All categories

2 years ago

A lack of precision in scientific measurement typically arises from?

Physics

1 answer:

PtichkaEL [24]2 years ago

6 0

Answer:

limitations of the measuring instrument.

Explanation:

Hope this helps

You might be interested in

A bead slides without friction around a loopthe-loop. The bead is released from a height 21.9 m from the bottom of the loop-the-

wariber [46]

Answer:

Part a)

$v = 12.45 m/s$

Part B)

$F_n = 0.05 N$

Explanation:

Part A)

As we know that the point A lies on the top of the loop

so we will have by energy conservation

$mgH = \frac{1}{2}mv^2 + mg(2R)$

so the speed at point A is given as

$mg(H - 2R) = \frac{1}{2}mv^2$

$v = \sqrt{2g(H - 2R)}$

$v = \sqrt{2(9.81)(21.9 - 2\times 7)}$

$v = 12.45 m/s$

Part B)

Now the force equation at point A is given as

$F_n + mg = \frac{mv^2}{R}$

$F_n = \frac{mv^2}{R} - mg$ [/tex]

$F_n = 0.004(\frac{12.45^2}{7} - 9.81)$

$F_n = 0.05 N$

6 0

3 years ago

Helppppppppppp im dieing

Otrada [13]

It would be 1. B 2. A 3. A

7 0

3 years ago

What does the principal quantum number determine?

morpeh [17]

<span>principal quantum number (n) </span>represents the relative overall energy of each orbital

Hope this helps!

4 0

2 years ago

what will be the gravitational force between two bodies if the mass of each is doubled and the distance between them is halved?

katen-ka-za [31]

Answer:

Gravitational force will be 16 times more.

Explanation:

we know;

Gravitational force (F) = (Gm1m2)/d^2

when mass of each is doubled and distance between them is halved;

F= (G2m1×2m2)/(d/2)^2

=(4Gm1m2)/(d^2/4)

=4×4(Gm1m2)/d^2

=16(Gm1m2)/d^2

=16F

3 0

2 years ago

A small candle is 34 cm from a concave mirror having a radius of curvature of 22 cm. Where will the image of the candle be locat

xxTIMURxx [149]

Answer:

16.26 cm in front of the mirror

Explanation:

Using,

1/f = 1/u+1/v....................... Equation 1

Where f = focal focal length of the concave mirror, u = object distance, v = image distance.

make v the subject of the equation

v = fu/(u-f)................... Equation 2

Note: The focal length of a concave mirror is positive

Using the real- is- positive convention

Given: f = 22/2 = 11 cm, u = 34 cm.

Substitute into equation 2

v = (34×11)/(34-11)

v = 374/23

v = 16.26 cm.

The image will be formed 16.26 cm in front of the mirror.

5 0

2 years ago