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HACTEHA [7]
4 years ago
8

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:

<h2>0.5J</h2>

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.

d. The speed of the particle at t=7.00\,\mathrm s is the magnitude of the velocity at this time:

\vec v(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)\,\vec\jmath

\vec v(7.00\,\mathrm s)=\left(8.00\,\vec\imath+14.0\,\vec\jmath\right)\dfrac{\rm m}{\rm s}

Then its speed at this time is

\|\vec v(7.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+(14.0\dfrac{\rm m}{\rm s}\right)^2}=16.1\dfrac{\rm m}{\rm s}

4 0
3 years ago
Hooke's law describes an ideal spring. Many real springs are better described by the restoring force (F Sp ) s =−kΔs−q(Δs) 3 (p)
Montano1993 [528]

Answer with Explanation:

We are given that

Restoring force,(FS_p)s=-k\Delta s-q(\Delta s)^3

k=350N/m

q=750 N/m^3

We have to find the work must you do to compress this spring 15 cm.

\Delta s=15 cm=0.15 m

Using 1 m=100 cm

Work done=\int_{0}^{0.15}-Fd(\Delta s)

W=-\int_{0}^{0.15}(-k\Delta s-q(\Delta s)^3))d(\Delta s)

W=k[\frac{(\Delta s)^2}{2}]^{0.15}_{0}+q[\frac{(\Delta s)^4}{4}]^{0.15}_{0}

W=0.01125k+0.000127q=0.01125\times 350+0.000127\times 750

W=4.033 J

Ideal spring work=0.5k(\Delta s)^2=0.5\times 350\times (0.15)^2=3.938 J

Percentage increase in work=\frac{4.033-3.938}{3.928}\times 100=2.4%

6 0
4 years ago
The force of attraction that a -40.0 μC point charge exerts on a +108 μC point charge has magnitude 4.00 N. How far apart are th
Karolina [17]

Answer:

F = k q1 q2 / r^2

r^2 = k q1 q2 / F  = 9E9 * 4E-5 * 10.8E-5 / 4

r^2 = 9 * 4 * 10.8 / 4 * E-1 = 9.72 m^2

r = 3.12 m

8 0
3 years ago
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