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3 years ago

What is the average velocity of the object between 14 and 22 seconds

Physics

2 answers:

anastassius [24]3 years ago

4 0

Use the slope equation. It’s 1-8 Meters which is 7. And it’s 22-14 seconds. So the answer is d

Natalka [10]3 years ago

4 0

Answer:

v = - 0.88 m/s

Explanation:

Average velocity is defined as the ratio of displacement and time

so here as we can see that the displacement is the change in the position of the object

initially object is placed at x = 8 m

final position of the object is x = 1

so the displacement is given as

$d = x_f - x_i$

$d = 1 - 8 = -7 m$

now the time taken by the object

$\Delta t = 22 - 14 = 8 s$

now by the formula of average velocity we know

$v = \frac{displacement}{time}$

$v = \frac{-7}{8} = - 0.88 m/s$

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How does charging by conduction occur?

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2 years ago

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What is the best definition of luminous?

vagabundo [1.1K]

Answer:

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2 years ago

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A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis

bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

$B=\frac{\mu_o I_r}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

$I_r=JA_r$

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by $A_r=\pi(r^2-R_1^2)$

The current density is given by the total area and the total current:

$J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}$

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current = 4 A

Then, the current in the wire for a distance r is:

$I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}$ (2)

You replace the last result of equation (2) into the equation (1):

$B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})$

Finally. you replace the values of all parameters:

$B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T$

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

8 0

2 years ago

Plan an experiment to measure the ideal mechanical advantage of a three-hole punch. (a) What materials would you need? (b) What

Vaselesa [24]

Answer:

A) Three hole punch and either a layered plastic or paper

B) Identify the lengths involved ,

Length of input arm / length of output arm = L1/ L2

Explanation:

<u>a) Materials involved includes :</u>

Three hole punch and either a layered plastic or paper

Identify the forces acting on the three-hole punch which are Input and output forces

Identify the points where they act

<u>B) procedures involved </u>

The mechanical advantage = output force / input force

step one: Identify the lengths involved

assuming no friction or relatively small friction \

mechanical advantage can be calculated as : Length of input arm / length of output arm = L1/ L2

7 0

2 years ago

a body weights 28N at a height of 3200km from the earth surface.What will be the gravitational force on that body if its lies on

alekssr [168]

Answer:

The object would weight 63 N on the Earth surface

Explanation:

We can use the general expression for the gravitational force between two objects to solve this problem, considering that in both cases, the mass of the Earth is the same. Notice as well that we know the gravitational force (weight) of the object at 3200 km from the Earth surface, which is (3200 + 6400 = 9600 km) from the center of the Earth:

$F_G=G\,\frac{M_E\,m}{d^2} \\28\,\,N=G\,\frac{M_E\,m}{9600000^2}$

Now, if the body is on the surface of the Earth, its weight (w) would be:

$F_G=G\,\frac{M_E\,m}{d^2} \\w=G\,\frac{M_E\,m}{6400000^2}$

Now we can divide term by term the two equations above, to cancel out common factors and end up with a simple proportion:

$\frac{w}{28} =\frac{9600000^2}{6400000^2} \\\frac{w}{28} =\frac{9}{4} \\\\ \\w=\frac{9\,*\,28}{4}\,\,\,N\\w=63\,\,N \\$

4 0

3 years ago