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3 years ago

13

The hardest known natural material is _____. jade quartz diamond talc

2 answers:

kakasveta [241]3 years ago

8 0

Its diamond i am 100% right!

Burka [1]3 years ago

4 0

The Natural material known to be the hardest is the Diamond.

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a person using a machine applies a force of 100 newton's over a distance of 10 Meters to raise a 500 n object 1.5 meters what is

Hoochie [10]

Answer:

Work done by the machine (W) = 500 × 1.5 = 750 J

Work supplied to the machine (W) = 100 × 10 = 1000 J

Here, work supplied to the machine is input work = 1000 J

7 0

3 years ago

Consider the following cyclic process carried out in two steps on a gas. Step 1: 44 J of heat is added to the gas, and 20. J of

notka56 [123]

Answer:37 J

Explanation:

Given

Step :1

Heat added Q=44 J

Work done=-20 J

$\Delta E_1=Q+W=44-20=24 J$

Step :2

Heat added Q=-61 J

work done $W_2$

$\Delta E_2=Q+W_2$

$\Delta E_2=61+W_2$

$\Delta E_1+\Delta E_2=0$

as the process is cyclic

$44-20-61+W_2=0$

$W_2=37 J$

work done in compression is 37 J

3 0

3 years ago

A pulley system has an efficiency of 74.2%. If you perform 200 J of work, how

Fudgin [204]

Answer:

C

Explanation:

If a pulley system has an efficiency of 74.2%, then only that fraction of the work performed will be useful. 74.2%=0.742. 0.742*200 is about 148J. Hope this helps!

8 0

3 years ago

When there is a change of state, such as a solid to liquid or liquid to gas, heat energy can be added without a temperature chan

Romashka-Z-Leto [24]

I’m pretty sure the answer is C. Any change of state or movement requires energy

3 0

3 years ago

Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless

drek231 [11]

Answer:

$v_f = 15 \frac{m}{s}$

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum $\vec{L}$ is

$\vec{L} = \vec{r} \times \vec{p}$

where $\vec{r}$ is the position and $\vec{p}$ the linear momentum.

We also know that the torque is

$\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt} ( \vec{r} \times \vec{p} )$

$\vec{\tau} = \frac{d}{dt} \vec{r} \times \vec{p} + \vec{r} \times \frac{d}{dt} \vec{p}$

$\vec{\tau} = \vec{v} \times \vec{p} + \vec{r} \times \vec{F}$

but, as the linear momentum is $\vec{p} = m \vec{v}$ this means that is parallel to the velocity, and the first term must equal zero

$\vec{v} \times \vec{p}=0$

so

$\vec{\tau} = \vec{r} \times \vec{F}$

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

$\vec{\tau}_{rod} = 0$

this means, for the angular momentum measure from the rod:

$\frac{d\vec{L}_{rod}}{dt} = 0$

that means :

$\vec{L}_{rod} = constant$

So, the magnitude of initial angular momentum is :

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)$

but the angle is 90°, so:

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|$

$| \vec{L}_{rod_i} | = r_i * m * v_i$

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

$| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}$

$| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}$

For our final angular momentum we have:

$| \vec{L}_{rod_f} | = r_f * m * v_f$

and the radius is 0.250 m and the mass is 2.00 kg

$| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f$

but, as the angular momentum is constant, this must be equal to the initial angular momentum

$7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f$

$v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}$

$v_f = 15 \frac{m}{s}$

8 0

3 years ago