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vampirchik [111]

3 years ago

10

Please Help!!!!! If you were in a completely weightless environment, would you need a force to make an object at rest start to m

ove? Explain you answer.

1 answer:

MrRissso [65]3 years ago

4 0

Answer:

Yes

Explanation:

Absolutely yes! Even in space objects have mass. And if they have mass, they have inertia.

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Water ﬂows through a horizontal nozzle in steady ﬂow at the rate of 10m3/s. The inlet and outlet diameters are d1 = 0.5m and d2

Dvinal [7]

Answer:

P₁ = 2.215 10⁷ Pa, F₁ = 4.3 106 N,

Explanation:

This problem of fluid mechanics let's start with the continuity equation to find the speed of water output

Q = A v

v = Q / A

The area of a circle is

A = π r² = π d² / 4

Let's look at the speeds at each point

v₁ = Q / A₁ = Q 4 /π d₁²

v₁ = 10 4 /π 0.5²

v₁ = 50.93 m / s

v₂ = Q / A₂

v₂ = 10 4 /π 0.25²

v₂ = 203.72 m / s

Now we can use Bernoulli's equation in the colon

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Since the tube is horizontal y₁ = y₂. The output pressure is P₂ = Patm = 1.013 10⁵ Pa, let's clear

P₁ = P2 + ½ rho (v₂² - v₁²)

P₁ = 1.013 10⁵ + ½ 1000 (203.72² - 50.93²)

P₁ = 1.013 10⁵ + 2.205 10⁷

P₁ = 2.215 10⁷ Pa

la definicion de presion es

P₁ = F₁/A₁

F₁ = P₁ A₁

F₁ = 2.215 10⁷ pi d₁²/4

F₁ = 2.215 10⁷ pi 0.5²/4

F₁ = 4.3 106 N

6 0

3 years ago

Is this true of false <br> Apples and Onions taste the same if you plug your nose?

xxMikexx [17]

I think yes because you won’t be able to smell

7 0

3 years ago

A 125kg bumper car going 12m/s hits a 235kg bumper car going -13m/s.if the first car bounces back at -12.5m/s what is the veloci

vovikov84 [41]

According to the law of conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$

m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision

Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.

m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$
$(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'$
$1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')$
$-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')$

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.

$-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')$
$7.5kg.m/s=(235kg)(v_{2}')$
$(7.5kg.m/s)/(235kg)=(v_{2}')$
$0.03m/s=(v_{2}')$

The velocity of the 2nd car after the collision is 0.03m/s.

5 0

3 years ago

In a physics lab, Asha is given a 10.7 kg uniform rectangular plate with edge lengths 67.3 cm by 53.5 cm . Her lab instructor re

Leya [2.2K]

Answer:

$I=2.6363\ kg.m^2$

Explanation:

Given:

dimension of uniform plate, $(0.673\times 0.535)\ m^2$

mass of plate, $m=10.7\ kg$

Now we find the moment of inertia about the center of mass of the rectangular plate is given as:

$I_{cm}=\frac{1}{12} \times m(L^2+B^2)$

where:

$L=$ length of the plate

$B=$ breadth of the plate

$I_{cm}=\frac{1}{12} \times 10.7\times(0.673^2+0.535^2)$

$I_{cm}=0.6591\ kg.m^2$

We know that the center of mass of the rectangular plane is at its geometric center which is parallel to the desired axis XX' .

Now we find the distance between the center of mass and the corner:

$s=\frac{\sqrt{ (0.673^2+0.535^2)}}{2}$

$s=0.4299\ m$

Now using parallel axis theorem:

$I=I_{cm}+m.s^2$

$I=0.6591+10.7\times 0.4299^2$

$I=2.6363\ kg.m^2$

6 0

3 years ago

PLEASE HELP ASAP WILL GIVE BRAINLIEST

stealth61 [152]

Explanation

(m) is measured in kilograms (kg)

<h2>(F) is measured in newtons (N)</h2>

<h3>acceleration (a) is measured in metres per second squared (m/s²)</h3>

4 0

2 years ago

Read 2 more answers