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3 years ago

Maritime climates can be expected in the middle of large continents. True False

Physics

1 answer:

kap26 [50]3 years ago

5 0

False maritime climates are found near the ocean so it wouldn’t make sense to be in the middle

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A helicopter flies over the arctic ice pack at a constant altitude, towing an airborne 130-kg laser sensor that measures the thi

umka21 [38]

Answer:T=1316.21 N

Explanation:

The tension has two components: Vertical and Horizontal. The

horizontal component is ma, the vertical component is mg. Using

Pythagoras theorem, we can find the tension as:

T=((ma)^2 (mg)^2)^(1/2)

T=((129*2.84)^2 (129*9.8)^2)^(1/2)

T=1316.21 N

8 0

3 years ago

I forgot this someone help

dexar [7]

Answer:

A. water + carbon dioxide + sunlight = oxygen + <u>glucose </u><u>!</u>

6 0

2 years ago

What did you notice about the angle of incidence and the angle of reflection?

Readme [11.4K]

Answer:

They are equal

Explanation:

angle of incidence = angle of reflection

6 0

2 years ago

The arrows in the chart below represent phase transitions.

Vikentia [17]

Answer:

1, 2, and 3.

Explanation:

Hello.

In this process, since the phase transitions that require energy are those that pass from a state with less energy or more molecular order to a state with more energy or less molecular order, say, from solid to liquid (melting), from liquid to gas (boiling) and from solid to gas (sublimation), we can conclude that the arrows representing heat energy gained are 1, 2, and 3 since 1 represents boiling, 2 melting and 3 sublimation.

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6 0

3 years ago

A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24

alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

$I=0.8*m*r^{2}$

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

$N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}$

$m=40kg$

moment of inertia

$M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}$

Kinetic energy of the rotation motion

$K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J$

Kinetic energy translational

$K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J$

Total kinetic energy

$K=3317.79J+4147.2J\\K=7464.99J$

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

$K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m$

6 0

4 years ago