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4 years ago

What are the units of measure for distance?

1 answer:

6 0

The most common unit is meters (m for short). It is the base unit for distance or displacement in the metric system. If you are dealing with larger distances, you might use kilometers (I'm for short) which is just 1000 meters. On the other hand, centimeter (cm) are used for small distances and are 1/100 of a meter. Another common unit is millimeters (mm) which is 1/1000 of a meter.

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Arrange the objects in order from greatst to least of potential energy assume that gravity is constant

aleksley [76]

Answer:

Water > Box of books > Stone > Ball

Explanation:

We'll begin by calculating the potential energy of each object. This can be obtained as follow:

For stone:

Mass (m) = 15 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 3 m

Potential energy (PE) =?

PE = mgh

PE = 15 × 10 × 3

PE = 450 J

For water:

Mass (m) = 10 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 9 m

Potential energy (PE) =?

PE = mgh

PE = 10 × 10 × 9

PE = 900 J

For ball:

Mass (m) = 1 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 20 m

Potential energy (PE) =?

PE = mgh

PE = 1 × 10 × 20

PE = 200 J

For box of books:

Mass (m) = 25 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 2 m

Potential energy (PE) =?

PE = mgh

PE = 25 × 10 × 2

PE = 500 J

Summary:

Object >>>>>>>> Potential energy

Stone >>>>>>>>> 450 J

Water >>>>>>>>> 900 J

Ball >>>>>>>>>>> 200 J

Box of books >>> 500 J

Arranging from greatest to least, we have:

Object >>>>>>>> Potential energy

Water >>>>>>>>> 900 J

Box of books >>> 500 J

Stone >>>>>>>>> 450 J

Ball >>>>>>>>>>> 200 J

Water > Box of books > Stone > Ball

4 0

3 years ago

An object of mass m is initially at rest. After a force of magnitude F acts on it for a time T, the object has a speed v. Suppos

adell [148]

Answer:

It takes $\frac{1}{2}T$ to accelerate the object from rest to the speed v.

Explanation:

From Newton's second law:

$F=m\cdot a$ (1)

and the definition of acceleration,

$\displaystyle{a = \frac{\Delta v}{\Delta t}}$ (2)

we can solve this problem. Putting (2) in (1) we have:

$\displaystyle{F = m\cdot \frac{\Delta v}{\Delta t}}$ and solving for $\Delta t$ and considering the initial time as zero ( $t_0=0$ ) and the initial velocity also zero ( $v_0=0$ ) we have:

$\displaystyle{T=\frac{mv}{F}}$

Now, for a mass $m^*= 2m$ and the $F^*=4F$ we can wrtie the same equation:

$\displaystyle{T^*=\frac{m^*v}{F^*}}$ and substituting $m^*$ and $F^*$ :

$\displaystyle{T^*=\frac{2m\cdotv}{4F}=\frac{2}{4}T=\boxed{\frac{1}{2}T}}$

So now, it only takes half the time to accelerate the object from rest to the speed v

5 0

3 years ago

Ariel's puppy weighs 15 lb. Heather's puppy weighs 50 N. Whose puppy weighs more?

yarga [219]

15 lb is nearly 6 kg amd 50 n is nearly 5 kg,so first puppy weighs more

5 0

3 years ago

A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con

Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5 x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

$E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}$

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

$r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \ \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm$

The magnitude of the electric field is now given as;

$E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m$

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0

3 years ago

What absorb emitted neutrons?

Karolina [17]

Nuclear energy is released during: fission. radioactive decay. man-induced splitting of atoms. Match the basic components of a nuclear reactor with their descriptions. 1. slows down neutrons -> moderator. 2. absorb emitted neutrons -> control rods.

5 0

3 years ago

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