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3 years ago

The valid digits in a measurement are called _____ digits.

Physics

2 answers:

Harman [31]3 years ago

8 0

Answer:

Significant

Explanation:

As the word suggests, significant figures or digits are numbers that are valid in measurement.

Sunny_sXe [5.5K]3 years ago

5 0

Answer:

SIGNIFICANT DIGITS

Explanation:

MARK ME BRAINLIEST PLZZZ

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If the moon were twice its present distance from earth, what would be the most noticeable effects on earth events

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There would be no light at night foe nocturnal animals

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4 years ago

A 0.144kg baseball is pitched horizontally at 38.0 m/s. After the baseball is hit by the bat, it moves at the same speed, but in

Hoochie [10]

Answer:

Given:

mass of the ball m = 0.144 kg

velocity v = 38 m/s

now, change in momentum

P = m v- ( - mv)

= 2 mv

=2 x (0.144) x (38)

= 10.944 kg-m/s

Impulse J= F. Δt

change in momentum is equal to impulse

J = 10.944 kg-m/s

we know force is equal to change in momentum per unit time

$F = \dfrac{P}{t}$

$F = \dfrac{10.944}{0.8\times 10^{-3}}$

F = 13.68 x 10³ N

F = 13.68 kN

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4 years ago

Calculate the potential difference between points x and y

aliya0001 [1]

Answer:

4.275v

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3 years ago

Read 2 more answers

A closed, rigid tank fitted with a paddle wheel contains 2 kg of air, initially at 300 K. During an interval of 5 minutes, the p

anzhelika [568]

Answer:

The final temperature of the air is $T_2= 605 K$

Explanation:

We can start by doing an energy balance for the closed system

$\Delta KE+\Delta PE+ \Delta U = Q - W$

where

$\Delta KE$ = the change in kinetic energy.

$\Delta PE$ = the change in potential energy.

$\Delta U$ = the total internal energy change in a system.

Q = the heat transferred to the system.

W = the work done by the system.

We know that there are no changes in kinetic or potential energy, so $\Delta KE = 0$ and $\Delta PE=0$

and our energy balance equation is $\Delta U = Q - W$

We also know that the paddle-wheel transfers energy to the air at a rate of 1 kW and the system receives energy by heat transfer at a rate of 0.5 kW, for 5 minutes.

We use this information to calculate the total internal energy change $\Delta U=W+Q$ using the energy balance equation.

We convert the interval of time to seconds $t = 5 \:min = 300\:s$

$\Delta \dot{U}=\dot{W}+ \dot{Q}\\=\Delta U=(W+ Q)\cdot t$

$\Delta U=(1 \:kW+0.5\:kW)\cdot 300\:s\\\Delta U=450 \:kJ$

We can use the change in specific internal energy $\Delta U = m(u_2-u_1)$ to find the final temperature of the air.

We are given that $T_1=300 \:K$ and the air can be describe by ideal gas model, so we can use the ideal gas tables for air to determine the initial specific internal energy $u_1$

$u_1=214.07\:\frac{kJ}{kg}$

Next, we will calculate the final specific internal energy $u_2$

$\Delta U = m(u_2-u_1)\\\frac{\Delta U}{m} =u_2-u_1$

$\frac{\Delta U}{m} =u_2-u_1\\u_2=u_1+\frac{\Delta U}{m}$

$u_2=214.07 \:\frac{kJ}{kg} +\frac{450 \:kJ}{2 \:kg}\\u_2= 439.07 \:\frac{kJ}{kg}$

With the value $u_2=439.07 \:\frac{kJ}{kg}$ and the ideal gas tables for air we make a regression between the values $u = 434.78 \:\frac{kJ}{kg},T=600 \:K$ and $u = 442.42 \:\frac{kJ}{kg}, T=610 \:K$ and we find that the final temperature $T_2$ is 605 K.