What are the answers to this

How many significant digits are measurement 0.00210 mg?

jek_recluse [69]

There are 3 significant figures, if that answers the question.

8 0

3 years ago

A tall cylinder contains 30 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until t

Anastaziya [24]

Answer:

The gauge pressure is $P_g = 2058 \ P_a$

Explanation:

From the question we are told that

The height of the water contained is $h_w = 30 \ cm = 0.3 \ m$

The height of liquid in the cylinder is $h_t = 40 \ cm = 0.4 \ m$

At the bottom of the cylinder the gauge pressure is mathematically represented as

$P_g = P_w + P_o$

Where $P_w$ is the pressure of water which is mathematically represented as

$P_w = \rho_w * g * h_w$

Now $\rho_w$ is the density of water with a constant values of $\rho_w = 1000 \ kg /m^3$

substituting values

$P_w = 1000 * 9.8 * 0.3$

$P_w = 2940 \ Pa$

While $P_o$ is the pressure of oil which is mathematically represented as

$P_o = \rho_o * g * (h_t -h_w )$

Where $\rho _o$ is the density of oil with a constant value

$\rho _o = 900 \ kg / m^3$

substituting values

$P_o = 900 * 9.8 * (0.4 - 0.3)$

$P_o = 882 \ Pa$

Therefore

$P_g = 2940 - 882$

$P_g = 2058 \ P_a$

6 0

3 years ago

Sal sprinted 40 m to the right in 5.5s what is his average velocity

PolarNik [594]

Answer: 7.27 m/s

Explanation:

6 0

3 years ago

One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a hel

Gekata [30.6K]

Answer:

a) $v_{U-235} = 2.68 \cdot 10^{5} m/s$

$v_{He-4} = -1.57 \cdot 10^{7} m/s$

b) $E_{He-4} = 8.23 \cdot 10^{-13} J$

$E_{U-235} = 1.41 \cdot 10^{-14} J$

Explanation:

Searching the missed information we have:

E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J

m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg

m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg

a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

Linear momentum:

$p_{i} = p_{f}$

$m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}$

Since the plutonium nucleus is originally at rest, $v_{Pu-239} = 0$ :

$0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}$

$v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}}$ (1)

Kinetic Energy:

$E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}$

$2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}$

$1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}$ (2)

By entering equation (1) into (2) we have:

$1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}$

$1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2}$

Solving the above equation for $v_{U-235}$ we have:

$v_{U-235} = 2.68 \cdot 10^{5} m/s$

And by entering that value into equation (1):

$v_{He-4} = -\frac{3.92 \cdot 10^{-25} kg*2.68 \cdot 10^{5} m/s}{6.68 \cdot 10^{-27} kg} = -1.57 \cdot 10^{7} m/s$

The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.

b) Now, the kinetic energy of each nucleus is:

For He-4:

$E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J$

For U-235:

$E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J$

I hope it helps you!

3 0

3 years ago

One hundred turns of insulated copper wire are wrapped into a circular coil of crosssectional area 1.20⇥103 m2. The two ends of

arsen [322]

Answer:

236.3 x $10^-^3$ C

Explanation:

Given:

B(0)=1.60T and B(t)=-1.60T

No. of turns 'N' =100

cross-sectional area 'A'= 1.2 x $10^-^3$ m²

Resistance 'R'= 1.3Ω

According to Faraday's law, the induced emf is given by,

ℰ=-NdΦ/dt

The current given by resistance and induced emf as

I = ℰ/R

I= -NdΦ/dtR

By converting the current to differential form(the time derivative of charge), we get

$\frac{dq}{dt}$ = -NdΦ/dtR

dq= -N dΦ/R

The change in the flux dФ =Ф(t)-Ф(0)

therefore, dq = $\frac{N}{R}$ (Ф(0)-Ф(t))

Also, flux is equal to the magnetic field multiplied with the area of the coil

dq = NA(B(0)-B(t))/R

dq= (100)(1.2 x $10^-^3$ )(1.6+1.6)/1.3

dq= 236.3 x $10^-^3$ C

5 0

4 years ago