What would be the orbital speed and period of a satellite in orbit 0.82 ✕ 108 m above the earth?

How much is the moon in the first quarter actually lit

grandymaker [24]

Half of the moon is illuminated.

8 0

2 years ago

A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back

elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity= $9.8 m/s^2$

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates= $\mu mg=0.202\times 9.8\times 35$

Force by which box accelerates= $62.286 N$

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

$ma=\mu mg$

$a=\mu g=9.8\times 0.202=1.9796 m/s^2$

Hence, the truck can have maximum acceleration before the box slide= $1.9796 m/s^2$

3 0

2 years ago

A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c

viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = $71\times 10^{- 9} C$

Q' = + 42 nC = $42\times 10^{- 9} C$

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

$\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}$

$\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}$

$\vec{E} = 708.03 N/C$

Now,

Electric field at the mid-point due to charge Q' is given by:

$\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}$

$\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}$

$\vec{E'} = 418.84 N/C$

Now,

The net Electric field is given by:

$\vec{E_{net}} = \vec{E} - \vec{E'}$

$\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C$

5 0

2 years ago

A 2.0 m tall person is standing in front of a flat vertical mirror. What minimum height must the mirror have if the person is to

dangina [55]

Answer:

100cm

Explanation:

Since the eyes are 6 cm below the top of her head, the point of incidence of the ray must be

200cm-3cm=197cm

Since the eyes are 194 cm from her feet, the point of incidence of this ray must be

194cm/2=97cm

So the lower edge of the mirror must be 97 cm from the ground and the vertical dimension of the mirror must be 197 cm - 97 cm = 100 cm, which is half the height of the person.

7 0

2 years ago

An object of mass kg is released from rest m above the ground and allowed to fall under the influence of gravity. Assuming the f

IgorLugansk [536]

Answer:

Explanation:

From, the given information: we are not given any value for the mass, the proportionality constant and the distance

Assuming that:

the mass = 5 kg and the proportionality constant = 50 kg

the distance of the mass above the ground x(t) = 1000 m

Let's recall that:

$v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}$

Similarly, The equation of mption:

$x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})$

replacing our assumed values:

where $v_=0 \ and \ g= 9.81$

$x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})$

$x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m$

$\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}$

So, when the object hits the ground when x(t) = 1000

Then from above derived equation:

$\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}$

$1000= 0.981t-0.981(1-e^{-(10)t}) \ m$

By diregarding $e^{-(10)t} \ m$

$1000= 0.981t-0.981$

1000 + 0.981 = 0.981 t

1000.981 = 0.981 t

t = 1000.981/0.981

t = 1020.36 sec

7 0

2 years ago