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-Dominant- [34]

3 years ago

12

A man lifts a 25.9 kg bucket from a well and does 5.92 kJ of work.The acceleration of gravity is 9.8 m/s^2 . How deep is the wel

l? Assume that the speed of the bucket remains constant as it is lifted. Answer in units of m.

1 answer:

Jet001 [13]3 years ago

4 0

Answer:

The well is 23.3 m

Explanation:

As the bucket is lifted out of the well, energy in the man is being transferred to the bucket as gravitational potential energy.

Work done against gravity = mass * height * acceleration due to to gravity

W = mgh

5 920 J = 25.9 kg * h * 9.8 m/s²

h = 23.3 m

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The density of a glass is 2.5 g/cm3. if the glass breaks, what is the density of the smaller pieces

sp2606 [1]

Answer:

See the explanation below.

Explanation:

Density will remain the same since density is the relationship between mass and volume. As we can see in the equation below.

$Ro=m/V$

where:

Ro = density = 2.5 [g/cm³]

m = mass [g]

V = volume [cm³]

In such a way that when the glass is broken the small fragments retain the same density ratio. That is, each fragment has a small mass and a small volume. That's why the density remains the same.

4 0

3 years ago

To test the performance of its tires, a car

velikii [3]

<u>Answer</u>:

The coefficient of static friction between the tires and the road is 1.987

<u>Explanation</u>:

<u>Given</u>:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

<u>To Find:</u>

The coefficient of static friction between the tires and the road = ?

<u>Solution</u>:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

8 0

3 years ago

A sample of a gas has a volume of 639 cm3 when the pressure is 75.9 kPa. What is the volume of the gas when the pressure is incr

const2013 [10]

Answer:

$388 cm^3$

Explanation:

For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:

$pV=const.$

which can also be rewritten as

$p_1 V_1 = p_2 V_2$

In our case, we have:

$p_1 = 75.9 kPa$ is the initial pressure

$V_1 = 639 cm^3$ is the initial volume

$p_2 = 125 kPa$ is the final pressure

Solving for V2, we find the final volume:

$v_2 = \frac{p_1 V_1}{p_2}=\frac{(75.9)(639)}{125}=388 cm^3$

7 0

3 years ago

What investigations are best for demonstrating cause and effect relationship

poizon [28]

The cause would be how the idea started or developed and the effect would be how they both effected you together

8 0

3 years ago

Read 2 more answers

A train whose proper length is 1200 m passes at a high speed through a station whose platform measures 900 m, and the station ma

TiliK225 [7]

Answer:

0.66c

Explanation:

Use length contraction equation:

L = L₀ √(1 − (v²/c²))

where L is the contracted length,

L₀ is the length at 0 velocity,

v is the velocity,

and c is the speed of light.

900 = 1200 √(1 − (v²/c²))

3/4 = √(1 − (v²/c²))

9/16 = 1 − (v²/c²)

v²/c² = 7/16

v = ¼√7 c

v ≈ 0.66 c

6 0

3 years ago