How many cm3 are there in 1 dm3?

Answer:

An amorphous solid does not have a definite melting point; instead, it melts gradually over a range of temperatures, because the bonds do not break all at once. This means an amorphous solid will melt into a soft, malleable state (think candle wax or molten glass) before turning completely into a liquid.

Explanation:

Hope this helps

May I get braineist pls?

        Control rods

____________________

Control rods are inserted into the core of a nuclear reactor and adjusted in order to control the rate of the nuclear chain reaction

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Answer:

Friction, normal force, and weight

Explanation:

If the book slows down, it means that there must be friction acting in the opposite direction of the direction the book is moving in.

Weight is caused by the gravitational pull of the Earth on the book, and normal force is the table pushing the book up because the book is pushing down on the table (3rd law.)

Note that weight and normal force is not the 3rd law action-reaction pair. The pair is the force of the book on the table and the force of the table on the book.

Answer:

Let L be the length of the wire.

N λ = M L      where N is the number of wavelengths λ. Also, M and are integers and L is the length of the wire

There must be nodes at the ends of the wire and anti-nodes at the points of zero displacement - λ/2, λ, 3λ/ 2, etc.

The simplest would be 2 λ = L   or L = λ / 2

In such a situation one would have

N-A-N    or node-antinode-node

There is a 180 degree phase change upon reflection of the wave creating the standing wave

The distance between nodes or anti-nodes must be λ/2 and there must  be nodes at the ends of the wire

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]

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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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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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:

1kW = 1000W

600W = 0.6kW

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

                               = $21.60

Answer:

5m/s

Explanation:

equation for momentum is mass times velocity

25 x 5 = 125

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:

69 meters

Explanation:

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

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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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

The motion of the package can be described as the motion of a projectile,

given that it has an horizontal velocity and it is acted on by gravity.

Reasons:

Velocity of the aircraft = 70 m/s

Direction of flight of the aircraft = Eastward

Height from which the aircraft drops the package, h = 800 m

a) The initial velocity of the package, [tex]\vec{v}[/tex] = 70·i

b) The time it will take the package to reach the ground, t, is given by the formula;

[tex]\displaystyle h = \mathbf{\frac{1}{2} \cdot g \cdot t^2}[/tex]

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore;

[tex]\displaystyle t = \mathbf{\sqrt{ \frac{2 \cdot h}{g} }}[/tex]

Which gives;

[tex]\displaystyle t = \sqrt{ \frac{2 \times 800}{9.81} } \approx \mathbf{12.77}[/tex]

The time it will take the package to reach the ground, t ≈ 12.77 seconds

c) The vertical velocity just before the package reaches the ground, [tex]v_y[/tex], is given as follows;

[tex]v_y^2[/tex] = 2·g·h

Therefore;

[tex]v_y[/tex] = √(2·g·h)

Which gives;

[tex]v_y[/tex] = √(2 × 9.81 × 800) ≈ 125.28

[tex]v_y[/tex] ≈ 125.28 m/s

Which gives; [tex]\vec{v}[/tex] = 70·i - 125.28·j

Therefore, |v| = √(70² + (-125.28)²) ≈ 143.51

The speed of the package as it lands, |v| ≈ 143.51 m/s

d) The motion of the package that includes both horizontal and vertical motion is a projectile motion.

Therefore;

The path of the package is the path of a projectile, which is a parabolic shape.

e) As seen by someone on the aeroplane, the horizontal velocity will be

zero, therefore, the package will appear as accelerating directly vertical

downwards.

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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

Answer:

384000

Explanation:

it would be 384000 in km

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:

3000N

Explanation:

divided to get answer

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

Answer:

You need a graphing calculator that implemented vectors. I don't know any that have it. Sorry.

Answer:

Provided an object traveled 500 meters in 3 minutes , to calculate the average velocity you should take the following steps:

Change minutes into seconds (so that the final result would be in meters per second). 3 minutes = 3 * 60 = 180 seconds ,

Divide the distance by time: velocity = 500 / 180 = 2.77 m/s .

Explanation:

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:

Answer:

1. 60 kg m/s

2. 2.4 kg

3. none they both have same momentum

Answer:

This question assumes that the car accelerates at the same rate as when it went from 0 to 60km/h

24.29m/s or 87.4km/h

Explanation:

Let's find the acceleration of the car:

let vi=0, vf=60km/h (16.67m/s), Δt = 8.0s

a = (vf-vi)/Δt

a = (16.67m/s-0)/8.0

a = 2.08m/s^2

Now we can use this acceleration to find vf in the second part:

50km/h is 13.89m/s

a = (vf-vi)Δt

vf = aΔt + vi

vf = 2.08m/s^2*5.0+13.89m/s

vf = 24.29m/s (87.4km/h)

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.

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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