All categories

4 years ago

The volume of an object is the amount of space it takes up

Physics

2 answers:

zimovet [89]4 years ago

8 0

Yes, the amount of space taken up, or occupied, by an object is known as the Objects volume.

soldier1979 [14.2K]4 years ago

6 0

Yes it is. Uh huh, uh huh, shore enuff. Mmm hmm. Yeah yeah yeah. Yah Mon ! Indubitably.

You might be interested in

A horizontal spring required a force of 1.0 N to compress it 0.1 m. How much work is required to stretch the spring 0.4 m?

Crank

Answer:

Explanation:

given data

Force applied F= 1N

extension e= 0.1m

let us find the spring constant first

applying

F=ke

k=F/e

k=1/0.1

k=10N/m

Step two:

Required is the work done

we know that the expression/formula for the work done by a spring is given as

Wd=1/2kx^2

x=0.4m

substitute

Wd= 1/2*10*0.4^2

Wd=0.5*10*0.16

Wd=0.5J

3 0

3 years ago

Three identical very dense masses of 7500 kg each are placed on the x axis. One mass is at x1 = -100 cm , one is at the origin,

sukhopar [10]

Answer:

0.00354 (N)

Explanation:

Convert to metric system:

$x_1 = -100 cm = 1 m$

$x_2 = 420 cm = 4.2 m$

Formula for gravitational force:

$F_g = G\frac{mM}{s^2}$

where s is the distance between 2 bodies masses m and M

Substitute the number to the formula above and since the 2 forces are acting in opposite direction, the total net gravitational force on the mass of origin be:

$F_g = F_{g1} - F_{g2}$

$F_g = G\frac{m_1M}{x_1^2} - G\frac{m_2M}{x_2^2}$

$F_g = GM(\frac{m_1}{x_1^2} - \frac{m_2}{x_2^2})$

$F_g = 6.67*10^{-11} * 7500 (\frac{7500}{1^2} - \frac{7500}{4.2^2})$

$F_g = 5*10^{-7}(7500 - 425.17)$

$F_g = 5*10^{-7} * 7074.83$

$F_g = 0.00354 (N)$

5 0

3 years ago

A particle initially located at the origin has an acceleration of = 2.00ĵ m/s2 and an initial velocity of i = 8.00î m/s. Find (a

sergejj [24]

a. The particle has position vector

$\vec r(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)t\right)\,\vec\imath+\left(\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\right)\,\vec\jmath$

$\vec r(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath$

b. Its velocity vector is equal to the derivative of its position vector:

$\vec v(t)=\vec r'(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath$

c. At $t=7.00\,\mathrm s$ , the particle has position

$\vec r(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(7.00\,\mathrm s)\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)^2\,\vec\jmath$

$\vec r(7.00\,\mathrm s)=\left(56.0\,\vec\imath+49.0\,\vec\jmath\right)\,\mathrm m$

That is, it's 56.0 m to the right and 49.0 m up relative to the origin, a total distance of $\|\vec r(7.00\,\mathrm s)\|=\sqrt{(56.0\,\mathrm m)^2+(49.0\,\mathrm m)^2}=74.4\,\mathrm m$ away from the origin in a direction of $\theta=\tan^{-1}\dfrac{49.0\,\mathrm m}{56.0\,\mathrm m}=41.2^\circ$ relative to the positive $x$ axis.