At the gym, a man pulls a bar on a machine that works the muscles of the upper back. It takes him 0.5 seconds to raise 30 kilograms of weights a vertical distance of 0.5 meters. Which of these exerts the same power output? (Estimate g as 10 m/s2.)
A) lifting 25 kilograms a distance of 2.4 meters in 2.0 seconds
B) lifting 45 kilograms a distance of 2.4 meters in 3.0 seconds
C) leg pressing 45 kilograms a distance of 0.5 meters in 0.5 seconds
D) bench pressing 30 kilograms a distance of 0.5 meter in 1.5 seconds

Answer:

i) 3.514 s, ii) 5.692 m/s

Explanation:

i) We can use Newton's second law of motion to find out how long does it take for the Eagle to touch down.

as the equation says for free-falling

h = ut +0.5gt^2

Here, h = 10 m, g = acceleration due to gravity = 1.62 m/s^2( on moon surface)

initial velocity u = 0

10 = 0.5×1.62t^2

t = 3.514 seconds

Therefore, it takes t = 3.514 seconds for the Eagle to touch down.

ii) use Newton's 1st equation of motion to calculate the velocity of the lunar module when it hits the surface of the moon

v = u + gt

v = 0+ 1.62×3.514

v= 5.692 m/s

Answer:

5.49×10¯⁴ m²

Explanation:

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

Distance (d) = 2.22×10¯⁴ m

Charge (Q) = 5.24×10¯⁹ C

Potential difference (V) = 240 V

Permittivity of free space (ε₀) = 8.85×10¯¹² F/m

Area (A) =?

Thus, the area of the plate can be obtained as follow:

Q = ε₀AV /d

5.24×10¯⁹ = 8.85×10¯¹² × A × 240 / 2.22×10¯⁴

5.24×10¯⁹ = 2.12×10¯⁹ × A / 2.22×10¯⁴

Cross multiply

2.12×10¯⁹ × A = 5.24×10¯⁹ × 2.22×10¯⁴

Divide both side by 2.12×10¯⁹

A = (5.24×10¯⁹ × 2.22×10¯⁴) / 2.12×10¯⁹

A = 5.49×10¯⁴ m²

Thus, the area of the plates is 5.49×10¯⁴ m².

Answer:

it's actually 5.48

Explanation:

Answer:ssssw

Explanation:

The

friction

force acting on the bag is 4.36µ.

What is friction force?

The

force

preventing sliding against one another of solid surfaces, fluid layers, and material components is known as friction.

Given,

Force =F=22.5 n

Mass=m= 10 kg

mu=µ=?                            

(1N= 1kg m/s²)

Gravitational force =g= 9.81 m/s²

As we know,

F= µ*m*g

µ= m*g/F

  = 10*9.81/ 22.5

µ= 4.36 .

Therefore, the friction force acting on the bag is 4.36.

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The bag is subject to a friction force of 4.36. A man applies a 22.5 N force at a 32° angle while dragging a 10.0 kilogram bag of mulch at a steady speed.

Friction refers to the force that resists hard surface, fluid layering, and materials ingredients from gliding past one another. Friction lessens a machine's mechanical advantage, or, to put it another way, friction decreases the output to input ratio.

Force = 22.5 N

Mass  = 10 kg

⇒ mu     = µ (1N= 1kg m/s²)

⇒ g        = 9.81 m/s²

As we know, F= µ × m × g

⇒ µ = (m × g)/F

⇒ (10 × 9.81)/ 22.5

⇒ µ = 4.36 .

Therefore, the friction force acting on the bag is 4.36.

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Answer:

when a plastic rod is rubbed with a duster, electrons are transferred from one material to the other. The material that gains electrons becomes negatively charged. The material that loses electrons becomes positively charged.

Answer:

F = I L X B = I L B sin theta

F = 5.2 * 1.2 * .87 sin 45 = 3.84 N   in the direction of (L X B)

Note: L and B must be at right angles for   F = I L B

The statement that best illustrates the application of the principles of behaviorism is -

"A parent is using "time out" as a consequence for bad behavior".

The theory that human and animal behaviour can be explained in terms of conditioning, without appeal to thoughts or feelings, and that some mental conditions are best treated by altering behaviour patterns.

Given are the applications of the principles of behaviorism.

The statement that best illustrates the application of the principles of behaviorism is -

"A parent is using "time out" as a consequence for bad behavior".

Therefore, the statement that best illustrates the application of the principles of behaviorism is -

"A parent is using "time out" as a consequence for bad behavior".

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Dispersion in wave theory is the phenomena where a wave's phase velocity changes depending on its frequency. The article can explore different definitions. This is the typical definition for dispersion (waves).

When sinusoidal waves with various wavelengths propagate at various speeds, a wave packet with mixed wavelengths has a tendency to disperse in space. Dispersion in wave theory is the phenomena where a wave's phase velocity changes depending on its frequency.

The article can explore different definitions. This is the typical definition for dispersion (waves).

A full set of approximate solutions to the one-electron Schrödinger equation are found in a three-dimensional atomic lattice that is randomly filled with atoms of different kinds and has, on average, the periodicity of a Bloch wave.

Equations are provided that enable the calculation of each wave's energy, average behavior, absent a phase factor, close to any sort of atom, and damping parameter.

A comparison is made between a periodic lattice with a complicated, energy-dependent potential and a random A-B lattice. The dispersion relation of the system is obtained by solving the eigenvalue problem in this potential using any of the established techniques for periodic potentials.

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Answer:

it is a

Explanation:

dermis,epidermis,hypodermis

Answer:

B. hypodermis, dermis, epidermis

this is because it is inside to outside

hypodermis is deeper tissue, dermis is beneath epidermis, and epidermis is the outermost layer

Answer:

TheThe charge applied on each ball is 2 X 10∧-9 C.

An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

The pitch balls repent each other with force

F= 4 X 10^-5N

Let the charge on each ball be Q.

4 X 10^-5= k(Q)(Q) /  (3 X 10^-2)^2

Q^2= 4*10^-5X 9*10^-4 / 9*10^9

Q^2 = 4 X 10^-18

Q= √4 X 10^-18

Q = 2 X 10^-9 C.

Therefore, the charge on each ball is 2X10^-9C.

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Each ball has been charged with 2 X 10⁹ C. When a product is held in electrostatic energy, its properties referred to as electric discharge allows it to feel force.

A force can compel a heavily armored item to increase or modify overall velocity. A push or a pull is a simple and effective method to explain force. That's because a force has both magnitude and direction it is a dimensionless number.

The pitch balls repent each other with force

F = 4 X 10⁻⁵N

4 X 10⁻⁵ = k(Q)(Q) /  (3 X 10⁻²)²

⇒ Q²      = 4 × 10⁻⁵ X   9 × 10⁻⁴ / 9 × 10⁹

⇒ Q²      = 4 X 10⁻¹⁸

⇒ Q        = √4 X 10⁻¹⁸

⇒ Q        = 2 X 10⁻⁹ C.

Therefore, the charge on each ball is 2 X 10⁻⁹C.

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Answer:

During its trajectory the cannon ball will have kinetic and potential energy

E = KE + PE

The maximum value of PE will occur when the cannon ball was fired straight up - for other trajectories PE will be less because of translational KE

So E = M g h = 1/2 M v^2    for maximum height

h = v^2 / (2 * g) = 400 / (2 * 9.8) = 20.4 m

Answer:

r = 1.63×10^5 mi

Explanation:

Let r = distance of object from earth

Rs = distance between earth and sun

Ms = mass of the sun

= 3.24×10^5 Me (Me = mass of earth)

At a distance R from earth, the force Fs exerted by the sun on the object is equal to the force Fe exerted by the earth on the object. Using Newton's universal law of gravitation,

Fs = Fe

GmMs/(Rs - r)^2 = GmMe/r^2

This simplifies to

Ms/(Rs - r)^2 = Me/r^2

(3.24×10^5 Me)/(Rs - r)^2 = Me/r^2

Taking the reciprocal and then its square root, this simplifies further to

Rs - r = (569.2)r ----> Rs = 570.2r

or

r = Rs/570.2 = (9.3×10^7 mi)/570.2

= 1.63×10^5 mi

Answer:

Conduction

Explanation:

Conduction is a process whereby heat is transfer through solid state.

Answer:

1: True

2: False

3: True

4: False

5: True

heating or pressing but heating is most appropriate, so the answer is heating.

Answer:

Explanation:

4

The magnitude and direction of the resultant force on q3 is;

F_net = 32.2 N in the direction of q3

We are given;

q1 = +2 μC = 2 × 10^(-6) C

q2 = -3 μC = -3 × 10^(-6) C

q3 = +4 μC = 4 × 10^(-6) C

Now,from the given image, charges q1 and q2 attract charge q3.

Formula for force on a charge is;

F = (kq1 * q2)/r²

Where;

k = 9 × 10^(9) N.m²/C²

Thus;

F_y = kq1*q3/(r1)²

We have;

q1 = 2 × 10^(-6) C

r1 = 5 cm = 0.05 m

q3 = 4 × 10^(-6) C

Thus;

F_y = (9 × 10^(9) × 2 × 10^(-6) × 4 × 10^(-6))/(0.05^(2))

F_y = 28.8 N

Similarly;

F_x = kq2*q3/(r2)²

r2 = 5/(tan 30)

r2 = 8.66 cm = 0.0866 m

F_x = (9 × 10^(9) × -3 × 10^(-6) × 4 × 10^(-6))/(0.0866^(2))

F_x = -14.4 N

Thus, magnitude of resultant force is;

F_net = √((28.8)² + (-14.4)²)

F_net = √1,036.8

F_net = 32.2 N

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Answer:

a) I = 464 kg m²,  b)  K = 631 .6 J, c)  v = 8.25 m / s

Explanation:

a) the moment of inertia of point particles is

          I = ∑ m_i r_i²

in this case

          I = 8 5² + 3 (-2) ² + 7 (-6) ²

          I = 464 kg m²

b) The kinetic energy is

          K = ½ I w²

          K = ½ 464 1.65²

          K = 631 .6 J

c) linear and angular velocity are related

          v = w r

          v = 1.65 5

          v = 8.25 m / s

Ans-The density of states function describes the number of states that are available in a system and is essential for determining the carrier concentrations and energy distributions of carriers within a semiconductor.

Answer:

density equal to mass upon volume

simple definition

Answer:

If freshwater consumption was greater than freshwater renewal.

Explanation:

Similar to another Brainly answer :O

The mass of the moving item is referred to as momentum. There is magnitude and direction to momentum. The object has a momentum of 7.2 g-m/s.

Momentum is a vector quantity that is created during motion by an object's mass and velocity. Thus, momentum will be expressed in units of kg-m/s.

Considering that item has motion. Likewise, 1 kilogram equals 1000 g. Therefore, the momentum will transform into units of grams when we enter the value of 1 kg.

Momentum = 7.2 x 10⁻³ 1000g-m/s

Momentum = 7.2 g-m/s.

Thus, we may deduce that the object's momentum, measured in grams per second, is 7.2 g-m/s.

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Answer:

Explanation:

All 3 are critical for habitability

Answer:

put the full thin g

Explanation:

Answer:

Opening stock details of raw material, semi-finished goods and finished goods.

Closing stock details of raw material, semi-finished goods, and finished goods.

Total purchases of goods fewer Purchase Returns.

Total sales of goods fewer Sales Returns.

Since the efficiency of the pulley is 75% and the single movable pulley is used to lift 4000N load to the height of 50 cm, the effort distance, effort applied, output work and mechanical advantage are as follows:

Effort Distance = 50 cm

Effort Applied = 5333.33 N

Output work = 2000 J

Mechanical Advantage = 0.75.

When a single movable pulley is used to lift a load, the effort applied is equal to the weight of the load being lifted, and the effort distance is the distance the effort is applied.

Load (output force) = 4000 N

Height lifted = 50 cm = 0.5 m

Efficiency of the pulley = 75%

Given the above information, we can calculate the effort applied, output work, and mechanical advantage as follows:

Effort applied (input force) = Load ÷ Efficiency

= 4000 N ÷ 0.75

= 5333.33 N

Output work = Load x Height lifted

= 4000 N x 0.5 m

= 2000 J

Effort distance = Height lifted

= 50 cm

Mechanical advantage = Load ÷ Effort applied

= 4000 N ÷ 5333.33 N

= 0.75

It is worth noting that pulley efficiency is the ratio of output work to the input work. And since the efficiency of the pulley is 75% so the input work is greater than the output work, which means some of the energy is wasted in overcoming friction and other losses.

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Answer:

The answer is "Option A".

Explanation:

Please find the complete question in the attached file.

The strength is equal due to the third law of Newton, and [tex]Fsp = Fps = \frac{Gmsmp}{r^2}\[/tex], therefore the force increases with decreased separation r.

Solution :

Mass of the sphere, m = 10 kg

Diameter, D = 300 mm = 0.3 m

Volume of the sphere is [tex]$V=\frac{4}{3} \pi \left(\frac{0.3}{2}\right)^3$[/tex]

                                       [tex]$V=0.01414 \ m^3$[/tex]

So the volume of displaced water, [tex]$V_w=V = 0.01414 \ m^3$[/tex]

Additional mass, [tex]$m_w = \frac{1}{2} \ \rho_w\times v_w$[/tex]

                                  [tex]$=\frac{1}{2} \times 1000 \times 0.01414$[/tex]

                                  [tex]$=7.0686 \ kg$[/tex]

So the total mass, [tex]$M = m_w+m$[/tex]

                                   = 7.0686 + 10

                                   = 17.0686

Force required, F = Ma

[tex]$F=17.0686 \times 10$[/tex]

   = 170.686 N

The force that is necessary to accelerate when the acceleration rate of 10 m/s2 in the horizontal direction should be 170.686 N.

Since Mass of the sphere, m = 10 kg

Diameter, D = 300 mm = 0.3 m

Now volume of the sphere is

V = 4/π (diameter/3)^3

V = 4/3π(0.3/2)^3

= 0.01414m^3

Now the additional mass is

= 1/2* 1000 * volume

= 1/2 *1000*0.01414

= 7.0686

So, the total mass is

= Additional mass + force

= 7.0686 + 10

= 17.0686

Now the force is

= 17.0686 * 10

= 170.686

Hence, The force that is necessary to accelerate when the acceleration rate of 10 m/s2 in the horizontal direction should be 170.686 N.

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Answer:

16.3 wave speed.

Explanation: