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posledela
3 years ago
6

Babies typically say their first words:

Physics
1 answer:
deff fn [24]3 years ago
3 0

Answer:

2nd one if you ask me

almost every baby stats to talk after their a year old

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A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for
nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block (W), measured in newtons, is:

W = m\cdot g (1)

Where:

m - Mass of the block of ice, measured in kilograms.

g  - Gravitational acceleration, measured in meters per square second.

If we know that m = 14\,kg and g = 9.807\,\frac{m}{s^{2}}, the magnitudes of the weight and normal force of the block of ice are, respectively:

N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

N = W = 137.298\,N

And the pull force is:

F_{pull} = 98\,N

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f} (2)

Where:

v_{o}, v_{f} - Initial and final speeds of the block, measured in meters per second.

\Sigma F - Horizontal net force, measured in newtons.

\Delta t - Impact time, measured in seconds.

Now we clear the final speed in (2):

v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}

If we know that v_{o} = 0\,\frac{m}{s}, m = 14\,kg, \Sigma F = 98\,N and \Delta t = 1.40\,s, then final speed of the ice block is:

v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}

v_{f} = 9.8\,\frac{m}{s}

The final speed of the block of ice is 9.8 meters per second.

6 0
2 years ago
What process is used to release energy in nuclear power plants
Anika [276]
.........Nucler fission.... .
5 0
3 years ago
Read 2 more answers
The mass of a coin is measured to be 12.5±0.1 g. The diameter is 2.8±0.1 cm and the thickness 2.1 ±0.1 mm. Calculate the average
Evgesh-ka [11]

The average density of the material from which the coin is made is 9.67 g/cm³.

<h3>Volume of the coin</h3>

The volume of the coin at the given diameter is calculated as follows;

V = Ah

where;

  • A is area of the coin
  • h is the thickness of the coin

V = πd²/4 x h

V = π(2.8)²/4  x (0.21 cm)

V = 1.293 cm³

<h3>average density of the coin</h3>

The average density of the material from which the coin is made is calculated as follows;

density = mass/volume

density = 12.5 g / (1.293 cm³)

density = 9.67 g/cm³

Thus, the average density of the material from which the coin is made is 9.67 g/cm³.

Learn more about average density here: brainly.com/question/1354972

#SPJ1

7 0
1 year ago
A ______ is something that can be placed between the sun and the subject to diffuse the light. *
Gekata [30.6K]

Answer:

scrim

Explanation:

A scrim is something that can be placed between the sun and the subject to diffuse the light.

An instance of a diffuser is a softbox that is put on its front side around a strobe containing diffusion content. The sun is a form of hard light that is often diffused through a scrim. The light rays are dispersed by putting a scrim between the sun and the object, and the harsh sun's rays is gentler.

6 0
3 years ago
Read 2 more answers
Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This
vova2212 [387]

Answer:

1.554\times 10^{32}\ \text{kg}

Explanation:

M = Mass of each star

T = Time period = 15.5 days

v = Orbital velocity = 230 km/s

G = Gravitational constant = 6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2

Radius of orbit is given by

R=\dfrac{vT}{2\pi}

We have the relation

\dfrac{Mv^2}{R}=\dfrac{GM^2}{(2R)^2}\\\Rightarrow M=\dfrac{4Rv^2}{G}\\\Rightarrow M=\dfrac{4\dfrac{vT}{2\pi}v^2}{G}\\\Rightarrow M=\dfrac{2v^3T}{\pi G}\\\Rightarrow M=\dfrac{2\times 230000^3\times 15.5\times 24\times 60\times 60}{\pi\times 6.674\times 10^{-11}}\\\Rightarrow M=1.554\times 10^{32}\ \text{kg}

The mass of each star is 1.554\times 10^{32}\ \text{kg}

6 0
3 years ago
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