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1 year ago

9

Random errors can be reduced by taking repeated measurements.Error and uncertainty are interchangeable words that describe the s

ame concept.are these statements true or false

1 answer:

fiasKO [112]1 year ago

5 0

repeated mesurement can reduce the error

it is true

if you take any mesurement repeatedly and the average is taken, the error will be less

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A 15,000 kg rocket traveling at +230 m/s turns on its engines. Over a 6.0 s period it burns 1,000 kg of fuel. An observer on the

lesya692 [45]

Answer:

a) $v = 312.791\,\frac{m}{s}$ , b) $a = 13.333\,\frac{m}{s^{2}}$

Explanation:

The problem is asking the rocket velocity and acceleration at t = 6 s.

a) The general equation of the rocket is:

$v=v_{o} -v_{ex}\cdot \ln \frac{m}{m_{o}}$

$v = 230\,\frac{m}{s}-(1200\,\frac{m}{s} )\cdot \ln \frac{14000\,kg}{15000\,kg}$

$v = 312.791\,\frac{m}{s}$

b) The acceleration experimented by the rocket is:

$a = \frac{v_{ex}}{m_{o}}\cdot \frac{dm}{dt}$

$a = \frac{1200\,\frac{m}{s} }{15000\,kg}\cdot \frac{1000\,kg}{6\,s}$

$a = 13.333\,\frac{m}{s^{2}}$

3 0

4 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.

3 0

3 years ago

An area of the sky where multiple meteors can be sometimes observed is referred to as a:

vodka [1.7K]

it would be a meteor shower

7 0

3 years ago

You extend an impromptu invitation to a friend for dinner. The only food you have is a couple of frozen steaks. You wish to defr

kakasveta [241]

Answer:

Microwave: Radiation

Water: Conduction, convection & radiation.

Explanation:

When we heat a food using microwave then the water content of the food is only heated by the microwave.

Microwaves make the molecules of water vibrate frequently with a frequency closer to the frequency of microwaves and this increases the kinetic energy of the molecules which produces heat in the water molecules this heat then propagates to the whole food by conduction and convection, but the heat enters the food only via radiation.

Now when the food item is kept into warm water then the molecules closer to the food heat the food by conduction through direct energy transfer by lattice vibrations and when they become cooler than the mass of water all around then due to density difference they settle down and their place is occupied by warmer molecules around in the fluid leading to convection.

Radiation of energy from a mass occurs continuously irrespective of the medium present there. So the heat of water also enters the food by radiation of energy from the water molecules.

3 0

3 years ago

A missile is moving 1350 m/s at 25.0° angle. It needs to hit in a 55.0° direction in 10.20 s. What is the direction of its final

77julia77 [94]

Answer:

final velocity = 3504 m/s

Explanation:

<em>Given data:</em>

velocity of missile = Vi = 1350m/s

angle at which missile is moving = 25degree

distance between missile and targets = 23500m

angle between target and missile=55degree

time=10.2s

<em>To find:</em>

Final velocity: ?

<em>Formula:</em>

x = Vx*t + ½*ax*t²

Let x be the horizontal component of distance

x = ertical component of distance

t-time

ax = horizontal component of acceleration

ay = Vertical component of acceleration

Vx = horizontal component of velocity

Vy = Vertical component of velocity

<em>Solution:</em>

x = Vx*t + ½*ax*t²

23500m * cos55.0º = 1350m/s * cos25.0º * 10.20s + ½ * ax * (10.20s)²

ax = 19.2 m/s²

V'x = Vx + ax*t = 1350m/s * cos25.0º + 19.2m/s² * 10.20s = 1419 m/s

<em>similarly vertically:</em>

y = Vy*t + ½*ay*t²

23500m * sin55.0º = 1350m/s * sin25.0º * 10.20s + ½ * ay * (10.20s)²

ay = 258 m/s²

V'y = Vy + ay*t

= 1350m/s * sin25.0º + 258m/s² * 10.20s = 3204 m/s

V = √(V'x² + V'y²)

= 3504 m/s

8 0

3 years ago