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marysya [2.9K]
2 years ago
9

Scientists have detected an asteroid that is 700,000,000 km from Earth. About how many astronomical units is that? 0. 5 AU 1 AU

3. 5 AU 4. 5 AU.
Physics
2 answers:
Elza [17]2 years ago
6 0

Answer:

it is 4.67921098559 astronomical units

Explanation:

nalin [4]2 years ago
3 0

An asteroid that is 700,000,000 km is equivalent to 4.67 AU.

Option D is the correct answer.

<h3>How do you convert KM into AU?</h3>

Given that an asteroid that is 700,000,000 km from Earth.

We know that 1 AU is equal to 149597870.691 KM. The expression can be given below.

1 \;\rm AU = 149597870.691\;\rm KM

1 \;\rm KM =  \dfrac {1}{149597870.691} \;\rm AU

700,000,000 \;\rm KM = \dfrac {1}{149597870.691} \times 700,000,000 \;\rm AU

700,000,000 \;\rm KM = 4.67 \;\rm AU

Hence we can conclude that 700,000,000 km is equivalent to 4.67 AU. Option D is the correct answer.

To know more about the astronomical units, follow the link given below.

brainly.com/question/21464262.

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You apply a horizontal force of 25N to push a shopping cart across the parking lot at a constant velocity. a) what is the net fo
AlekseyPX

(a) The net force on the shopping cart is zero.

(b) The the force of friction on the shopping cart is 25 N.

(c) When same force is applied to the shopping cart on a wet surface, it will move faster.

<h3>Net force on the shopping cart</h3>

The net force on the shopping cart is calculated as follows;

F(net) = F - Ff

where;

  • F is the applied force
  • Ff is the frictional force

ma = F - Ff

where;

  • a is acceleration of the cart
  • m is mass of the cart

at a constant velocity, a = 0

0 = F - Ff

F(net)  = 0

F = Ff = 25 N

Net force is zero, and frictional force is equal to applied force.

<h3>On wet surface</h3>

Coefficient of kinetic friction of solid surface is greater than that of wet surface.

Since frictional force limit motion, when the frictional force is smaller, the object tends to move faster.

Thus, the cart will move faster on a wet surface due to decrease in friction.

Learn more about frictional force here: brainly.com/question/24386803

#SPJ1

4 0
2 years ago
Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105
Dovator [93]
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating). 
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
3 0
3 years ago
Read 2 more answers
A small metal object is tied to the end of a string and whirled round a circular path of radius 30cm. The object makes 20 oscill
LUCKY_DIMON [66]

Answer:

(i) The angular speed of the small metal object is 25.133 rad/s

(ii) The linear speed of the small metal object is 7.54 m/s.

Explanation:

Given;

radius of the circular path, r = 30 cm = 0.3 m

number of revolutions, n = 20

time of motion, t = 5 s

(i) The angular speed of the small metal object is calculated as;

\omega = \frac{20 \ rev}{5 \ s} \times \frac{2 \pi \ rad}{1 \ rev} = \frac{40\pi \ rad}{5 \ s} = 8\pi \ rad/s = 25.133 \ rad/s

(ii) The linear speed of the small metal object is calculated as;

v = \omega r\\\\v = 25.133 \ rad/s \ \times \ 0.3 \ m\\\\v = 7.54 \ m/s

6 0
3 years ago
I need so much help 50 points!!!!!!<br><br>complete the graph
motikmotik

<u>First Symbol </u>: Cobalt (Co)

Its Group Number - 9

Its Period Number - 4

Its Family Name - Transition Metal

<u>Second Symbol</u> : Silicon (Si)

Its Group Number - 14

Its Period Number - 2

Its Family Name - Semiconductor

<u>Third Symbol</u> : Astatine (At)

Its Group Number - 17

Its Period Number - 6

Its Family Name - Halogen

<u>Fourth Symbol </u>: Magnesium (Mg)

Its Group Number - 2

Its Period Number - 3

Its Family Name - Alkaline Earth Metal

<u>Fifth Symbol</u> : Xenon (Xe)

Its Group Number - 18

Its Period Number - 5

Its Family Name - Noble Gas

6 0
3 years ago
Which law is used to find the magnitude of a magnetic force?
Talja [164]

Answer:

The Flemings left hand rule is used to find the magnitude of a magnetic force

Explanation:

Fleming's left hand rule states that if the first three fingers are held mutually at right angles to one another, then the fore finger points into the direction of magnetic field the middle finger in the direction of current while the thumb points in the direction of force.

Mathematically

Magnetic Force F= BILsinθ

Where

B= magnetic field density Tesla

I= current

L= length of conductor

θ= angle of conductor with field

3 0
3 years ago
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