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3 years ago

Name 4 COARSE grained rocks

Physics

2 answers:

MrRa [10]3 years ago

8 0

Answer:

Diorite, Gabbro, and Granite

Explanation:

frosja888 [35]3 years ago

6 0

Granite gabbro diorite and igneous

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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

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

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

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

you know that there are 1609 meter in a mile. the number of feet in a mile is 5280. what is the speed snail from problem 7 per m

Zanzabum

Answer:

We know that 1 meter = 100 centimeters, and 1 foot = 12 inches.

So (1,609 meters) x (100 centimeters/meter) = (5,280 feet) x (12 inches/foot)

The second fraction on each side of the equation is equal to ' 1 ', because

the numerator is equal to the denominator, so sticking it in there doesn't

change the value of that side of the equation. But now we can cancel some

units,and wind up with the units we need.

(1,609 meters) x (100 centimeters/meter) = (5,280 feet) x (12 inches/foot)

(1,609 x100) centimeters = (5,280 x 12) inches

160,900 centimeters = 63,360 inches

Divide each side by 63,360 : 2.54 centimeters = 1 inch

Explanation:

5 0

2 years ago

A ball traveling at 15 m/s hits a bat with a force of 200N. How much force does the bat (moving at 20m/s)

just olya [345]

Answer:

200 N

Explanation:

Given that,

A ball traveling at 15 m/s hits a bat with a force of 200 N.

We need to find the force that the bat moving at 20 m/s hit the ball with.

We know that, this probelm is based on Newton's third law of motion. The force that the ball exerting on bat should be equal to the force that the bat exerting in the ball but in opposite direction.

It would mean that the ball hits the ball with a force of 200 N. Hence, the correct option is (a).

8 0

4 years ago

When a 5.0kg cart undergoes a 2.2m/s increase in speed, what is the impulse of the cart

Karolina [17]

Answer:

11.0 kg m/s

Explanation:

The impulse exerted on the cart is equal to its change in momentum:

$I=\Delta p=m\Delta v$