All categories

3 years ago

in positive numbers less than 1, the zeros between the decimal point and a non-zero number are _______ significant?

Physics

1 answer:

DedPeter [7]3 years ago

8 0

Answer:

Explanation:

If a number of less than 1, then the number has a decimal point like

0.085, 0.008 e.t.c.

The zeros before the none zero digit are insignificant. The significant figure is 8 and 5.

But if there a zero between the none zero e.g. 0.0087056

Here the zero between 7 and 5 is significant, then the significant numbers are 8,7,0,5,6

But if the zero is not in between the none zero digit, then the zero is insignificant

E.g 0.05800

The last two zero is insignificant, the significant number is 5 and 8

So, If a positive numbers less than 1, the zeros between the decimal point and a non-zero number are NOT significant.

You might be interested in

The difference in electric potential between a thunder cloud and the ground is 1.53 108 V. Electrons move from the ground which

Kryger [21]

Answer:

=−2.451 330 152 1*10^27J

Explanation:

The electric potential=the Voltage * Charge:

E = VQ

V = 1.53x10^8 V (positive, because the cloud has a higher potential)

Q = -1.60217657 x10^19 C (the charge of an electron)

E = (1.53x10^8 V )* (-1.60217657 x10^19 C)

E=−2.451 330 152 1*10^27J

The negative sign indicates that the potential energy is decreased by the movement of the electron.

8 0

3 years ago

Particle q1 has a positive 6 µc charge. particle q2 has a positive 2 µc charge. they are located 0.1 meters apart. recall that k

Dmitrij [34]

(a) Force between the two charges

The electrostatic force between the two charges is given by:

$F=k\frac{q_1 q_2}{r^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, r their separation.

In this problem:

$q_1 =6 \mu C=6 \cdot 10^{-6}C$

$q_2=2 \mu C=2 \cdot 10^{-6}C$

$r=0.1 m$

Substituting into the equation, we find

$F=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(6 \cdot 10^{-6}C)(2 \cdot 10^{-6}C)}{(0.1 m)^2}=10.8 N$

(b) direction of particle q2

Particle q2 wants to move in the direction of the force acting on it. The direction of the force depends on the relative sign of the two charges: like charges attract each other, opposite charges repel each other. In this case, the two charges are both positive, so they repel each other and q2 tends to move away from particle q1.

7 0

3 years ago

Read 2 more answers

A 75.0 Ohm resistor uses 0.285 W of power. What is the voltage across the resistor?

Brilliant_brown [7]

Answer:4.62

Explanation:

Acellus

6 0

2 years ago

You have a 35X objective lens in place, and the numerical aperture of the objective lens is 0.75. The numerical aperture of the

olasank [31]

Answer:

350x

Explanation:

In a microscope the objective has higher magnification than the eyepiece so, this is a microscope

The magnification of a microscope is given by the product of the magnifications of the eyepiece and and the objective.

Objective lens magnification = 35x = $m_o$

Eyepiece magnification = 10x = $m_e$

Total magnification

$M=m_o\times m_e\\\Rightarrow M=35\times 10\\\Rightarrow M=350$

Total magnification is 350x

8 0

3 years ago

A 1kg cart slams into a stationary 1kg cart at 2 m/s. The carts stick together and move forward at a speed of 1 m/sl. Determine

finlep [7]

Answer:

No, it is not conserved

Explanation:

Let's calculate the total kinetic energy before the collision and compare it with the total kinetic energy after the collision.

The total kinetic energy before the collision is:

$K_i = K_1 + K_2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2=\frac{1}{2}(1 kg)(2 m/s)^2+\frac{1}{2}(1 kg)(0)^2=2 J$

where m1 = m2 = 1 kg are the masses of the two carts, v1=2 m/s is the speed of the first cart, and where v2=0 is the speed of the second cart, which is zero because it is stationary.

After the collision, the two carts stick together with same speed v=1 m/s; their total kinetic energy is

$K_f = \frac{1}{2}(m_1+m_2)v^2=\frac{1}{2}(1 kg+1kg)(1 m/s)^2=1 J$

So, we see that the kinetic energy was not conserved, because the initial kinetic energy was 2 J while the final kinetic energy is 1 J. This means that this is an inelastic collision, in which only the total momentum is conserved. This loss of kinetic energy does not violate the law of conservation of energy: in fact, the energy lost has simply been converted into another form of energy, such as heat, during the collision.

3 0

3 years ago