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2 years ago

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47. What is necessary for differentiation to occur in a planet? a. It must be made of metal and rock. b. It must be made of a mi

x of materials of different density. c. Material inside must be able to flow. d. all of the above e. B and C

1 answer:

Arlecino [84]2 years ago

3 0

Answer:

e. B and C

Explanation:

When the sun ignited in our solar system evaporation of the material closest to the sun took place. The low density material were pushed further. The rocks that remained formed the protoplanets. As, the early solar system was very hot parts of the protoplanets melted. This caused the melted material to flow towards the center of the planet. As the mass of the planet increases this causes more accretion which causes materials of different densities to mix.

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A uniform rectangular plate is hanging vertically downward from a hinge that passes along its left edge. By blowing air at 11.0

Serjik [45]

Answer:

The airspeed must be 7.78 m/s for the rectangular plate kept at 30°.

Explanation:

By looking at the images below wee see that the airspeed on one side of the rectangular plate decreases the statical pressure over this side. Since over the downside, the pressure still bein the atmospheric pressure. This difference in pressure produces a lift force in the plate. The list force is the net force obtained between the difference of the forces that produce the pressure over the upside and the downside:

$F_{lift}=F_{up} - F_{dw}=0.5*p*V^2$

Where up and down relate to what movement the forces produce. And p and V are the respective air density and velocity.

When the plate is kept horizontal the lift force balance the moment due to the weight of the plate and considering that both forces act at the same point:

$F_{lift}=0.5*p*V^2=W$

By replacing the known values it is possible to find the plate's weight:

$F_{lift}=0.5*1.2 \frac{kg}{m^{3}}*(11 m/s)^2=W$

$W=72.6 N$

When the plate kept to 30° from the vertical the moment equation balance is written as:

$F_{lift}=0.5*p*V^2=W*sen(30\°)$

The sine of 30° is due to the weight is 30° oriented, therefore the new value for the airspeed is:

$V=\sqrt(W*sen(30\°)/0.5p)$

$V=\sqrt(\frac{72.6 N * 0.5}{0.5*1.2 kg/m^3})$

$V=\sqrt(60.5 \frac{N}{kg/m^3})$

$V=\sqrt(60.5 \frac{kg.m/s^2}{kg/m^3})$

$V=\sqrt(60.5 \frac{m^2}{s^2})$

$V= 7.78 m/s$

7 0

2 years ago

If you stand next to a wall on a frictionless skateboard and push the wall with a force of 40n, how hard does the wall push on y

pentagon [3]

The wall pushes back with the same force you exert so 40N.

to show your acceleration is .5m/s^2 use newtons second law F=ma

so plugging in numbers gives 40N=80kg*a knowing that a newton is equal 1kgm/s^2 we could write 40kg*m/s^2= 80kg*a so solving for a gives

40kg*m/s^2/80kg = a we see the kg's cancel and we're left with
(40m/s^2)/80=a which gives .5m/s^2=a

4 0

2 years ago

Car 1 goes around a level curve at a constant speed of 65 km/h . The curve is a circular arc with a radius of 95 m . Car 2 goes

Arte-miy333 [17]

Answer:

The radius of the curve that Car 2 travels on is 380 meters.

Explanation:

Speed of car 1, $v_1=65\ km/h$

Radius of the circular arc, $r_1=95\ m$

Car 2 has twice the speed of Car 1, $v_2=130\ km/h$

We need to find the radius of the curve that Car 2 travels on have to be in order for both cars to have the same centripetal acceleration. We know that the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

According to given condition,

$\dfrac{v_1^2}{r_1}=\dfrac{v_2^2}{r_2}$

$\dfrac{65^2}{95}=\dfrac{130^2}{r_2}$

On solving we get :

$r_2=380\ m$

So, the radius of the curve that Car 2 travels on is 380 meters. Hence, this is the required solution.

4 0

2 years ago

Gbenga needs to get glasses to correct his farsightedness. His eyes currently cannot focus on objects that are within 2 ft (or 6

Scilla [17]

Answer:

Focal Length = 38.61cm, Power = 2.59 Diopter, Converging lens.

Explanation:

When an object is placed 25cm from Gbenga's eye, the glasses lens must produce an image 61cm away (Gbenga's eye near point).

An image 61cm from the eye will be (61cm - 1.6cm) from the glasses.

i.e. $d_{i}=61cm-1.6cm=59.4cm$

and $d_{o} = 25cm - 1.6cm = 23.4cm$

note $d_{i}$ will be negative because the image is formed on the same side as the object.

finally, $d_{i}=-59.4cm\\d_{o}=23.4cm$

the formula for finding the focal length $f$ is given as

$f=\frac{d_{i}d_{o} }{d_{o}+d_{i} }$

$f=\frac{-59.4*23.4}{23.4-59.4} \\$

$f=\frac{-1389.96}{-36}$

$f=38.61cm$

The focal length is positive which indicates converging lens

power $p=\frac{1}{f}$

but $f\\$ must be in metres

Therefore, $f=38.61cm=0.3861m$

$p=\frac{1}{0.3861}$

$p=2.59 Diopter$

3 0

2 years ago

A refrigerator with a weight of 1,127 newtons needs to be moved into a house using a ramp. The length of the ramp is 2.1 meters,

lina2011 [118]

496/1127 = 0.44 = 44%

<span>sin A = 0.85/2.1. </span>
<span>A = 23.9o. </span>

<span>Fp = 1127 sin23.9 = 457 N. = Force parallel to the ramp. </span>

<span>Fn = 1127 Cos23.9 = 1,030 N. = Force </span>
<span>perpendicular to the ramp = Normal force. </span>

<span>Eff. = Fp/Fap = 457/496 = 0.92 = 92%
Correct answer: 92%</span>

3 0

2 years ago

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