Ask question

Ask question

All categories

3 years ago

9

Which planet shown (Venus, Mercury, or Mars) is most likely able to support life now or was able to in the past? Explain your re

asoning.

1 answer:

nadezda [96]3 years ago

8 0

Gfdcbzvzv sh d d u runs eh. Uewneumeu ru

You might be interested in

The magnitude of a force vector is 89.6 newtons (N). The x component of this vector is directed along the +x axis and has a magn

insens350 [35]

Answer:

(a) θ = 33.86°

(b) Ay = 49.92 N

Explanation:

You have that the magnitude of a vector is A = 89.6 N

The x component of such a vector is Ax = 74.4 N

(a) To find the angle between the vector and the x axis you use the following formula for the calculation of the x component of a vector:

$A_x=Acos\theta$ (1)

Ax: x component of vector A

A: magnitude of vector A

θ: angle between vector A and the x axis

You solve the equation (1) for θ, by using the inverse of cosine function:

$\theta=cos^{-1}(\frac{A_x}{A})=cos{-1}(\frac{74.4N}{89.6N})\\\\\theta=33.86\°$

the angle between the A vector and the x axis is 33.86°

(b) The y component of the vector is given by:

$A_y=Asin\theta\\\\A_y=(89.6N)sin(33.86\°)=49.92N$

the y comonent of the vecor is Ay = 49.92 N

3 0

3 years ago

if a spring has a spring constant of 2 N/m and it is stretched 5 cm, what is the force of the spring?

djyliett [7]

Answer:

0.1 N

Explanation:

Considering the relationship between force,

spring constant and extension as defined by Hook's law

The force F=xk as from Hooke's law where F is the force of the spring, k is spring constant and x is extension or compression. Substituting 2 N/m for k and 5cm which is equivalent to 0.05 m for extention x then the force will be

F=2*0.05=0.1 N

4 0

3 years ago

The instantaneous velocity of an object is the blank of the object with a blank

777dan777 [17]

The instantaneous velocity of the object is its speed and direction at that instant.

7 0

3 years ago

Read 2 more answers

An object has a height of 0.066 m and is held 0.210 m in front of a converging lens with a focal length of 0.140 m. (Include the

Anvisha [2.4K]

Explanation:

Given that,

Size of object, h = 0.066 m

Object distance from the lens, u = 0.210 m (negative)

Focal length of the converging lens, f = 0.14 m

If v is the image distance from the lens, we can find it using lens formula as follows :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{0.14 }+\dfrac{1}{(-0.21)}\\\\v=0.42\ m$

(a) Magnification,

$m=\dfrac{v}{u}\\\\m=\dfrac{0.42}{(-0.21)}\\\\m=-2$

(b) Magnification, $m=\dfrac{h'}{h}$

h' is image height

$-2=\dfrac{h'}{0.066}\\\\h'=-2\times 0.066\\\\h'=-0.132\ m$

Hence, this is the required solution.

4 0

2 years ago

A hollow cylinder with an inner radius of and an outer radius of conducts a 3.0-A current flowing parallel to the axis of the cy

Artemon [7]

Complete Question:

A hollow cylinder with an inner radius of 4.0mm and an outer radius of 30mm conducts a 3.0-A current flowing parallel to the axis of the cylinder. If the current density is uniform throughout the wire, what is the magnitude of the magnetic field at a point 12mm from its center?

Answer:

The magnitude of the magnetic field = 7.24 μT

Explanation:

Inner radius, a = 4.0 mm = 0.004 m

Outer radius, b = 30 mm = 0.03 m

Radius, r = 12 mm = 0.012 m

let h² = b² - a²

h² = 0.03² - 0.004²

h² = 0.000884

Let d² = r² - a²

d² = 0.012² - 0.004²

d² = 0.000128

Current I = 3A

μ = 4π * 10⁻⁷

The magnitude of the magnetic field is given by:

$B = \frac{\mu I d^{2} }{2\pi r h^{2} } \\B = \frac{4\pi * 10^{-7} * 3* 0.000128^{2} }{2\pi *0.012* 0.000884^{2} }$

B = 7.24 * 10⁻⁶T

B = 7.24 μT

7 0

3 years ago