Hey I need help can someone help me out, please

Runs 4 m North and then runs 9 m West. Calculate Distance and Displacement.

mixas84 [53]

Answer:

Take north as y and west as -x.

distance=√(x^2+y^2)

displacement= tan theta(tan^-1) (y/x)

answers should be, (9.85m, 66.04°)

8 0

2 years ago

The magnetic field strength at the north pole of a 2.0-cm-diameter, 8-cm-long Alnico magnet is 0.10 T. To produce the same field

frosja888 [35]

Answer:

N = 3032 turns

Explanation:

The magnetic field produced by a solenoid is described by

B = μ₀ n I

Where is the permittivity in a vacuum with a value of 4π 10⁻⁷ N /A², n is the turn density and I the current

Let's apply this equation to the problem, the turn density is the number of turns per unit length, in this case it is the same magnet length

L = 8 cm = 0.08 m

Let's calculate

B = μ₀ N/L I

N = B L / μ₀ I

N = 0.10 0.08 / (4π 10⁻⁷ 2.1)

N = 3,032 103 turns

4 0

3 years ago

A loop of area 0.250 m^2 is in a uniform 0.020 0-T magnetic field. If the flux through the loop is 3.83 × 10-3T· m2, what angle

dangina [55]

Answer:

40.0⁰

Explanation:

The formula for calculating the magnetic flux is expressed as:

$\phi = BAcos\theta$ where:

$\phi$ is the magnetic flux

B is the magnetic field

A is the cross sectional area

$\theta$ is the angle that the normal to the plane of the loop make with the direction of the magnetic field.

Given

A = 0.250m²

B = 0.020T

$\phi$ = 3.83 × 10⁻³T· m²

3.83 × 10⁻³ = 0.020*0.250cosθ

3.83 × 10⁻³ = 0.005cosθ

cosθ = 0.00383/0.005

cosθ = 0.766

θ = cos⁻¹0.766

θ = 40.0⁰

<em>Hence the angle normal to the plane of the loop make with the direction of the magnetic field is 40.0⁰</em>

4 0

2 years ago

The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas

Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 $m,/sec^2$

(c) Escape velocity on earth is $3.563\times 10^4m/sec$

Explanation:

We have given radius of mars $R_{mars}=6.9\times 10^3km=6.9\times 10^6m$ and radius of earth $R_{E}=1.3\times 10^4km=1.3\times 10^7m$

Mass of earth $M_E=5.972\times 10^{24}kg$

So mass of mars $M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg$

Volume of mars $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3$

So density of mars $d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3$

Volume of earth $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3$

So density of earth $d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3$

(A) So the ratio of mean density $\frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735$

(B) Value of g on mars

g is given by $g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2$

(c) Escape velocity is given by

$v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec$

5 0

3 years ago

Read 2 more answers

During the process of psychotherapy, Elaine recovered some long-forgotten and painful memories from her childhood. This experien

Alekssandra [29.7K]

In psychoanalytic therapies, this gives credence and value to an individual's childhood experiences and the unconscious. This would suggest that childhood experiences are highly involve in later life.

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6 0

3 years ago