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Was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s? Why?/Why Not?

(was not able to hit the target, but I don't know the reason)

Answers

Answer:

The cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

Explanation:

From the question given above, the following data were obtained:

Height to which the target is located = 50 m

Initial velocity (u) = 20 m/s

To know whether or not the cannon ball is able to hit the target, we shall determine the maximum height to which the cannon ball attained. This can be obtained as follow:

Initial velocity (u) = 20 m/s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = 10 m/s²

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 20² – (2 × 10 × h)

0 = 400 – 20h

Collect like terms

0 – 400 = – 20h

– 400 = – 20h

Divide both side by – 20

h = – 400 / – 20

h = 20 m

Thus, the the maximum height to which the cannon ball attained is 20 m.

From the calculations made above, we can conclude that the cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

A 0N
B 6N
C 10 N
D 12 N

Answers

Answer:

The net force acting on the object is 0 N

Explanation:

Newton's Second Law of Forces

The net force acting on a body is proportional to the mass of the object and its acceleration.

The net force can be calculated as the sum of all the force vectors in each rectangular coordinate separately.

The image shows a free body diagram where four forces are acting: two in the vertical direction and two in the horizontal direction.

Note the forces in the vertical direction have the same magnitude and opposite directions, thus the net force is zero in that direction.

Since we are given the acceleration a =0, the net force is also 0, thus the horizontal forces should be in equilibrium.

The applied force of Fapp=10 N is compensated by the friction force whose value is, necessarily Fr=10 N in the opposite direction.

The net force acting on the object is 0 N

Which of the following is true about thoughts

Answers

ya but what’s the following. there’s no options to choose between

Answer:

they can be unpredictable until you think it

Explanation:

calculate the magnitude of the net force on a 4.5 kg mass if the mass moves straight up at constant speed

Answers

Answer:

0

Explanation:

as Newton 1st Law said that every object will remain at rest or in uniform motion (constant speed) in a straight line unless compelled to change its state by the action of an external force

Net Force = 0

If an elevator is moving at a speed of 5 m/s and the person in it is 82 kg what is the person’s kinetic energy?

Answers

Answer:

KE 1025J

[tex] \frac{1}{2} m {v}^{2} \\ [/tex]

Explanation:

[tex] \frac{1}{2} \times 82 \times {5}^{2} [/tex]

1025J

concave lens has negative focal length why​

Answers

if it does do it then yeah it will for it
For focal length, f in lens is always taken as negative for concave and positive for convex. ... It is taken as negative in Concave lens since image is formed in -X side of the Cartesian. In Concave mirror V is taken as positive and negative for convex mirror.

marked as brainiest if correct

Answers

Answer:

I think the the answer is creating and layout and template style     ( C )

Explanation:

I did it before and i was checking my notes and i wrote that down , Hope this Helps :)

pls I will give brainliest for correct answer



find the distance travelled im the first 8 hours



Answers

Answer

I think it is 112 havent done this in a while but im pretty sure thats it Good luck

Explanation:

Sometimes you can see a faint reflection in the surface of a shiny plate or cup.Why ?

Answers

Answer:

The degree of reflection whether faint or bright you see on the surface of an object is an indication that light particles had hit the surface. Since light is a wave and as part of its characteristics can get reflected. However, the amount of light reflected by a surface is dependent on the smoothness of the surface which can be shiny or dull, it can also be dependent on the nature of the surface which can be glass, water, and so on. So, from the question, you can see a faint reflection on the surface of a shiny plate or cup because of the smoothness of the surface which reflects the lights that hit it from a particular direction at the same angle.

If red and blue light rays fall with the same angle of incidence on the separating
surface between two different transparent media, then the ratio between the refraction
angle of the red light and the refraction angle of the blue light (.) is
a) greater than 1
b)equal to 1
c )indeterminable
d)less than 1​

Answers

Answer:

I'm gonna say it's D

Explanation:

but when u do the experiment on in u head you'll actually find out that it is actually , indeterminable

The diagram shows a state of matter in a closed system before and after undergoing a
change.
System before change
System after change
SEE
Which statement best explains the change in the system?
Particle motion decreases when thermal energy is added to the system
O Particle motion increases when solid particles are added to the system.
O Particle motion increases when thermal energy is added to the system.
O Particle motion decreases when gas particles are added to the system.

Answers

Answer:

the second one

Explanation:

Particle motion increases when solid particles are added to the system.

A ball is thrown directly upward with an initial speed of 72 m/s. How much time passes before the ball reaches the highest point it will ever reach

Answers

Answer:

7.35 seconds

Explanation:

The computation of the time passed prior to the ball reached to the highest point is as follows;

As we know that the acceleration because of gravity on earth is 9.8 m/s

As the ball is thrown directly in an upward direction with a speed of 72 m/s

So, the time passed prior to the ball reached to the highest point is

= 72 ÷ 9.8

= 7.35 seconds

Hence, the above represents the answer

Free electromagnetic oscillations occur in an electrical circuit. Knowing the characteristic dimensions of the elements, L=20mH și C=0,005mF, determine: a)The period of free electromagnetic oscillations: b)Frequency of free electromagnetic oscillations: c)Electromagnetic energy in the circuit when the maximum value of the current intensity is equal to 100 mA.

Answers

Answer:

a. 0.199 ms b. 5.03 kHz c. 0.1 mJ

Explanation:

a. The period of oscillation of an L-C circuit is T = 2π√(LC) where L = inductance = 20 mH = 20 × 10⁻³ H and C = capacitance = 0.005 mF = 5 × 10⁻⁶ F.

So, T = 2π√(LC)

= 2π√(20 × 10⁻³ H × 5 × 10⁻⁶ F)

= 2π√(100 × 10⁻¹¹)

= 2π√(10 × 10⁻¹⁰)

= 2π(3.16 × 10⁻⁵)

= 19.87 × 10⁻⁵

= 1.987 × 10⁻⁴ s

= 1.99 × 10⁻⁴ s

= 0.199 × 10⁻³ s

= 0.199 ms

b. frequency , f = 1/T where T = period = 0.199 × 10⁻³ s.

So, f = 1/0.199 × 10⁻³ s

= 5.03 × 10³ Hz

= 5.03 kHz

c. The electromagnetic energy E = 1/2Li² where L = inductance = 20 × 10⁻³ H and i = current = 100 mA = 0.1 A

So, E = 1/2Li²

= 1/2 × 20 × 10⁻³ H × (0.1 A)²

= 0.1 × 10⁻³ J

= 0.1 mJ

How is an emulsion different from other colloids?

Answers

The difference between colloid and emulsion is that a colloid can form when any state of matter (solid, liquid or gas) combine with a liquid whereas an emulsion has two liquid components which are immiscible with each other.

Answer:

wrthfnhdghkyulyf

Explanation:

zdfhdfhghdhddddddddddddddddddddd

Choose the element that has the larger atomic radius: francium or sodium.

Answers

Answer:

Francium.

I hope it will be useful.

Longitudinal seismic waves are known as
a primary waves.
b. secondary waves.
surface waves.
d. transverse waves.
Please select the best answer from the choices provided
О А
ОВ
ОС
D

Answers

Its “A“ primary waves!!!!!!

Answer:

A. primary waves

Explanation:

calculate the resistance of a wire 150cm long and diameter 2.0mm constructed from an alloy of resistivity 44*10-⁸Ωm​

Answers

Answer:

R = 0.21 Ω

Explanation:

the formula:

R = r x l/A

R = (44 x 10-⁸ Ωm) x 1.5 / (π x (1 x 10-³ m)²)

R = 6.6 x 10-⁷ / 3.14 x 10-⁶

R = 0.21 Ω

Plzz Help easy quesion for 20 points Why does someone get burned if they touch an old filament bulb that has been on for a few hours? Plzz Help

Answers

incandescent lights get hot very quickly and therefore can easily burn u or catch fire

Answer:

In reality, the filament gets so hot it in a real sense bubbles off molecules and electrons. Now and again this material gathers as a dull spot at the highest point of the bulb. Eventually, the filament falls apart, gets frail, and breaks, subsequently finishing the life of the light. Lights radiate light by siphoning an electric flow through a dainty tungsten fiber. The filament warms and emits light. Over the long haul, the filament oxidizes and turns out to be increasingly fragile, until it splits up and the bulb goes out. ... Tungsten picks up obstruction as it warms.

Hope this helped :)

If six moles of hydrogen chloride (HCl) react with plenty of aluminum, how many moles of aluminum chloride (AlCl3) will the reaction produce?
2Al + 6HCl → 2AlCl3 + 3H2

Answers

Answer:

Two moles of aluminum chloride [tex](AlCl_3)[/tex] are produced when six miles of hydrogen Chloride [tex](HCl)[/tex] react with plenty of aluminum

Explanation:

6 Moles of [tex]HCl[/tex] will only react with 2 moles of [tex]Al[/tex] irrespective of the number of moles of each compound present. The reaction wiil take place in this ratio only. The products produced will be 2 moles of [tex]AlCl_3[/tex] and 3 moles of [tex]H_2[/tex] this ratio will also be constant.

So, six moles of hydrogen chloride [tex](HCl)[/tex] will react with plenty of aluminum to produce many 2 moles of aluminum chloride [tex](AlCl_3)[/tex].

How will you show that metals form basic oxides?​

Answers

Answer:

when Metallic oxide reacts with water they form metallic hydroxide which are basic in nature

Standing waves are created in the four strings shown in Figure 25. All strings have the same mass per unit length and are under the same tension The lengths of the strings are given. Rank the frequencies of the oscillations, from largest to smallest

Answers

Answer:

The rank of the frequencies from largest to smallest is

The largest frequency of oscillation is given by the string in option D

The second largest frequency of oscillation is given by the string in option B

The third largest frequency of oscillation is given by the string in option A

The smallest frequency of oscillation is given by the string in option C

Explanation:

The given parameters are;

The mass per unit length of all string, m/L = Constant

The tension of all the string, T = Constant

The frequency of oscillation, f, of a string is given as follows;

[tex]f = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L}[/tex]

Where;

T = The tension in the string

m = The mass of the string

L = The length of the string

n = The number of overtones

[tex]Therefore, \ {\sqrt{\dfrac{T}{m/L} } } = Constant \ for \ all \ strings = K[/tex]

For the string in option A, the length, L = 27 cm, n = 3 we have;

[tex]f_A = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(3 + 1) \times K }{2 \times 27} = \dfrac{2 \times K}{27} \approx 0.07407 \cdot K[/tex]

For the string in option B, the length, L = 30 cm, n = 4 we have;

[tex]f_B = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(4 + 1) \times K }{2 \times 30} = \dfrac{ K}{12} \approx 0.08 \overline 3\cdot K[/tex]

For the string in option C, the length, L = 30 cm, n = 3 we have;

[tex]f_C = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(3 + 1) \times K }{2 \times 30} = \dfrac{K}{15} \approx 0.0 \overline 6 \cdot K[/tex]

For the string in option D, the length, L = 24 cm, n = 4 we have;

[tex]f_D = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(4 + 1) \times K }{2 \times 24} = \dfrac{5 \times K}{48} \approx 0.1041 \overline 6 \cdot K[/tex]

Therefore, we have the rank of the frequency of oscillations of th strings from largest to smallest given as follows;

1 ) [tex]f_D[/tex] 2) [tex]f_B[/tex] 3) [tex]f_A[/tex] 4) [tex]f_C[/tex]

                                         

The order of the frequencies is  [tex]f_D>f_B>f_A>f_C[/tex]

Standing waves:

The frequency of the standing wave in a string tied at both ends is given by:

[tex]f=\frac{nv}{2L}[/tex]

where n is the mode of frequency

v is the velocity of the wave

and L is the length of the string.

Now the velocity of a wave in a string tied at both ends is given by

[tex]v=\sqrt{\frac{T}{\mu}}[/tex]

where T is the tension and μ is the mass per unit length.

Since T and μ are the same for all the strings, velocity [tex]v[/tex] will be the same for all.

Now to find the mode of frequency we can calculate the number of nodes (including the nodes at the ends) in the given figure and subtract by 1. Nodes are the point where the amplitude of the wave is zero.

[tex]f_A=\frac{3v}{2\times27}=\frac{v}{18}\;s^{-1}\\\\f_B=\frac{4v}{2\times30}=\frac{v}{15}\;s^{-1}\\\\f_C=\frac{3v}{2\times30}=\frac{v}{20}\;s^{-1}\\\\f_D=\frac{4v}{2\times24}= \frac{v}{12}\;s^{-1}[/tex]

Hence, [tex]f_D>f_B>f_A>f_C[/tex]

Learn more about standing waves:

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