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2 years ago

Can someone help me name the layers.

Physics

2 answers:

BARSIC [14]2 years ago

7 0

Answer:

A. Crust

B. Mantle

C. Outer Core

D. Inner Core

Explanation:

Nadya [2.5K]2 years ago

4 0

Answer:

A. Crust

B. mantle

C. outer core

D. inner core

Explanation:

because it is

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A 0.37-kg object connected to a light spring with a force constant of 23.2 N/m oscillates on a frictionless horizontal surface.

mars1129 [50]

Answer:

a) v = 31.67 cm / s , b) v = -29.36 cm / s , c) v= 29.36 cm/s, d) x = 3.46 cm

Explanation:

The angular velocity in a simple harmonic movement is

w = √ K / m

w = √ 23.2 / 0.37

w = 7,918 rad / s

a) the expression against the movement is

x = A cos (wt + Ф)

Speed is

v = dx / dt = - A w sin (wt + Ф)

The maximum speed occurs for cos = ± 1

v = A w

v = 4.0 7,918

v = 31.67 cm / s

b) as the object is released from rest

0 = -A w sin (0+ Фi)

sin Ф = 0

Ф = 0

The equation is

x = 4.0 cos (7,918 t)

v = -4.0 7,918 sin (7,918 t)

v = - 31.67 sin (7.918t)

Let's look for the time for a displacement of x = 1.5 cm, remember that the angles must be in radians

7,918 t = cos⁻¹ 1.5 / 4.0

t = 1,186 / 7,918

t = 0.1498 s

We look for speed

v = -31.67 sin (7,918 0.1498)

v = -29.36 cm / s

c) if the object passes the equilibrium equilibrium position again at this point the velocity has the same module, but the opposite sign

v = 29.36 cm / s

d) let's look for the time for the condition v = v_max / 2

31.67 / 2 = 31.67 sin ( 7,918 t)

7.918t = sin⁻¹ 0.5

t = 0.5236 / 7.918

t = 0.06613

With this time let's look for displacement

x = 4.0 cos (7,918 0.06613)

x = 3.46 cm

6 0

2 years ago

A commuter airplane starts from an airport and takes theroute. The plane first flies to city A, located 175 km away in a directi

bearhunter [10]

Answer:

245.45km in a direction 21.45° west of north from city A

Explanation:

Let's place the origin of a coordinate system at city A.

The final position of the airplane is given by:

rf = ra + rb + rc where ra, rb and rc are the vectors of the relative displacements the airplane has made. If we separate this equation into its x and y coordinates:

rfX = raX+ rbX + rcX = 175*cos(30)-150*sin(20)-190 = -89.75km

rfY = raY + rbY + rcT = 175*sin(30)+150*cos(20) = 228.45km

The module of this position is:

$rf = \sqrt{rfX^2+rfY^2} = 245.45km$

And the angle measure from the y-axis is:

$\alpha =atan(rfX/rfY) = 21.45\°$

So the answer is 245.45km in a direction 21.45° west of north from city A

6 0

2 years ago

A uniform electric field with a magnitude of 5750 N/C points in the positive x direction. Find the change in electric potential

castortr0y [4]

Explanation:

Given that,

Electric field = 5750 N/C

Charge $q=+10.5\times10^{-6}\ C$

Distance = 5.50 cm

(a). When the charge is moved in the positive x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot d$

$\Delta U=-q(E\cdot d)$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times5.50\times10^{-2}$

$\Delta U=-3.32\times10^{-3}\ J$

The change in electric potential energy is $-3.32\times10^{-3}\ J$

(b). When the charge is moved in the negative x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot (-d)$

$\Delta U=-q(E\cdot (-d))$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times(-5.50\times10^{-2})$

$\Delta U=3.32\times10^{-3}\ J$

The change in electric potential energy is $3.32\times10^{-3}\ J$

Hence, This is the required solution.

3 0

2 years ago

A hammer has a mass of 1 kg. What is its weight (i) on Earth (ii) on the

solong [7]

Given mass= 1kg

Weight on earth = mg(gravity of earth) = 9.8N

weight on moon = mg(gravity of moon)= 1.62N

weight on outer space mg(gravity outer space = 0) = 0N

4 0

2 years ago

What is the magnitude of the angular momentum (in kgm2/s) of a 40 g golf ball flying through the air and spinning at 4300 rpm af

sladkih [1.3K]

Answer:

$L=0.0045\ kg-m^2/s$

Explanation:

Given that,

The mass of a golf ball, m = 40 g = 0.04 kg

Its angular velocity, $\omega=4300\ rpm=450.29\ rad/s$

The radius of the sphere is 2.5 cm or 0.025 m

We need to find the magnitude of the angular momentum of the ball. It is given by the formula as follows:

$L=I\omega$

Where I is moment of inertia

For sphere, $I=\dfrac{2}{5}mr^2$

$L=\dfrac{2}{5}mr^2\omega\\\\L=\dfrac{2}{5}\times 0.04\times (0.025)^2\times 450.29\\\\L=0.0045\ kg-m^2/s$

So, the magnitude of the angular momentum of the sphere is $0.0045\ kg-m^2/s$ .

4 0

2 years ago