A solid Cute hug sides 0.50m

You have to run 2.2 miles in track. How far is this in feet? Note: There are

Romashka [77]

Answer: B

Explanation: This can be easily done by inputting 5280 * 2.2.

3 0

3 years ago

Read 2 more answers

What is the magnitude of the momentum change of two gallons of water (inertia about 7.3 kg ) as it comes to a stop in a bathtub

aliya0001 [1]

We know that the change in momentum is equals to the product of force and time that is impulse ( $F \times t$ ). Therefore, we need to determine the value of that the water is in air by using the second equation of motion,

$s=ut+\frac{1}{2} gt^2$

Here, u is initial velocity which is zero.

$s= \frac{1}{2} gt^2 \\\\ t = \sqrt{\frac{2s}{g} }$ .

Thus, impulse

$= F \times \sqrt{\frac{2s}{g} }$

From Newton`s second law,

$F =mg$

Therefore, impulse

$= mg \times \sqrt{\frac{2s}{g} } = m \sqrt{2gs}$

Given, $m = 7.3 kg$ and $s = 2.0 m$

Substituting these values, we get

Change in momentum = impulse

$= 7.3 \ kg \sqrt{2 \times 9.8 \ m/s^2 \times 2.0 \ m } = 45 .7 \ Ns$ .

8 0

3 years ago

an object is placed 15.8 cm in front of a thin converging lens with an unknown focal length. if a real image forms behind the le

storchak [24]

On what:

f (is the focal length of the lens) = ?
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm

Using the Gaussian equation, to know where the object is situated (distance from the point).

$\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}$
$\frac{1}{f} = \frac{1}{15.8} + \frac{1}{4.2}$
$\frac{1}{f} = \frac{2.1}{33.18} + \frac{7.9}{33.18}$
$\frac{1}{f} = \frac{10}{33.18}$
Product of extremes equals product of means:
$10*f = 1*33.18$
$10f = 33.18$
$f = \frac{33.18}{10}$
$\boxed{\boxed{f = 3.318\:cm}}\end{array}}\qquad\quad\checkmark$

6 0

4 years ago

Pre-Test Active

vlada-n [284]

n is the answer I guess

not so sure

7 0

2 years ago

What problem do refractor telescopes have that reflectors don't? Group of answer choices bad seeing light loss from secondary el

Paul [167]

chromatic aberration problem do refractor telescopes have that reflectors don't

Explanation:

Chromatic aberration is a phenom in which light rays crossing through a lens focus at various points, depending on their wavelength. Chromatic aberration is a dilemma in which lens or refracting, telescopes undergo from. The various image distances for the respective colors affect various image sizes for them.

This involves the creation of disturbing color fringes in the image. Chromatic aberration can be pretty well adjusted by the use of an achromatic doublet. Here, a positive biconvex lens is coupled with a negative lens placed backward with greater dispersion. Thus partly compensates for the chromatic aberration.

8 0

3 years ago