A ball is thrown downward with an initial speed of 6m/s. the ball's velocity after 4 seconds is m/s. (g=-9.8m/s^2)

A solar system has the five planets shown below. The mass of each planet is proportion

slava [35]

Answer:

Planet C

Explanation:

The figure of the problem is missing: find it in attachment.

The magnitude of the gravitational force between two objects is given by the equation:

$F=\frac{Gm_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

In this problem, we have four planets around planet X, and the mass of each planet is proportional to its size in the figure.

As we can see from the previous equation, the magnitude of the gravitational force is proportional to the mass of the planets: therefore, the planet with largest mass will exert the largest gravitational force on planet X.

From the figure, we see that planet C has the largest size, so the largest mass: therefore, planet C exerts the greatest gravitational force on planet X.

6 0

4 years ago

Why is it necessary to enrich the soil?

enyata [817]

Plants rely on proper soil conditions to give them nutrients and minerals.

5 0

4 years ago

Read 2 more answers

A 10-kg object is dropped from rest. after falling a distance of 50 m, it has a speed of 26 m/s. what is the change in mechanica

likoan [24]

The change in mechanical energy caused by the dissipative resistance force is equal to, difference between the potential energy and kinetic energy of the object.

Potential energy of the object, P.E = mgh

m is mass of the object = 10 kg

g is acceleration due to gravity = 9.8 m/s²

h= height from which it is dropped =50 m

Substituting the value we get,

P.E = 10×9.8×50 = 4900 J

Kinetic energy of the object, K.E = $\frac{1}{2}mv^{2}$

v is the velocity of the object = 26 m/s²

K.E = (1/2)×10×(26)²

= 3380 J

Change in mechanical energy caused by dissipative force = P.E ₋ K.E

= 4900 ₋ 3380 = 1520 J

4 0

3 years ago

A common cylindrical copper wire used in a lab is 841 m long. Find the radius (in mm) of a wire necessary to have 0.5 Ohms of re

slava [35]

The relationship between the resistance R of a wire and its resistivity $\rho$ is given by
$R= \frac{\rho L}{A}$
where L is the length of the wire and A is its cross sectional area.

In the problem, we have $R=0.5 \Omega$ , $\rho = 1.68 \cdot 10^{-8} \Omega m$ and $L=841 m$ . So we can solve the find the area A:
$A= \frac{\rho L}{R}=2.83 \cdot 10^{-5} m^2$

For a cylindrical wire, the cross sectional area is given by
$A= \pi r^2$
where r is the radius. We know the value of the area A, so now we can find the radius of the wire:
$r= \sqrt{ \frac{A}{\pi} }= \sqrt{ \frac{2.83 \cdot 10^{-5}m^2}{\pi} }=0.003 m=3.0 mm$

3 0

3 years ago

The cornea behaves as a thin lens of focal lengthapproximately 1.80 {\rm cm}, although this varies a bit. The material of whichi

Keith_Richards [23]

Answer:

Explanation:

a )

from lens makers formula

$\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})$

f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face

putting the values

$\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})$

1.462 = 2 - 1 / r₂

1 / r₂ = .538

r₂ = 1.86 cm .

= 18.6 mm .

b )

object distance u = 25 cm

focal length of convex lens f = 1.8 cm

image distance v = ?

lens formula

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}$

$\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}$

.5555 - .04

= .515

v = 1.94 cm

c )

magnification = v / u

= 1.94 / 25

= .0776

size of image = .0776 x size of object

= .0776 x 10 mm

= .776 mm

It will be a real image and it will be inverted.

5 0

3 years ago