All categories

3 years ago

Hey I'm Chloe Can you Help Me.

Physics

1 answer:

Viefleur [7K]3 years ago

3 0

Answer:

Hello There!!

Explanation:

The answer is High-Fibre foods they take longer to digest. For example vegetables,fruits,nuts and seeds and potatoes with skin.

hope this helps,have a great day!!

~Pinky~

hope this helps,

You might be interested in

A 20-kg block slides down a fixed rough curved track The block has a speed of 5 0 m/s after its height above a horizontal surfac

Crank

Answer:

U = 102.8 J (100 J to two significant digits)

Explanation:

potential energy converted = 20(9.8)(1.8) = 352.8 J

kinetic energy at base of track = ½(20)5.0² = 250 J

energy (work) of friction 352.8 - 250 = 102.8 J

8 0

3 years ago

Find the energy released when there is a decrease of 0.3kg of material in a nuclear reaction

MrRissso [65]

Answer:

E = 2.7 x 10¹⁶ J

Explanation:

The release of energy associated with the mass can be calculated by Einstein's mass-energy relation, as follows:

$E = mc^2$

where,

E = Energy Released = ?

m = mass of material reduced = 0.3 kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

$E = (0.3\ kg)(3\ x\ 10^8\ m/s)^2$

4 0

2 years ago

The wavelength of red light is 7 x 10^-7 meter. Express this value in nanometers

3241004551 [841]

The wavelength of the red light in "nanometer" is 7× $10^{2}$

Wavelength is given as : 7× $10^{-7}$ meter

1 nanometer = ( $10^{-9}$ meter)

Let X= value of the wavelength in nanometer.

1 nanometer = $10^{-9}$ meter

X nanometer = 7× $10^{-7}$ meter

<em>If we Cross multiply</em>

X nanometer = ( $\frac{ 7* 10^{-7} }{ 10^{-9} }$ )

X= 7× $10^{2}$ nanometer

Therefore, the wavelength in "nanometer" is 7× $10^{2}$

Learn more at :brainly.com/question/12924624?referrer=searchResults

6 0

2 years ago

A 4.0-m-diameter playground merry-go-round, with a moment of inertia of

HACTEHA [7]

Answer:

$7.1$ ms⁻¹

Explanation:

$d$ = diameter of merry-go-round = 4 m

$r$ = radius of merry-go-round = $\frac{d}{2}$ = $\frac{4}{2}$ = 2 m

$I$ = moment of inertia = 500 kgm²

$w_{i}$ = angular velocity of merry-go-round before ryan jumps = 2.0 rad/s

$w_{f}$ = angular velocity of merry-go-round after ryan jumps = 0 rad/s

$v$ = velocity of ryan before jumping onto the merry-go-round

$m$ = mass of ryan = 70 kg

Using conservation of angular momentum

$Iw_{i} - m v r = (I + mr^{2})w_{f}$

$(500)(2.0) - (70) v (2) = (I + mr^{2})(0)$

$1000 = 140 v$

$v = 7.1$ ms⁻¹

5 0

2 years ago

3) If a ball launched at an angle of 10.0 degrees above horizontal from an initial height of 1.50 meters has a final horizontal

Irina-Kira [14]

Answer:

35.6 m

Explanation:

3 0

3 years ago