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3 years ago

What is the netforce?

Physics

2 answers:

Kipish [7]3 years ago

3 0

Answer:

6 Newtons to the left.

Explanation:

We can convert this into a generic algebra equation by giving directions positive and negative values.

The 6 will be positive, and the 10 and 2 will be negative.

Add 10 and 2 to have 12.

6-12 = -6.

Therefore you have 6 newtons to the left (negative).

777dan777 [17]3 years ago

3 0

Take the force towards the right as positive (+6N) and towards the left as negative (-12N)
Add both as net force is the sum of all forces acting on an object
6N+(-12N)= -6N

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We can estimate the age of an open cluster of stars by examining _________.

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4 years ago

What creates the energy of the Sun?

Svet_ta [14]

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3 years ago

Calculate the density of the football. Use the formula D = m/V where D is the density, m is the mass, and V is the volume. Recor

s344n2d4d5 [400]

Answer:

Detailed explanation:

Density of water=1000kg/m³

Hence mass of water displaced is:

m=d×v

=1000kg/m³×(4.3×10^-3)m³ (volume of water displaced converted to L)

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=mass÷volume

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4 0

3 years ago

Read 2 more answers

2. An 873 kg dragster accelerates at a rate of 44.6 m/s during a race.

Ira Lisetskai [31]

Answer:

Explanation:

According to newton's second law:

Force = mass * acceleration

Given

Mass = 873kg

acceleration = 44.66m/s²

Magnitude of the force is expressed as;

F = ma

F = 873 * 44.6

F = 38,673.9N

<em>Hence the magnitude of the net force exerted on the dragster during this time is 38,673.9N</em>

8 0

3 years ago

Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol

klasskru [66]

The current flowing in silicon bar is 2.02 $\times$ 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

J = (1.6 $\times$ 10^-19) / 0.001 (104 $\times$ 1200 + 1016 $\times$ 500)

J = 1012480 $\times$ 10^-16 A / m^2.

J = 1.01 $\times$ 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 $\times$ 10^-9 $\times$ 0.002 = 2.02 $\times$ 10^-12

So, the current flowing in silicon bar is 2.02 $\times$ 10^-12 A.

6 0

4 years ago