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3 years ago

PLEASE HELP!!!

Physics

2 answers:

NNADVOKAT [17]3 years ago

8 0

Answer:

sodium

Explanation:

yulyashka [42]3 years ago

3 0

A. Sodium . . . Element #11 on the Periodic Table; when combined with other stuff as a chemical compound, it can make salt, Drano, and Alka-Seltzer

B. Carbon Dioxide . . . compound, made of elements Carbon and Oxygen ; produced when wood or gasoline burns; comes out of your lungs when you breathe out; air has a little bit of it, which is getting to be too much

C. Dihydrogen Monoxide . . . compound, made of elements Hydrogen and Oxygen, usually known by its popular name: "Water".

D. Magnesium Sulfate . . . compound, made of elements Magnesium and Sulfur; I don't know anything about this compound.

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A cart of mass 20 kg, initially at rest, is pushed with a net force of 40 N on a flat surface. If the cart is pushed 5 m, what i

bixtya [17]

Using the theorem of kinetic energy
1/2mVf² - 1/2mVi²= WF + Wp, Wp=0
WF = F. AB, AB=5m and F= 40N, m=20kg
so the final kinetic is KEf= 1/2mVf² = WF =F. AB= 40*5=200J

the final velocity is 1/2mVf² =200, implies Vf= sqrt(20)=2sqrt(5)m/s

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3 years ago

Cómo puedo saber si el agua destilada que nos vendieron es destilada de verdad?

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Answer:

Caliente el agua del grifo hasta que se convierta en vapor. Cuando el vapor se vuelve a condensar en agua

Explanation:

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3 years ago

A boy sleds down a hill and onto a frictionless ice- covered lake at 10.0 m/s. In the middle of the lake is a 1000-kg boulder. W

mina [271]

Answer:

The speed of the sled is 9.2 m/s

The speed of the boulder is 0.82 m/s

Solution:

As per the question:

Mass of the boulder, $m_{B} = 1000\ kg$

Mass of the sled, $m_{S} = 2.50\ kg$

Mass of the boy, $m_{b} = 40\ kg$

Initial Velocity, v = 10.0 m/s

Now,

To calculate the speed of both the sled and the boulder after the occurrence of the collision:

m = $m_{b} + m_{S} = 40 + 2.50 = 42.50\ kg$

Initial velocity of the boulder, $v_{B} = 0\ m/s$

Since, the collision is elastic, both the energy and momentum rem,ain conserved.

Now,

Using the conservation of momentum:

$mv + m_{B}v_{B} = mv' + m_{B}v'_{B}$

where

v' = final velocity of the the system of boy and sled

$v'_{B}$ = final velocity of the boulder

$42.50\times 10 + m_{B}.0 = 42.50v' + 1000v'_{B}$

$42.50v' + 1000v'_{B} = 425$ (1)

Now,

Using conservation of energy:

$\frac{1}{2}mv^{2} + \frac{1}{2}m_{B}v_{B}^{2} = \frac{1}{2}mv'^{2} + \frac{1}{2}m_{B}v'_{B}^{2}$

$42.50\times 10^{2} + m_{B}.0 = 42.50v'^{2} + 1000v'_{B}^{2}$

$42.50v'^{2} + 1000v'_{B}^{2} = 4250$ (2)

Now, from eqn (1) and (2):

$v' = \frac{m - m_{B}}{m + m_{B}}\times v$

$v' = \frac{42.50 - 1000}{42.5 + 1000}\times 10 = - 9.2\ m/s$

Now,

$v'_{B} = \frac{2m}{m + m_{B}}\times v$

$v'_{B} = \frac{2\times 42.50}{42.5 + 1000}\times 10 = 0.82\ m/s$

5 0

3 years ago

Read 2 more answers

An arrow is launched from a bow with an initial horizontal velocity of 40

Ivan

Answer:

V = (Vx^2 + Vy^2)^1/2 = (40^2 + 62^2)^1/2

V = 73.8 m/s

tan theta = Vy / Vx = 62/40 = 1.55

theta = 57.2 deg

4 0

3 years ago

A 5.0-kg box is pulled by a horizontal force F applied to the top of the box. When the box meets a low doorstep, it begins to ro

NARA [144]

Answer:

the required minimum magnitude of the force F is 21 N

Explanation:

Given the data in the question,

m = 5 kg

width = 60 cm

height = 80 cm

Let force is F represent in the image below,

so when the block about to rotate normal shifted to edge of cube

mg(w/2) = Fh

F = mg(w/2) / h

we know that g = 9.8 m/s²

we substitute

F = (5 × 9.8 ( 60/2)) / 70

F = (5 × 9.8 × 30 ) / 70

F = 1470 / 70

F = 21 N

Therefore, the required minimum magnitude of the force F is 21 N

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3 years ago