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3 years ago

When energy from the sun hits the air above

Physics

1 answer:

valentina_108 [34]3 years ago

3 0

Answer: C I think.

Explanation:

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A toy car is given a quick push so that it moves up an inclined ramp. After it is released, it moves up, reaches its highest poi

Alexeev081 [22]

Answer:

7. Net constant force down the ramp

Explanation:

After the car is released and starts moving up the ramp, the only force that is applied on the car is weight because of the gravity, we were told that the friction force is neglected. the force because of the weight is given by:

$F=m*g*sin(\theta)$

where θ is the angle of the ramp.

as you can see those values won't change, so the force remains constant down the ramp.

4 0

3 years ago

A 0.5 kg block slides down a frictionless semicircular track from a height, h, and collides with a second 1.5 kg block at the bo

Ede4ka [16]

Answer:

$h' = 0.062\cdot h$

Explanation:

The speed of the 0.5 kg block just before the collision is found by the Principle of Energy Conservation:

$U_{g} = K$

$m\cdot g \cdot h = \frac{1}{2}\cdot m \cdot v^{2}$

$v = \sqrt{2\cdot g \cdot h}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot h}$

$v \approx 4.429\cdot \sqrt{h}$

Knowing that collision is inelastic, the speed just after the collision is determined with the help of the Principle of Momentum Conservation:

$(0.5\,kg) \cdot (4.429\cdot \sqrt{h}) = (2\,kg)\cdot v$

$v = 1.107\cdot \sqrt{h}$

Lastly, the height reached by the two blocks is:

$K = U_{g}$

$\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot h'$

$h' = \frac{v^{2}}{2\cdot g}$

$h' = \frac{(1.107\cdot \sqrt{h})^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$h' = 0.062\cdot h$

3 0

3 years ago

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 49 m in front of you. You reaction time

Leno4ka [110]

Answer:

v = 26.7 m/s

Explanation:

Given,

speed of the car = 20 m/s

distance between the car and the deer = 49 m

time taken to press the brake = 0.50 s

maximum deceleration of the car = 10 m/s²

Now,

distance travel by the car in 0.5 s = u x t = 20 x 0.5 = 10 m

distance travel by the car after the break is pressed

Using equation of motion

v² = u² + 2 a s

0² = 20² - 2 x 10 x s

s = 20 m

Total distance travel by the car = 20 + 10 = 30 m

Distance between deer and car = 49-30 = 19 m.

b. Maximum speed a car could have

Distance travel by the car in reaction time = v' x 0.5

v' is the maximum speed of the car.

maximum distance car can cover = 49 - 0.5 v'

Now, Using equation of motion

v² = u² + 2 a s

0² =v'² - 2 x 10 x (49- 0.5 x v')

v'² +10 v' -980 = 0

By solving

v = 26.7 m/s

Hence, maximum speed of the car can be 26.7 m/s

4 0

4 years ago

Please help with last equation i don't understand at all!!!

GalinKa [24]

What equation?
please comment on this and type it in

5 0

3 years ago

The nucleus of an atom has a positive charge.<br>True<br>False<br>

xxTIMURxx [149]

The answer is True they have a positive charge

4 0

3 years ago

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