All categories

3 years ago

PLZ ANSWER ASAP!!!!!!!

Physics

1 answer:

jenyasd209 [6]3 years ago

4 0

Explanation:

It is given that,

Mass of platinum bar, m = 750 g

Length of the bar, l = 5 cm

Breadth of the bar, b = 4 cm

Width of the bar, h = 1.5 cm

Volume of the bar, $V=l\times b\times h$

$V=5\times 4\times 1.5=30\ cm^3$

We need to find the density of the platinum bar. The density of any substance is given by :

$d=\dfrac{m}{V}$

$d=\dfrac{750\ g}{30\ cm^3}$

$d=25\ g/cm^3$

So, the density of the platinum bar $25\ g/cm^3$ . Hence, this is the required solution.

You might be interested in

An Elevator of mass 200 kg travels upwards at constant velocity. What is the tension in the cables?

I am Lyosha [343]

The tension in the cables as the elevator travel upwards is 1,960 N.

The given parameters:

Mass of the elevator, m = 200 kg

<h3>Newton's second law of motion;</h3>

Newton's second law of motion states that the force applied to an object is directly proportional to the product of mass and acceleration of the object.

The tension in the cables as the elevator travel upwards is calculated by applying Newton's second law of motion as shown below;

T = ma + mg

where;

a is the acceleration of the elevator
g is the acceleration due to gravity

At constant velocity, acceleration is zero (a = 0)

T = m(0) + mg

T = mg

T = 200 x 9.8

T= 1,960 N

Thus, the tension in the cables as the elevator travel upwards is 1,960 N.

Learn more about Newton's second law here: brainly.com/question/3999427

8 0

2 years ago

As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis

klio [65]

Hi there!

We can begin by finding the acceleration of the block.

Use the kinematic equation:

$d = v_0t + \frac{1}{2}at^2$

The block starts from rest, so:

$d = \frac{1}{2}at^2\\\\12 = \frac{1}{2}a(4^2)\\\\\frac{24}{16} = a = 1.5 m/s^2$

Now, we can do a summation of forces of the block using Newton's Second Law:

$F = ma = m_bg - T$

mb = mass of the block

T = tension of string

Solve for tension:

$T = m_bg - ma = 8.2(9.8) - 8.2(1.5) = 68.06 N$

Now, we can do a summation of torques for the wheel:

$\Sigma \tau = rF\\\\\Sigma\tau = rT$

Rewrite:

$I\alpha = rT$

We solved that the linear acceleration is 1.5 m/s², so we can solve for the angular acceleration using the following:

$\alpha = a/r\\\\\alpha = 1.5/.42= 3.57 rad/sec^2$

Now, plug in the values into the equation:

$I(3.57) = (0.42)(68.06)\\\\I = (0.42)(68.06)/(3.57) = \boxed{8.00 kgm^2}$

8 0

3 years ago

Which statements describe the sun? Check all that apply

lesya692 [45]

Answer:

All but 4 I believe

Explanation:

8 0

3 years ago

Read 2 more answers

HELP PLS!!!! Light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is th

maw [93]

You've already told us the speed in ft/s . It's right there in the question. You said that light travels about 982,080,000 ft/s.

We don't know how accurate that number is, but for purposes of THIS question, that's the number we're going with.

In scientific notation, it's written . . . <em>9.8208 x 10⁸ ft/s .</em>

We don't know where you were going with the number of seconds in a year. But to answer the question that you eventually asked, it turned out that we don't even need it.

6 0

3 years ago

b) A satellite is in a circular orbit around the Earth at an altitude of 1600 km above the Earth's surface. Determine the orbita

asambeis [7]

Explanation:

The orbiting period of a satellite at a height h from earth' surface is

T=2πr32gR2

where r=R+h.

Then, T=2π(R+h)R(R+hg)−−−−−−−−√

Here, R=6400km,h=1600km=R/4

T=2πR+R4−−−−−−√R(R+R4g)−−−−−−−−−⎷=2π(1.25)32Rg−−√

Putting the given values,

T=2×3.14×(6.4×106m9.8ms−2)−−−−−−−−−−−−√(1.25)32=7092s=1.97h

Now, a satellite will appear stationary in the sky over a point on the earth's equator if its period of revolution around the earthh is equal to the period of revolution of the earth up around its own axis whichh is 24h. Let us find the height h of such a satellite above the earth's suface in terms of the earth,'s radius.

Let it be nR.Then

T=2π(R+nR)R(R+nRg)−−−−−−−−−−√

=2π(Rg)−−−−−√(1+n)32

=2×3.14(6.4×106m/s9.8m/s2)−−−−−−−−−−−−−−−⎷(1+n)32

(5075s)(1+n)32=(1.41h)(1+n)32

For T=24h, we have (24h)=(1.41h)(1+n)32

or (1+n)32=241.41=17

or 1+n(17)23=6.61

or n=5.61

The height of the geostationary satellite above the earth's surface is nR=5.61×6400km=6.59×104km.

3 0

3 years ago