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2 years ago

Can I pass 6th grade with these grades:

Physics

2 answers:

Reika [66]2 years ago

7 0

Answer:

No you can't so sorry

Explanation:

I hope this helps.

coldgirl [10]2 years ago

5 0

Answer:

You can't

Explanation:

I'm sorry but with a 25% and a 50% it's not enough. Hope this helps you out and your teachers are nice enough to work with you.

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Answers please!!!!!!!!!!!!!!!!

kaheart [24]

I only know #2 and #4.
2.) cells
3.) cells, life , existing

Sorry that i dont know the rest but i took a test on this not to long ago, and i tend to forget stuff once i take a test on it.

3 0

3 years ago

URGENT

Advocard [28]

Answer:

$give \\ mass(m) = 1350kg \\ acceleration(a) = 1ms {}^{ - 2} \ \\ sln\ \\ from \: our \: formula \\ \: f = ma \\1350kg \times 1ms {}^{ - 2} \\ f = 1350newton$

the force you applied to your car =1350N

5 0

3 years ago

A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the

andreyandreev [35.5K]

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and how far away from the overpass is the roadrunner when the coyote drops the net.

$d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s$

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

d = 35.5 m

5 0

3 years ago

The vapor pressure of ethanol at 293 K is 5.95 kPa and at 336.5 K it is 53.3 kPa. Calculate the enthalpy of vaporization of etha

denis-greek [22]

Answer:

$H=41.3kJmol^{-1}$

Explanation:

The equation relating the the enthalphy, pressure and temperature is expressed as

$ln(\frac{P_{2}}{P_{1}} )=\frac{H}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}} ) \\$

Where P is the pressure, H is the enthalphy, and T is the temperature.

since the given values are

$T_{1}=293k, \\T_{2}=336.5k\\P_{1}=5.95kPA\\p_{2}=53.3kPA\\and R=8.314J.K^{-1}mol_{-1}$

if we insert values, we arrive at

$ln(\frac{53.3}{5.95} )=\frac{H}{8.314}(\frac{1}{293}-\frac{1}{336.5} )\\2.19=\frac{H}{8.314}(0.00044)\\H=(2.19*8.314)/0.00044\\H=41,268.8Jmol^{-1}\\H=41.3kJmol^{-1}$

4 0

3 years ago

Hey guys i need help (potential energy)

Vlad1618 [11]

Explanation:

Michael should put the vase at the bottom of the shelf to reduce the potential energy because the height of the vase to the floor is nearly zero.

8 0

3 years ago