A block and tackle consisting of 5 pulleys is used to raised a load of 400N through a height of 10m.If the work done against friction is 100j.
Calculate;
1)The work done by the effort
2)The efficiency of the system
3)The effort applied​

Answer:

The net force acting on the wagon is 63 N

Explanation:

Net Force

According to the second Newton's law, the net force exerted by an external agent on an object of mass (m) is:

F = m.a

Where (a) is the acceleration of the object.

The wagon has a mass of m=42 Kg is accelerated at [tex]a=1.5\ m/s^2[/tex]. The net force is:

F = 42*1.5

F = 63 N

The net force acting on the wagon is 63 N

Answer:

A subduction zone is also generated when two oceanic plates collide — the older plate is forced under the younger one — and it leads to the formation of chains of volcanic islands known as island arcs.

Explanation:

Answer:

Index of refraction.

Answer:

index of refraction

Explanation:

I just took the k12 quiz.

Answer:

yeet yeet yeet yeet

Explanation:

Kinetic energy (K.E):-

So, the Mass of the Squirrel is 0.51 Kg (or) 510 grams.

A squirrel runs at a speed of 9.9 m/s with 25 J of kinetic energy.

What is the squirrel’s mass?

Answer: 0.51 kg

Answer:

The water hits the wall at a height of 5.38 m

Explanation:

Projectile Motion

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

The object describes a parabolic path given by the equation:

[tex]{\displaystyle y=\tan(\theta )\cdot x-{\frac {g}{2v_{0}^{2}\cos ^{2}\theta }}\cdot x^{2}}[/tex]

Where:

y   = vertical displacement

x   = horizontal displacement

θ   = Elevation angle

vo = Initial speed

The hose projects a water current upwards at an angle of θ=40° at a speed vo=20 m/s.

The height at which the water hits a wall located at x=8 m from the hose is:

[tex]{\displaystyle y=\tan40^\circ\cdot 8-{\frac {9.8}{2*20^{2}\cos ^{2}40^\circ }}\cdot 8^{2}}[/tex]

Calculating:

y = 5.38 m

The water hits the wall at a height of 5.38 m

Answer: Number 6 is Periods

Explanation:

Answer:

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 15 × 2

We have the final answer as

Hope this helps you

Answer:mechanical waves.

Explanation:

Mechanical waves require the particles of the medium to vibrate in order for energy to be transferred. For example, water waves, earthquake/seismic waves, sound waves, and the waves that travel down a rope or spring are also mechanical waves.

Explanation:

We need to draw position-time graph when the speed is increasing.

The slope of position-time graph gives the speed of an object.

Position means distance covered.

When the speed of an object is increasing with time. It means it is moving with increasing speed.

The attached figure shows the position -time graph when speed is increasing​.

Answer:

1.1 lbs

Explanation:

To convert kg to lbs you multiply kilograms by 2.2. So 0.5kg × 2.2 equals to 1.1 lbs

Answer: The work done on the flagstone is 20178 J

Explanation:

Power is the rate at which work is done . It is equal to the amount of work done divided by the time it takes to do the work.

[tex]Work=Power\times time[/tex]

Given : work = ?

Power = 342 W = 342J/s

Time = 59 s

[tex]Work=342J/s\times 59s=20178J[/tex]

Thus the work done on the flagstone is 20178 J

Answer:

heat

Explanation:

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 50 × 0.6

We have the final answer as

Hope this helps you

Answer: Dynamic - Static Flexibility

Explanation:

Answer:

The final velocity of the train and the moose after collision is approximately 19.51 m/s

Explanation:

The given mass of the moose, m₁ = 250 kg

The velocity of the moose, v₁ = 0

The mass of the oncoming train, m₂ = 10,000 kg

The velocity of the train, v₂ = 20 m/s

The velocity of the moose and the train after collision = v₃

By the principle of conservation of linear momentum, the total initial momentum before the collision = The total final momentum after collision

m₁·v₁ + m₂·v₂ = (m₁ + m₂)·v₃

Therefore, by substitution, we have;

250×0 + 10,000× 20 = (10,000 + 250) × v₃

200,000 = 10,250 × v₃

v₃ = 200,000/10,250 ≈ 19.51 m/s

The final velocity of the train and the moose after collision = v₃ ≈ 19.51 m/s

Answer:

The boat won't be able to move if the oars were out and there was no thruster. If there was a flow of the water then yes there would be a moving boat.

Answer:

Give me some time okkkk

Solution:

The magnitude of A vector is 84.9 m in the positive y-axis direction.

So the X component of A =0

     the Y component of A = 84.9 m

Now the magnitude of B vector is 195 m and it makes an angle of 41° in the direction from the positive x-axis direction.

So  the X component of B = B cos 41°

                                           = 195 x cos 41°

                                           = 195 x 0.75   = 146.25 m

   the Y component of B = B sin 41°

                                           = 195 x sin 41°

                                           = 195 x 0.65   = 126.75 m

Now it is given that vector C has a magnitude of 126.2 m and it makes a direction towards the positive x-axis.

So the X component of C =126.2 m

     the Y component of C = 0

Comparing all these, we get

1. B vector has the largest X component

2. B vector has the largest Y component

Answer:

70 m/s.

Explanation:

Given that,

Initial speed, u = 40 m/s

Acceleration = 2.5 m/s²

Time, t = 12 s

We need to find criminal's new  speed. Let it is v. Using equation of motion to find it as follows :

v = u +at

Substitute all the values

v = 40 + 2.5(12)

v = 70 m/s

So, the new speed is 70 m/s.

Answer:

33,333.33 N

Explanation:

Given that :

Initial kinetic energy = 500,000 J

Final kinetic energy = 100,000 J

Using the relation :

Force * time = change in momentum (Newton's law)

Force (F) * 0.12 = (500,000 - 100,000)

0.12F = 400,000 J

Force = (400,000 J) / 0.12s

Force = 33333.333

Force = 33,333.33 N

Answer:

a) La piedra es lanzada desde una altura de 0,785 metros.

b) La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.

Explanation:

a) Dado que la piedra es lanzada horizontalmente, tenemos que la piedra experimenta un movimiento horizontal a velocidad constante y uno vertical uniformemente acelerado debido a la gravedad. La altura de la que fue lanzada la piedra se puede determinar mediante la siguiente ecuación cinemática:

[tex]y = y_{o}+v_{o,y}\cdot t +\frac{1}{2}\cdot g\cdot t^{2}[/tex] (1)

Donde:

[tex]y[/tex] - Altura final, medida en metros.

[tex]y_{o}[/tex] - Altura inicial, medida en metros.

[tex]v_{o,y}[/tex] - Componente vertical de la velocidad inicial, medida en metros por segundo.

[tex]t[/tex] - Tiempo, medido en segundos.

[tex]g[/tex] - Aceleración gravitacional, medida en metros por segundo cuadrado.

Si sabemos que [tex]y = 0\,m[/tex], [tex]v_{o,y} = 0\,\frac{m}{s}[/tex], [tex]t = 0,4\,s[/tex] y [tex]g = -9,807\,\frac{m}{s^{2}}[/tex], entonces la altura inicial de la piedra es:

[tex]y_{o} = y-v_{o,y}\cdot t -\frac{1}{2}\cdot g\cdot t^{2}[/tex]

[tex]y_{o} = 0\,m-\left(0\,\frac{m}{s} \right)\cdot (0,4\,s)-\frac{1}{2}\cdot \left(-9,807\,\frac{m}{s^{2}} \right) \cdot (0,4\,s)^{2}[/tex]

[tex]y_{o} = 0,785\,m[/tex]

La piedra es lanzada desde una altura de 0,785 metros.

b) Ahora, obtenemos el componente horizontal de la velocidad inicial a partir de la siguiente ecuación cinemática:

[tex]v_{o,x} = \frac{x-x_{o}}{t}[/tex] (2)

Donde:

[tex]x_{o}[/tex], [tex]x[/tex] - Posiciones horizontales iniciales y finales, medidas en metros.

[tex]t[/tex] - Tiempo, medido en segundos.

Si tenemos que [tex]x_{o} = 0\,m[/tex], [tex]x = 2,5\,m[/tex] y [tex]t = 0,4\,s[/tex], entonces el componente horizontal de la velocidad inicial es:

[tex]v_{o,x} = \frac{2,5\,m-0\,m}{0,4\,s}[/tex]

[tex]v_{o,x} = 6,25\,\frac{m}{s}[/tex]

La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.

Answer:

c i kn now it is

Explanation:

Answer:

go up the street as you exit the house and make a right and keep going up for 3 blocks and you should see the school

Answer:7800

work=force x distance

Force in Newtons

Distance in Meters

Work in Joules

Answer:

B. 1200

Explanation:

60 sec in one min in 2 min there will be 120 sec. 10x120=1200

The forces constant (spring constant) of the spring will be 41.09 N/m.

The force required to extend or compress a spring by some distance scales linearly concerning that distance is known as the spring force. Its formula is;

F = kx

The given data in the problem is;

F is the spring force = 30 N

K is the spring constant= ?

x is the displacement of spring = 0.73 m

The spring constant is;

K =F/x

K=30/0.73

K=41.09 N/m

Hence the force constant (spring constant) of the spring will be 41.09 N/m.

To learn more about the spring force refer to the link;

https://brainly.com/question/4291098

#SPJ1

Answer:

Explanation:

Power = Workdone/Time

Since workdone = Force * distance = mgh

Power = mgh/t

Given

Power = 10watts

speed = 2m/s

Power = Weight * velocity

Substitute

10 = Weight * 2

Weight = 10/2

Weight = 5N

Hence the weight of the object is 5N

Answer:

9.8×104Nm−2,0.025,3.92×106Nm−2

Solution :

Here, L=10cm=10×10−2m

F=100kgf=100×9.8N

ΔL=0.25cm=0.25×10−2m,

Shearing stress =FL2=100×9.8(10×10−2) Sheraing strain =ΔLL=0.25×10−210×10−2 = 0.025 Shear Modulus of elasticity, G=Shearing stressShearing strain=9.8×1040.025

=3.92×106Nm−2

Explanation:

Answer:

It is B

Explanation:

Answer: 2nd answer

Explanation: took exam

Answer:

5 sig figures

Explanation: