All categories

1 year ago

Which of the following is an example of a SOLUTION?

Physics

1 answer:

Mila [183]1 year ago

7 0

Answer:

Salt water

Explanation:

This is the answer

You might be interested in

Conduction, convection, and radiation are ways by which,

Amiraneli [1.4K]

the answer is THERMAL

6 0

2 years ago

Read 2 more answers

A family shares the same hairdryer.

IceJOKER [234]

Answer:

Idk I’m sorry

Explanation:

Sorry

7 0

3 years ago

Young's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured carefully on a s

PSYCHO15rus [73]

Answer:

Explanation:

wave length of light λ = 502 nm

screen distance D = 1.2 m

width of one fringe = 10.2 mm / 20

= .51 mm

fringe width = λ D / a , a is separation of slits

Puting the values given

.51 x 10⁻³ = 502 x 10⁻⁹ x 1.2 / a

a = 502 x 10⁻⁹ x 1.2 / .51 x 10⁻³

= 1181.17 x 10⁻⁶ m

1.18 x 10⁻³ m

= 1.18 mm .

8 0

2 years ago

Who water rocket starts from rest and roses straight up with an acceleration of 5 m/s until it runs out of water 2.5 seconds lat

Kitty [74]

Answer:

23. 4375 m

Explanation:

There are two parts of the rocket's motion

1 ) accelerating (assume it goes upto h1 height )

using motion equations upwards

$s = ut+\frac{1}{2}*a*t^{2} \\h_1=0+\frac{1}{2}*5*2.5^{2} \\=15.625 m$

Lets find the velocity after 2.5 seconds (V1)

V = U +at

V1 = 0 +5*2.5 = 12.5 m/s

2) motion under gravity (assume it goes upto h2 height )

now there no acceleration from the rocket. it is now subjected to the gravity

using motion equations upwards (assuming g= 10m/s² downwards)

V²= U² +2as

0 = 12.5²+2*(-10)*h2

h2 = 7.8125 m

maximum height = h1 + h2

= 15.625 + 7.8125

= 23. 4375 m

3 0

2 years ago

A light, flexible cable is wrapped around a solid cylinder with mass 3.3 kg and a radius of 0.8 meters. The cylinder rotates on

kari74 [83]

Answer:

$9.16rad/s^2$

Explanation:

We are given that

Mass, $m_1=3.3 kg$

Radius,r=0.8 m

$m_2=4.9 kg$

Height,h=2.9 m

We have to find the angular acceleration of the cylinder.

According to question

$4.9g-T=4.9a$

$Tr=I\alpha$

Where

$\alpha=\frac{a}{r}$

$Tr=\frac{1}{2}m_1ra$

$T=\frac{1}{2}m_1a=\frac{1}{2}(3.3)a$

Substitute the value

$4.9g-\frac{1}{2}(3.3a)=4.9a$

$4.9\times 9.8=4.9a+\frac{3.3a}{2}$

Where $g=9.8 m/s^2$

$48.02=a(4.9+1.65)=6.55a$

$a=\frac{48.02}{6.55}=7.33m/s^2$

Angular acceleration, $\alpha=\frac{a}{r}=\frac{7.33}{0.8}=9.16rad/s^2$

7 0

2 years ago