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3 years ago

13

If the distance separating the objects is the same for each pair which pair will have the greatest gravitational force between T

hem

1 answer:

hjlf3 years ago

7 0

Answer:

ur 2 sqyares out of the cicle. how many squares to get back in for example.

Explanation:

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A current of 17.0 mA is maintained in a single circular loop of 2.00 \mathrm{~m} circumference. A magnetic field of 0.800T is di

poizon [28]

The magnetic moment is -

M = 5.5 x $10^{-3}$ A $m^{2}$ .

We have current carrying single circular loop placed in a magnetic field

parallel to the plane of the loop.

We have to determine the Magnetic moment of the loop.

<h3>What is the formula to calculate the magnetic moment of loop?</h3>

The formula to calculate the magnetic moment of the loop is -

M = NIA

where -

N - Number of turns

I - Current in the loop

A - Area of the loop

According to the question, we have -

I = 17mA = 0.017A

Circumference (C) = 2 meters

B = 0.8 T

Now -

C = 2

2 $\pi$ r = 2

$\pi$ r = 1

r = $\frac{1}{\pi }$

r = 0.32

Using the formula -

M = NIA

M = $\pi$ NI $r^{2}$

M = 3.14 x 1 x 0.017 x 0.32 x 0.32

M = 3.14 x 4 x 0.017

M = 5.5 x $10^{-3}$ A $m^{2}$

Hence, the magnetic moment is -

M = 5.5 x $10^{-3}$ A $m^{2}$

To solve more questions on Magnetic moments, visit the link below-

brainly.com/question/13933686

#SPJ4

3 0

1 year ago

Determine the moment of inertia Ixx of the mallet about the x-axis. The density of the wooden handle is 860 kg/m3 and that of th

Yuki888 [10]

Complete Question

Diagram for this shown on the first uploaded image

Answer:

The moment of inertia Ixx of the mallet about the x-axis is $I{xx}= 0.119 kg \cdot m^2$

Explanation:

From the question we are told that

The density `of wooden handle is $\rho_w = 860 kg/m^3$

The density `of soft-metal head is $\rho_s =8000kg/m^3$

Generally the mass of the wooden can be mathematically obtained with this formula

$m_w = \rho_w A_w l_w$

Where $A_w$ is mass of wooden handle which is mathematically obtain with the formula

$A_w = \frac{\pi}{4} d^2_w$

Where $d_w$ is the diameter of the wooden handle which from the diagram is

$27mm = \frac{27}{1000} = 0.027m$

So $A_w = \frac{\pi}{4} * 0.027^2$

$l_w$ is the length of the the wooden handle which is given in the diagram as $l_w = 315mm = \frac{315}{1000} = 0.315m$

Substituting these value into the formula for mass

$m_w = 860 * (\frac{\pi}{4} * 0.027^2 ) *0.315$

$= 0.155kg$

Generally the mass of the soft-metal head can be mathematically obtained with this formula

$m_s = \rho_s A_s l_s$

Where $A_s$ is mass of soft-metal head which is mathematically obtain with the formula

$A_s = \frac{\pi}{4} d^2_s$

Where $d_s$ is the diameter of the soft-metal head which from the diagram is

$36mm = \frac{36}{1000} = 0.036m$

So $A_s = \frac{\pi}{4} * 0.036^2$

$l_s$ is the length of the the soft-metal head which is given in the diagram

as $l_s = 90mm = \frac{90}{1000} = 0.090m$

Substituting these value into the formula for mass

$m_s = 8000 * (\frac{\pi}{4} * 0.036^2 ) *0.090$

$=0.733kg$

Generally the mass moment of inertia about x-axis for the wooden handle is

$(I_{xx})_w = [\frac{1}{3}m_w + l_w^2 ]$

Substituting values

$(I_{xx})_w = [\frac{1}{3}*0.155 + 0.315^2 ]$

$=5.12*10^{-3}kg \cdot m^2$

Generally the mass moment of inertia about x-axis for the soft-metal head is

$(I_{xx})_s = [\frac{1}{12}m_s l_s ^2 + b^2]$

Where b is the distance from the centroid to the axis of the head which is mathematically given as

$b=l_w +\frac{d_s}{2}$

Substituting values

$b = 0.315 + \frac{0.036}{2}$

$= 0.336m$

Now substituting values into the formula for mass moment of inertia about x-axis for soft-metal head

$(I_{xx})_s = [\frac{1}{12} *0.733* 0.090^2 + 0.336^2]$

$=0.113 kg \cdot m^2$

Generally the mass moment of inertia about x-axis is mathematically represented as

$I_{xx} = (I_{xx})_w + (I_{xx})_s$

$= [\frac{1}{3}m_w + l_w^2 ] + [\frac{1}{12}m_s l_s ^2 + b^2]$

Substituting values

$I_{xx} = 5.12*10^{-3} +0.113$

$I{xx}= 0.119 kg \cdot m^2$

8 0

3 years ago

What is dark energy?

olga2289 [7]

Explanation:

Dark Energy. Dark Energy is a hypothetical form of energy that exerts a negative, repulsive pressure, behaving like the opposite of gravity. It has been hypothesised to account for the observational properties of distant type Ia supernovae, which show the universe going through an accelerated period of expansion

6 0

3 years ago

A gull is flying horizontally 10.80 m above the ground at 6.00 m/s. The bird is carrying a clam in its beak and plans to crack t

UkoKoshka [18]

Answer:

v_y = 14.55 m/s

Explanation:

given,

height at which gull is flying = 10.80 m

speed of the gull = 6 m/s

acceleration due to gravity = 9.8 m/s²

Relative to the seagull, the x-speed is 0,

because the seagull has the same x-speed.

Only the y-speed counts:

$v_y^2 = u^2 + 2 g h$

$v_y^2 = 0^2 + 2 g h$

$v_y = \sqrt{2gh}$

$v_y = \sqrt(2\times 9.8 \times 10.8)$

$v_y = \sqrt(211.68)$

v_y = 14.55 m/s

hence, the speed at which the clam smash the rock is v_y = 14.55 m/s

8 0

3 years ago

Please show work : A particle with mass 2.00 μg and a charge of – 200 nC has a velocity of 3000 m/s in the x-direction. There is

irga5000 [103]

Answer:

x =4.5 10⁴ m

Explanation:

To find the distance that the particle moves we must use the equations of motion in one dimension and to find the acceleration of the particle we will use Newton's second law

m = 2.00 mg (1 g / 1000 ug) (1 Kg / 1000g) = 2.00 10-6 Kg

q = -200 nc (1C / 10 9 nC) = -200 10-9 C

Let's calculate the acceleration

F = ma

F = q E

a = qE / m

a = -200 10⁻⁹ 1000 / 2.00 10⁻⁶

a = 1 10² m / s²

Let's use kinematics to find the distance traveled before stopping, where it has zero speed (Vf = 0)

Vf² = Vo² -2 a x

0 = Vo² - 2 a x

x = Vo² / 2a

x = 3000²/ 2100

x =4.5 10⁴ m

This is the distance the particule stop, after this distance in the field accelerates in the opposite direction of the initial

Second part

In this case Newton's second law is applied on the y axis

F -W = 0

F = w = mg

E q = mg

E = mg / q

E = 2.00 10⁻⁶ 9.8 / 200 10⁻⁹

E = 9.8 10⁵ C

The direction of the field is such that the force on the particle is up, as the particle has a negative charge, the field must be directed downwards F = qE = (-q) E

7 0

4 years ago