I couldn't find one .

The acceleration of the first block (4 kg) is -9.8 m/s².

The given parameters:

The net force on the system of the two blocks is calculated as follows;

[tex]m_2 g - T = m_1 a[/tex]

where;

[tex]m_2 g - m_1 g = m_1 a\\\\a = \frac{m_2 g - m_1g}{m_1} \\\\a = \frac{g(m_2 - m_1)}{m_1} \\\\a = \frac{9.8(2-4)}{2} \\\\a = -9.8 \ m/s^2[/tex]

Thus, the acceleration of the first block (4 kg) is -9.8 m/s².

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Answer:

Below in bold.

Explanation:

Newton's Second Law:

Force = mass * acceleration

So

Force = 24.5 * 2.3

= 56.35 Newtons.

Answer:

I believe it might be point A since the question ask what will result in the ln a largest increase in potential energy

Explanation:

point A since the question ask what will result in the ln a largest increase in potential energy

The energy can be defined as  ability to work or produce action and / or movement and manifests itself in many different ways, such as body movement, heat, electricity, etc.

The different types of energy include Kinetic energy which is associated with the movement of bodies. Potential energy which is  stored by virtue of a body's position relative to its surface is also called gravitational potential energy.

Thermal Energy or Heat can be defined as the energy  associated with the kinetic energy of the molecules that make up an element, it can be manifested if there is a temperature difference between two bodies.

Chemical energy released or formed from chemical reactions, Solar energy from sunlight. This form of energy is used to generate electricity through photovoltaic plates, for example.

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Answer:

397 j

Explanation:

Because 5.0kg yuh

The kinetic energy of the block with a mass of 5.0 kg is pushed on a frictionless surface by applying a horizontal force of 80.0 N. The block starts from rest, and its final velocity is 12.6 m/s is 396.9 J.

A moving item or particle has a certain kind of power called kinetic energy. When an item exerts a net force to perform work, which involves the transfer of energy, the object accelerates and as a result, gains kinetic energy.

The amount of kinetic energy that a moving object or particle possesses is determined by both its mass and rate of motion. Any combination of vibration, axis rotation, translation (or movement along a path from one location to another), and translation are all possible motion types.

Given:

The mass of the block, m = 5 kg,

The horizontal force, F = 80 N,

The velocity of the block, v = 12.6 m / s,

Calculate the kinetic energy by the formula given below,

[tex]KE = 1/2mv^2[/tex]

KE = 1 / 2 × 5 × 12.6²

KE =  1 / 2 × 5 × 158.6

KE = 396.9 J

Therefore, the kinetic energy of the block with a mass of 5.0 kg is pushed on a frictionless surface by applying a horizontal force of 80.0 N. The block starts from rest, and its final velocity is 12.6 m/s is 396.9 J.

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The complete question is A block with a mass of 5.0 kg is pushed on a frictionless surface by applying a horizontal force of 80.0 N. The block starts from rest, and its final velocity is 12.6 m/s. ​Find its kinetic energy?

Both masses will have the same acceleration. The cart accelerates to the right with a magnitude of 4.9 m/[tex]s^{2}[/tex]. The correct answer is 4.90 m/[tex]s^{2}[/tex]

Given that a massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface.

Let M = 1kg and m = 3 kg

Since the horizontal surface is frictionless, the tension in the string will be the same. when the mass is hanged over a frictionless pulley, the tension will also be the same.

When the mass is released, the cart accelerates to the right can be calculated  from Newton' second law of motion. That is,

M( g + a) = m(g - a)

1(9.8 + a) = 3( 9.8 - a)

9.8 +a = 29.4 - 3a

collect the like terms

4a = 19.6

a = 19.6/4

a = 4.9 m/[tex]s^{2}[/tex]

Therefore, the cart accelerates to the right with a magnitude of 4.9 m/[tex]s^{2}[/tex]. The correct answer is 4.90 m/[tex]s^{2}[/tex]

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Answer:

1. 60 kg m/s

2. 2.4 kg

3. none they both have same momentum

Both carts experience the same force but Cart A has a greater speed after the recoil.

The given parameters;

Apply the principle of conservation of linear momentum to determine the velocity of the carts after collision;

[tex]m_Av_0_A\ + m_Bv_0_B = m_Av_f_A \ + m_Bv_f_B\\\\the \ initial \ velocity \ of \ both \ carts = 0\\\\0.4(0) + 0.8(0) = 0.4v_f_A + 0.8v_f_B\\\\0 = 0.4v_f_A + 0.8v_f_B\\\\0.4v_f_A = -0.8v_f_B\\\\v_f_A= \frac{-0.8 v_f_B}{0.4} \\\\v_f_A = - 2 \ v_f_B \ \ m/s[/tex]

According to Newton's third law of motion, action and reaction are equal and opposite. The force exerted on cart A is equal to the force exerted on cart B but in opposite direction.

[tex]F_A = -F_B[/tex]

Thus, the correct statement that compares the motion and forces acting on the two carts is "Both carts experience the same force but Cart A has a greater speed after the recoil."

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Answer:

a = v^2 / R      centripetal acceleration

R = v^2 / a     a at the surface of earth is 9.8 m/s^2

R = (90 m/s)^2 / 9.8 m/s^2 = 827 m

2.1 kg of sodium bicarbonate is equal to the 2.1 x 10¹² ng of sample. Option D is correct.

Mass is the quantity of the substance in the body or object. The SI unit of mass is Kilogram.

There are other units of measure,

First convert kg to gram,

Since, 1 Kg = 1000 g

2.1 kg = grams of sample

So,

Do the cross multiplication,

[tex]\rm mass\ of\ sample = \dfrac {2.1\ kg \times 1000\ g }{ 1 kg}\\\\\rm mass\ of\ sample =2100 g[/tex]

Now, convert 2100 g to nano-grams

Since, 1 g = 1 x 10⁹ ng

2100 g = ng of sample

So,

Do the cross multiplication,

[tex]\rm mass\ of\ sample = \dfrac {2100 g \times 1 \times 10^9 ng }{1\ g}\\\\\rm mass\ of\ sample = 2.1 \times 10^1^2 ng[/tex]

Therefore, 2.1 kg of  sodium bicarbonate is equal to the 2.1 x 10¹² ng of sample.

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Answer:

I hope this helps:)

Explanation:

good luck:)

For the front glass of the car to get wet, [tex]V_c \geq 10 \ m/s[/tex].

The given parameters:

The relative velocity of the car with respect to the falling rain is calculated as;

[tex]V_{C/R} = V_C- V_R[/tex]

Thus, for the front glass of the car to get wet, [tex]V_c \geq 10 \ m/s[/tex].

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Answer:

3000N

Explanation:

divided to get answer

the force needed to accelerate the 1000kg car by 3m/s2 is 3000N

An example of the use of expansion of materials when they are heated is in

Thermometers. Thermometer is an instrument which is used to measure the

temperature of a place.

It contains mercury which expands as temperature increases. This is used

to accurately determine the temperature of an object.

The disadvantage of expansion of materials when they are heated can be

seen in roads as there is expansion of the surface as a result of heat

exposure leads to roads being rough thereby incurring more costs for

repairs.

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Time taken by the skydiver to land on the ground is 4.57 seconds

The applied force, F = 700 N

The mass of the skydiver, m = 80 kg

The velocity, v = 40 m/s

To calculate the time taken by the skydiver to fall to the ground will be calculated using the formula

[tex]F=\frac{mv}{t}[/tex]

Substitute F = 700, m = 80, and v = 40 to solve for t

[tex]700=\frac{80(40)}{t}\\\\700t=3200\\\\t=\frac{3200}{700}\\\\t=4.57 s[/tex]

Time taken by the skydiver to land on the ground is 4.57 seconds

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Answer:

384000

Explanation:

it would be 384000 in km

Answer:

1. Spray Drying

Explanation:

it is the most modern

Answer:8 AWG

Explanation: 6 current carrying conductors with 2 - 3 phase branch circuits. Divide 27000 by (480v × 1.732 or square root of 3) and you get 32.46 amps. Now using the code book, divide 32.46 by the adjustment factor of 6 CCC's which is .8 and you get 40.61 amps. Now go to Table 310.16 in the 2020 NEC book and you'll see that in the 90° column that 40.61 amps is above #10 AWG so you size up to #8 AWG.

By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s

Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:

mass m =  0.170 kg

initial speed u = 6 m/s.

Distance covered s = 61 m

To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V

To do this, let us first calculate the kinetic energy at which the ball move.

K.E = 1/2m[tex]U^{2}[/tex]

K.E = 1/2 x 0.17 x [tex]6^{2}[/tex]

K.E = 3.06 J

The work done on the ball is equal to the kinetic energy. That is,

W = K.E

But work done = Force x distance

F x S = K.E

F x 61 = 3.06

F = 3.06/61

F = 0.05 N

From here, we can calculate the acceleration of the ball from Newton second law

F = ma

0.05 = 0.17a

a = 0.05/0.17

a = 0.3 m/[tex]s^{2}[/tex]

To calculate the final velocity, let us use third equation of motion.

[tex]V^{2}[/tex] = [tex]U^{2}[/tex] + 2as

[tex]V^{2}[/tex]  = [tex]6^{2}[/tex] + 2 x 0.3 x 61

[tex]V^{2}[/tex] = 36 + 36

[tex]V^{2}[/tex] = 72

V = [tex]\sqrt{72}[/tex]

V = 8.485 m/s

Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.

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Hi there!

Recall the conservation of momentum for an inelastic collision:

[tex]\large\boxed{m_1v_1 + m_2v_2 = (m_1 + m_2)v_f}}[/tex]

Remember, velocity is a VECTOR and direction must be accounted for. Let the 1000 kg car have a positive velocity and the 1500 kg car have a negative velocity (opposite direction).

Plug in the given values:

[tex](1000)(20) + (1500)(-10) = (1000 + 1500)v_f}}[/tex]

Solve:

[tex]20000 -15000 = 2500v_f}}\\\\5000 = 2500v_f\\\\v_f =\boxed{ 2 m/s}[/tex]

Answer:

0.784m

Explanation:

This is a projectile motion problem so we analyze the horizontal and vertical motions separately. Since the question is asking for the height of the desk, it shows we need to analyze the vertical motion.

Answer:

5m/s

Explanation:

equation for momentum is mass times velocity

25 x 5 = 125

Hi there!

We can use the same equation for Gravitational Force:

[tex]\large\boxed{F_g = G\frac{m_1m_2}{r^2}}[/tex]

Fg = force due to gravity (N)

G = gravitational constant

m1,m2 = masses of objects (kg)

r = distance between objects (m)

Plug in the values provided:

[tex]F_g = (6.67*10^{-11})\frac{(2500)(3000)}{8^2} = \large\boxed{7.814 * 10^{-6}N }[/tex]

[tex]\huge\bf\underline{\underline{\pink{A}\orange{N}\blue{S}\green{W}\red{E}\purple{R:-}}}[/tex]

Here we've been given,

We have to find the gravitational attraction force (Fg) = ?

The standard formula to solve is given by,

[tex]:\implies\tt{F_g = g \frac{m_1m_2}{ {r}^{2} } } [/tex]

[tex]:\implies\tt{F_g = 6.67 \times {10}^{ - 11} \times \frac{(2500 )(3000)}{ {8}^{2} } }[/tex]

[tex]:\implies\tt{F_g = 6.67 \times {10}^{ - 11} \times \frac{7500000}{64} }[/tex]

[tex]:\implies\tt{F_g = 7.814 \times {10}^{ - 6} }[/tex]

Answer:

1kW = 1000W

600W = 0.6kW

Cost of electric bill = 0.6kWh × 24 × 30 × $0.05

                               = $21.60

Answer:

Explanation:

Height(h) = 4m; depth of the half filled tank 4m/2 =2m

Acceleration due to Gravity (g)=9.8m/s^2

Density (d)=1000kg/m^3

Pressure = hdg

Pressure = 2m × 9.8m/s^2 × 1,000kg/m^3

Pressure = 19,600Pa

Answer:

Below!

Explanation:

Referring to the picture, we can conclude that the picture J will have the most effect on ocean tides.

Whichever model is similar to model J in the picture will be your answer.

Hoped this helped.

Hi there!

II. Linear momentum of the system is zero.

This is an example of a RECOIL collision. With the Law of Conservation of Momentum, momentum remains constant before and after the collision.

Thus, the total momentum would also be equivalent to zero after the collision.

Answer:

because when an object approaches the speed of light, it's mass starts to increase exponentially, and would be infinite at the speed of light. It would therefore require MORE than an infinite amount of energy to accelerate even a single electron to the speed of light

Answer:

a 5 kilogram mass, at a height of 3 meters, while acted on by Earth's gravity would have 147.15 Joules of potential energy, PE = 3kg * 9.81 m/s2 * 5m = 147.15 J.

Explanation: