If you mix water that is 80° with the same amount of water that is 20° what will the resulting temperature be

A girl stand 5m away from a large plane mirror.How far must she walk ti be 2m away from her image?

KonstantinChe [14]

Answer:

3m

Explanation:

If a girl is 5m away from a mirror, she needs to walk a further 3m to be 2m away from her image.

5m-3m = 2m

4 0

2 years ago

A reactor operating at 1 MW is scrammed by instantaneous insertion of $5.00 of negative reactivity. Approximately how long does

Novosadov [1.4K]

Answer:

time is 3333.33 min or 55.55 hr

Explanation:

given data

reactor operating = 1 MW

negative reactivity = $5

power = 1 miliwatt

to find out

how long does it take

solution

we know here power coefficient that is

power coefficient = $\frac{10^{6} }{10^{6} }$

power coefficient = 1

so time required to reach power is

power = reactivity × time / power coefficient + reactor operating

1 × $10^{-3}$ = -5 t / 1 + 1 × $10^{6}$

5t = $10^{6}$ - $10^{-3}$

t = 199999.99 sec

so time is 3333.33 min or 55.55 hr

4 0

3 years ago

If you have a 1.0 m aqueous solution of NaCl, by how much will it increase the water’s boiling point, if KB = 0.512 °C/m? In oth

fgiga [73]

Answer: The elevation in boiling point is 1.024°C.

Explanation:

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=ik_b\times m$

where,

i = Van't Hoff factor = 2 (for NaCl)

$\Delta T_b$ = change in boiling point = ?

$k_b$ = boiling point constant = $0.512^oC/m$

m = molality = 1.0 m

Putting values in above equation, we get:

$\Delta Tb=2\times 0.512^oC/m\times 1.0m\\\\\Delta Tb=1.024^oC$

Hence, the elevation in boiling point is 1.024°C.

5 0

4 years ago

At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f

jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp = change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s, 0.65 kg m/s)

The magnitude of this vector is calculated as follows:

$|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s$

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

4 0

3 years ago

A small object carrying a charge of -2.50 nc is acted upon by a downward force of 18.0 nn when placed at a certain point in an e

Vesna [10]

Missing question in the text:
"A.What are the magnitude and direction of the electric field at the point in question?

B.What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?"

Solution:

A) A charge q under an electric field of intensity E will experience a force F equal to:

$F=qE$

In our problem we have $q=-2.5 nC=-2.5\cdot 10^{-9} C$ and $F=18 nN = 18 \cdot 10^{-9} N$ , so we can find the magnitude of the electric field:

$E= \frac{F}{q}= \frac{18\cdot 10^{-9}N}{2.5\cdot 10^{-9}C}=7.2 V/m$

The charge is negative, therefore it moves against the direction of the field lines. If the force is pushing down the charge, then the electric field lines go upward.

B) The proton charge is equal to

$e=1.6\cdot 10^{-19} C$

Therefore, the magnitude of the force acting on the proton will be

$F=eE=1.6\cdot 10^{-19} C \cdot 7.2 V/m=1.15 \cdot 10^{-18} N$

And since the proton has positive charge, the verse of the force is the same as the verse of the field, so upward.

7 0

3 years ago