All categories

3 years ago

Come si esprime il numero 0,00123 in notazione scientifica?

Physics

1 answer:

Inessa [10]3 years ago

3 0

Answer:

0,00123 = 1,2*10^{-3}

Explanation:

To write down correctly the number 0,00123 in scientific notation, you take into account that the scientific notation demands that there in only one number after the comma ( , ). Furthermore, it is necessary that you move the comma to the right of the first number different of zero, in this case the number 1. To do this you move the comma three positions.

Then, you have to multiply the expresion 1.23 by 10 with an exponential -3 (because of the movement of the comma in three positions). That is:

0,00123 = 1,23*10^{-3}

But it is mandatory that nly one number can stay after the comma, so, you approximate the number three. In this case, the number is lower than 5, hence, you approximate 3 to 0.

Finally, you have:

0,00123 = 1,2*10^{-3}

You might be interested in

At constant pressure, the volume of a fixed mass of gas and its kelvin temperature are said to be

juin [17]

They are said to be directly related.

a) directly related.

This is Charles' Law.

3 0

3 years ago

Read 2 more answers

What is the difference between a 4x100 and a 4x400 relay

qaws [65]

A 4x100 relay is where 4 people run 100 meters and a 4x400 relay is where 4 people run 400 meters

6 0

3 years ago

1. If an object of mass m collides and velocity v collides inelastically with an object of mass 3m that is initially at rest, th

Archy [21]

Answer:

1a). 4 times. 1b) 4. 1c) 1/4.

2a) 5 times. 1b) 5 1c) 1/5

3a) 7/3 times 3b) 7/3 3c) 3/7

4a) 8/5 times 4b) 8/5 4c) 5/8

Explanation:

1) Assuming no external forces acting during the collision, total momentum must be conserved, so the following general expression applies:

m₁*v₁₀ + m₂*v₂₀ = m₁*v₁f + m₂*v₂f (1)

If we assume that the collision is perfectly inelastic, this means that both masses stick together after the collision, so v₁f = v₂f.

If m₂ is initially at rest, ⇒ v₂₀ = 0.

Replacing in (1) we get the expression of vf as a function of v₁₀, as follows:

vf = v₁₀*(m₁/(m₁+m₂)

So, for the four cases we have the following:

1) initial mass = m

final mass = m+3m = 4 m

⇒final mass / initial mass = 4

vf = v₀* (m/4m) = v₀/4 ⇒v₀/vf = 4

So, the velocity of the system will decrease by a factor of 4. The new velocity will be vf= v₀/4.

2) Applying the same considerations, we get:

2a) final mass / initial mass = 5

2b) vf = v₀* (m/5m) = v₀/5 ⇒v₀/vf = 5

2c) vf = v₀/5

3) Applying the same considerations, we get:

3a) final mass / initial mass = 7/3

3b) vf = v₀* (3m/7m) =3/7* v₀ ⇒v₀/vf = 7/3

3c) vf = 3/7*v₀

4) Applying the same considerations, we get:

4a) final mass / initial mass = 8/5

4b) vf = v₀* (5m/8m) = 5/8*v₀ ⇒v₀/vf = 8/5

4c) vf = 5/8*v₀

3 0

3 years ago

A"car"initially"at"rest"experiences"a" constant"acceleration"along"a"horizontal" road."the"position"of"the"car"at"several" succe

marysya [2.9K]

In the process of peppering the question with those forty (40 !) un-necessary quotation marks, you neglected to actually show us the illustration. So we have no information to describe the adjacent positions, and we're not able to come up with any answer to the question.

7 0

3 years ago

A rock is thrown off a 50.0 m high cliff. How fast must the rock leave the cliff top to land on level ground below, 90 m from th

blagie [28]

Answer:

The rock must leave the cliff at a velocity of 28.2 m/s

Explanation:

The position vector of the rock at a time t can be calculated using the following equation:

r = (x0 + v0x · t, y0 + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the cliff so that x0 and y0 = 0.

When the rock reaches the ground, the position vector will be (see r1 in the figure):

r1 = (90 m, -50 m)

Then, using the equation of the vector position written above:

90 m = x0 + v0x · t

-50 m = y0 + 1/2 · g · t²

Since x0 and y0 = 0:

90 m = v0x · t

-50 m = 1/2 · g · t²

Let´s use the equation of the y-component of the vector r1 to find the time it takes the rock to reach the ground and with that time we can calculate v0x:

-50 m = 1/2 · g · t²

-50 m = -1/2 · 9.81 m/s² · t²

-50 m / -1/2 · 9.81 m/s² = t²

t = 3.19 s

Now, using the equation of the x-component of r1:

90 m = v0x · t

90 m = v0x · 3.19 s

v0x = 90 m / 3.19 s

v0x = 28.2 m/s

8 0

3 years ago