All categories

4 years ago

What is systematic error

Physics

2 answers:

julsineya [31]4 years ago

4 0

Answer: an error

Explanation:

IgorC [24]4 years ago

4 0

Answer:

A consistent, repeatable error that isn't random. Usually associated with faulty equipment and the like.

You might be interested in

An engine extracts 441.3kJ of heat from the burning of fuel each cycle, but rejects 259.8 kJ of heat (exhaust, friction,etc) dur

Lilit [14]

Answer:

The thermal efficiency is 0.4113.

Explanation:

We know that the thermal efficiency is the ratio of work done by the engine over the heat taken

$\eta = \frac{W_{made}}{Q_{in}}$

Now, how much work the engine do in a cycle?

We know that the work done in a cycle must be equal to the heat taken minus the heat rejected

$W{made} = Q_{in} - Q_{rejected}$

So, the thermal efficiency will be:

$\eta = \frac{Q_{in} - Q_{rejected}}{Q_{in}}$

$\eta = \frac{Q_{in}}{Q_{in}} - \frac{Q_{rejected}}{Q_{in}}$

$\eta = 1 - \frac{Q_{rejected}}{Q_{in}}$

Putting the values of the problem

$\eta = 1 - \frac{259.8 kJ }{441.3kJ}$

$\eta = 0.4113$

7 0

4 years ago

A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction

yawa3891 [41]

Answer:

$F'=\dfrac{F}{4}$

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

$F=\dfrac{mv^2}{r}$

v = 2v

$F=\dfrac{m(2v)^2}{r}$

$F=4\dfrac{m(v)^2}{r}$ ..........(1)

For the slower car on the road,

$F'=\dfrac{mv^2}{r}$ ............(2)

Equation (1) becomes,

$F=4F'$

$F'=\dfrac{F}{4}$

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

8 0

4 years ago

A motorboat travels 92 km in 2 hours going upstream. It travels 132 km going downstream in the same amount of time. What is the

taurus [48]

Answer:

The speed on boat in still water is $56 \frac{km}{h}$ and the rate of the current is $10 \frac{km}{h}$

Explanation:

Since speed , $v= \frac{Distance\, traveled(D)}{Time\, taken(t)}$

Therefore speed of motor boat while traveling upstream is

$v_{upstream}=\frac{92}{2}\frac{km}{h}=46\frac{km}{h}$

and speed of motor boat while traveling downstream is

$v_{downstream}=\frac{132}{2}\frac{km}{h}=66\frac{km}{h}$

Let speed of boat in still water be $v_b$ and rate of current be $v_w$

Therefore $v_{upstream}=v_b-v_w=46\frac{km}{h}$ ----(A)

and $v_{downstream}=v_b+v_w=66\frac{km}{h}$ ------(B)

Adding equation (A) and (B) we get

$2v_b= (46+66) \frac{km}{h}=112 \frac{km}{h}$

=> $v_b= 56 \frac{km}{h}$ ------(C)

Substituting the value of $v_b$ in equation (A) we get

$v_w= 10 \frac{km}{h}$

Thus the speed on boat in still water is $56 \frac{km}{h}$ and the rate of the current is $10 \frac{km}{h}$

5 0

4 years ago

Which three statements about electromagnetic radiation are true?

grandymaker [24]

I think that numbers one, three, and four are true

3 0

3 years ago

Read 2 more answers

For the last part of the lab, you should have found the mass of the meter stick. So if a mass of 85 g was placed at the 2 cm MAR

Bond [772]

Answer:

272.89g

Explanation:

Find the diagram to the question in the attachment below;.

Using the principle of moment to solve the question which states that the sum of clockwise moment is equal to the sum of anticlockwise moment.

Moment = Force * Perpendicular distance

Taking the moment of force about the pivot.

Anticlockwise moment:

The 85g mass will move in the anticlockwise

Moment of 85g mass = 85×36.6

= 3111gcm

Clockwise moment.

The mass of the metre stick M situated at the centre (50cm from each end) will move in the clockwise direction towards the pivot.

CW moment = 11.4×M = 11.4M

Equating CW moment to the ACW moment we will have;

11.4M = 3111

M = 3111/11.4

M = 272.89g

The mass of the metre stick is 272.89g

5 0

4 years ago

What is systematic error​

What is systematic error