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oee [108]
3 years ago
11

I Need help with this problem i don’t know what to do

Physics
1 answer:
Mazyrski [523]3 years ago
8 0

Answer:

The density of the sample is 36 g/cm³

Explanation:

m= 972g

l=3cm

V = l³ = 3³ = 27 cm³

density = mass/volume

= 972/27

= 36 g/cm³

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Explanation:

A common theory suggests that desert pavements are formed through gradual removal of sand and other fine particles by the wind and intermittent rains leaving behind the large fragments. The larger rock particles are shaken into place by actions of different agents such as rain, wind, gravity, and animals

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Most split-brain surgeries are preformed on ___________. A. neurology patients B. intractable epilepsy patients C. abnormal psyc
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B. Intractable epilepsy patients
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You are leaving for a party at your cousin’s house in a city that is 150 km away. You will travel at an average rate of 50 km/hr
choli [55]
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A rock is dropped from the edge of a cliff into a pool of water. Assume free-fall acceleration is 10 m/s per second, and air res
il63 [147K]

Answer:

20 m

Explanation:

From the equation of motion,

S = ut+1/2gt²................................. Equation 1

Where S = Height, u = initial velocity, t = time, g = acceleration due to gravity.

Note: Because the rocked is being dropped from a height, acceleration due to gravity is positive (g), and initial velocity (u) is negative

Given: t = 2.0 s, g = 10 m/s², u = 0 m/s (dropped from height)

Substituting into equation 1

S = 0(2) + 1/2(10)(2)²

S = 5(4)

S = 20 m

Hence the height of the the cliff above the pool is 20 m

7 0
3 years ago
Consult interactive solution 2.22 before beginning this problem. a car is traveling along a straight road at a velocity of +30.0
Inessa05 [86]

Let a_1 be the average acceleration over the first 2.46 seconds, and a_2 the average acceleration over the next 6.79 seconds.

At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let v be the velocity of the car after the first 2.46 seconds.

By definition of average acceleration, we have

a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}

and we're also told that

\dfrac{a_1}{a_2}=1.66

(or possibly the other way around; I'll consider that case later). We can solve for a_1 in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:

1.66a_2=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

\implies1.66\left(\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}\right)=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

Now we can solve for v. We find that

v=20.8\,\dfrac{\mathrm m}{\mathrm s}

In the case that the ratio of accelerations is actually

\dfrac{a_2}{a_1}=1.66

we would instead have

\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}=1.66\left(\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}\right)

in which case we would get a velocity of

v=24.4\,\dfrac{\mathrm m}{\mathrm s}

6 0
3 years ago
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