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Troyanec [42]
3 years ago
6

Which of the following measurements is used to compare the distance between objects in the solar system? Distance between Earth

and Saturn Distance between the sun and Earth Speed of rotation of Saturn Speed of rotation of Earth
Physics
2 answers:
xxMikexx [17]3 years ago
8 0
Answer is: <span>Distance between the sun and Earth

Explanation:
</span><span>Distance between Earth and Saturn is not correct answer because this distance varies greatly. When comparing distance we need some distance that is pretty much the same all the times. This is why distance between the sun and Earth is chosen to meassure distances in solar system.
Answers "</span>Speed of rotation of Saturn" and "<span>Speed of rotation of Earth" are not correct because they represent speed which can not measure distance.</span>
hammer [34]3 years ago
5 0

Answer:

Distance between earth and sun.

Explanation:

The distance between earth and sun is used to compare the distance between the objects in our solar system. It is because the distance between earth and sun is called as one Astronomical unit.

1 Astronomical unit (AU) is equal to 149,597,870.7 Km.

It is the mean distance between earth and sun.

The time taken by light to cover 1 Astronomical Unit is 499 seconds.

This is a big unit of distance.

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A student places blocks on a 100cm long see-saw as shown/
Hunter-Best [27]

Answer:

Part 1)

\tau_1 = 5 \times (0.50) = 2.5 N m

Part 2)

\tau_2 = 14 \times (0.30) = 4.2 N m

Part 3)

\tau_3 = 1.4 N m

Part 4)

Since torque on right side is more so here it will turn and slip over it

Explanation:

As we know that the block A is placed at distance

d = 50 cm from the hinge at 70 cm mark

So torque due to weight of A is given as

\tau_1 = 5 \times (0.50) = 2.5 N m

the block B is placed at distance

d = 30 cm from the hinge at 70 cm mark

So torque due to weight of B is given as

\tau_2 = 14 \times (0.30) = 4.2 N m

Now torque due to weight of the scale is given as

\tau_3 = 7(0.20)

\tau_3 = 1.4 N m

now torque on left side of scale is given as

\tau_{left} = \tau_1 + \tau_3

\tau_{left} = 2.5 + 1.4 = 3.9 N m

Torque on right Side is given as

\tau_{right} = \tau_2 = 4.2 Nm

Since torque on right side is more so here it will turn and slip over it

8 0
3 years ago
Which of the following is known as the following cycle because it's reservoir is a rock.
inessss [21]

Letra a obrigadoaijahahavqvqgqgqg

3 0
3 years ago
Objects are lighter on the moon than they are on earth. if an object A weighs 25lbs on the Moon and another object B weighs 25 N
solong [7]

Answer:

a. Object A

Explanation:

The mass of an object implies the quantity of matter in it, while the weight is the amount of gravitational force applied on an object.

The object A has a mass of 25 lbs, but object B on the earth has a weight, W, of 25 N.

So that,

For object A on the moon, mass = 25 lbs

For object B on the earth, W = 25 N,

W = m x g

25 = m x 10                (g = 10 m/s^{2})

m = \frac{25}{10}

   = 2.5 lbs

Mass of object B is 2.5 lbs.

Therefore, the mass of the object A is more than that of B.

5 0
3 years ago
Traumatic brain injury such as a concussion results when the head undergoes a very large acceleration. Generally an acceleration
eimsori [14]

The complete text of the problem is:

<em>"Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.43 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.1 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. "</em>

<em />

<u>Solution:</u>

1) Acceleration: -2336 m/s^2 on the hardwood floor, -382 m/s^2 on the carpeted floor

First of all, we need to calculate the speed of the child just before he hits the floor. This can be done by using the equation

v^2 - u^2 = 2ad

where

v is the final speed

u = 0 is the initial speed (the child starts from rest)

a = g = 9.8 m/s^2 is the acceleration of gravity

d = 0.43 m is the distance covered by the child as he falls from the bed

Solving for v,

v=\sqrt{2ad}=\sqrt{2(9.8)(0.43)}=2.9 m/s

Now we can analyze the moment of the collision. The child hits the floor with an initial speed of v = 2.9 m/s, and he comes to a stop, so the final speed is v' = 0. If the floor is hardwood, the stopping distance is

d = 1.8 mm = 0.0018 m

So we can find the acceleration by using again the equation

v'^2 - v^2 = 2ad

Solving for a,

a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.0018)}=-2336 m/s^2

For the carpeted floor instead,

d=1.1 cm = 0.011 m

therefore the acceleration is

a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.011)}=-382 m/s^2

2) Duration: 1.24 ms for the hardwood floor, 7.59 ms for the carpeted floor

We can find the duration of the collision in both cases by using the equation of the acceleration

a=\frac{v'-v}{t}

where

v' = 0

v = 2.9 m/s

For the hardwood floor,

a=-2336 m/s^2

So the duration of the collision is

t = \frac{v'-v}{a}=\frac{0-2.9}{-2336}=0.00124 s = 1.24 ms

For the carpeted floor,

a=-382 m/s^2

So the duration of the collision is

t = \frac{v'-v}{a}=\frac{0-2.9}{-382}=0.00759 s = 7.59 ms

We can now comment the results using the initial statement of the problem:

"Generally an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1ms will cause injury"

Therefore, the fall on the hardwood floor can result in injury (since the acceleration is greater than 1,000 m/s2 for more than 1 ms), while the fall on the carpeted floor is not dangerous (much less than 1000 m/s^2).

8 0
3 years ago
If you put a total of 8.05×106×106 electrons on an intially electrically neutral wire of length 1.03 m, what is the magnitude of
olga_2 [115]

Answer:

The magnitude of the electric field is 0.1108 N/C

Explanation:

Given;

number of electrons, e = 8.05 x 10⁶

length of the wire, L = 1.03 m

distance of the field from the center of the wire, r = 0.201 m

Charge of the electron;

Q = (1.602 x 10⁻¹⁹ C/e) x (8.05 x 10⁶ e)

Q = 1.2896 x 10⁻¹² C

Linear charge density;

λ = Q / L

λ = (1.2896 x 10⁻¹² C) / (1.03 m)

λ = 1.252 x 10⁻¹² C/m

The magnitude of electric field at r = 0.201 m;

E = (\frac{1}{4 \pi \epsilon_o} )\frac{ 2 \lambda}{r} \\\\E = k \frac{ 2 \lambda}{r}\\\\E = (8.89*10^9)*\frac{2*1.252*10^{-12}}{0.201}  \\\\E = 0.1108 \ N/C

Therefore, the magnitude of the electric field is 0.1108 N/C

6 0
3 years ago
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