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3 years ago

An airplane cabin is pressurized to 570 mmhg. what is the pressure inside the cabin in atmospheres?

2 answers:

8 0

1 atm corresponds to 760 mmHg, so we can set up a simple proportion to find how many atmospheres correspond to 570 mmHg:
$1 atm: 760 mmHg = x: 570 mmHg$
and from this, we find
$x= \frac{1 atm \cdot 570 mmHg}{760 mmHg} =0.75 atm$

OLEGan [10]3 years ago

7 0

Answer:

0.75 atmospheric pressure

Explanation:

Pressure, P = 570 mm of Hg

We know that 1 atmospheric pressure = 760 mm of Hg

So, 760 mm of Hg pressure = 1 atmospheric pressure

1 mm of Hg = 1 / 760 atm pressure

570 mm of Hg = 570 / 760 atm

= 0.75 atm

Thus, 570 mm of Hg pressure is equal to 0.75 atmospheric pressure.

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A small wooden block with mass 0.750 kg is suspended from the lower end of a light cord that is 1.72 m long. The block is initia

Ierofanga [76]

Answer:

$v_{0}=319.2 m/s$

Explanation:

We need to use the momentum and energy conservation.

$p_{0}}=p_{f}$

$mv_{0}=(m+M)V_{1}$

Where:

m is the mass of bullet (m=0.01 kg)
M is the mass of the wooden (M=0.75 kg)
v(0) initial velocity of bullet
V(1) is the velocity of the combined object in the moment the bullet hist the block

Conservation of energy.

We have kinetic energy at first and kinetic and potential energy at the end.

$(1/2)(m+M)V_{1}^{2}=(1/2)(m+M)V_{2}^{2}+(m+M)gh$

Here:

V(1) is the velocity of the combined at the initial position
h is the vertical height ( h = 0.800 m)

We can find V(2) using the definition of force at this point:

$\Sigma F=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)$

$T-(m+M)gcos(\theta)=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)$

$cos(\theta) =(L-h)/L=(1.72-0.8)/1.72$

$\theta =57.66$

Now, we can solve the equation to find V(2)

$V_{2}=\sqrt{\frac{R*(T-(m+M)*g*cos(\theta))}{(m+M)}}$

$V_{2}=\sqrt{\frac{1.72*(4.86-(0.01+0.75)*9.81*cos(57.66))}{(0.01+0.75)}}$

$V_{2}=1.40 m/s$

Now we can find V(1) using the conservation of energy equation

$(1/2)V_{1}^{2}=(1/2)V_{2}^{2}+gh$

$V_{1}=\sqrt{V_{2}^{2}+2gh}$

$V_{1}=\sqrt{1.40^{2}+2*9.81*0.8}$

$V_{1}=4.20 m/s$

Finally, using the momentum equation we find v(0)

$v_{0}=\frac{(m+M)V_{1}}{m}$

$v_{0}=\frac{(0.01+0.75)*4.20}{0.01}$

$v_{0}=319.2 m/s$

I hope it helps you!

7 0

4 years ago

9. A 0.500 kg ball with a speed of 4.0 m/s strikes a stationary 1.0 kg target. The ball and target

Virty [35]

Answer:

D. 2.0 kg m/s

Explanation:

Momentum after = momentum before

p = (0.500 kg) (4.0 m/s) + (1.0 kg) (0 m/s)

p = 2.0 kg m/s

4 0

4 years ago

A boy is pulling a wagon with a force of 70.0 N directed at an angle of 30 degrees above the horizontal. What are the x and y co

bogdanovich [222]

Answer:

x component is 60.60 N and y component is 35 N ⇒ answer A

Explanation:

A boy is pulling a wagon with a force of 70.0 N directed at an angle

of 30 degrees above the horizontal then,

$F_{x}$ is the horizontal component of the force in the direction

of positive part of x-axis

$F_{y}$ is the vertical component of the force force in the direction

of positive part of y-axis

The x component of the force $F_{x}$ is the force multiplied by

cos(30)°

The y component of the force $F_{y}$ is the force multiplied by

sin(30)°

→ The x component $F_{x}$ = F cos(30)

→ The x component $F_{x}$ = 70 cos(30) = 60.60 N

→ The y component $F_{y}$ = F sin(30)

→ The y component $F_{y}$ = 70 sin(3) = 35 N

<em>x component is 60.60 N and y component is 35 N</em>

7 0

4 years ago

Why do marine west coast climates have much precipitation

denis-greek [22]

The fronts of the water create clouds so it has a lot of precipitation

4 0

3 years ago

The law of conservation of matter states that during a chemical reaction, the amount of matter

mixas84 [53]

Answer:

The law of conservation of matter says that in chemical reactions, the total mass of the products must equal the total mass of the reactants.

7 0

3 years ago