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zheka24 [161]
3 years ago
12

What is one factor that helps keep the planets and other solar system objects in orbit around the sun?

Physics
2 answers:
pshichka [43]3 years ago
7 0
The answer is D<span>.the gravitational force between the sun and each object in the solar system</span>
zhuklara [117]3 years ago
3 0
D the sun`s gravitational force and the planet`s gravitational force keeps us in our spots
You might be interested in
A small object with a 5.0-mC charge is accelerating horizontally on a friction-free surface at 0.0050 m/s2 due only to an electr
kolbaska11 [484]

Answer:

0.002 N/C

Explanation:

Parameters given:

Charge of object, q = 5 mC = 5 * 10^{-3} C

Acceleration of object, a = 0.005 m/s^2

Mass of object, m = 2.0 g

The Electric field exerts a particular force on the object, causing it to accelerate (Electrostatic force).

We know that Electrostatic force, F, is given in terms of Electric field, E, as:

F = qE

This means that the object exerts a force of -qE on the Electric force (Action with equal and opposite reaction).

The object also has a force, F, due to its acceleration a. This force is the product of its mass and acceleration. Mathematically:

F = ma

Equating the two forces of the object, we get:

-qE = ma

=> E = \frac{-ma}{q}

Solving for E, we have:

E = \frac{-2 * 10^{-3} * 0.005}{5 * 10^{-3}} \\\\\\E = -0.002 N/C

The magnitude will be:

|E| = |-0.002| N/C = 0.002 N/C

The electric field has a magnitude of 0.002 N/C.

4 0
3 years ago
Why do objects move at constant speed
kati45 [8]
Most objects move at a constant speed because of friction and acceleration. The constant speed keeps them in place, and keeps a balance. 
7 0
3 years ago
Under the assumption that the beam is a rectangular cantilever beam that is free to vibrate, the theoretical first natural frequ
BartSMP [9]

Answer:

a) Δf = 0.7 n , e)   f = (15.1 ± 0.7) 10³ Hz

Explanation:

This is an error about the uncertainty or error in the calculated quantities.

Let's work all the magnitudes is the SI system

The frequency of oscillation is

        f = n / 2π L² √( E /ρ)

where n is an integer

Let's calculate the magnitude of the oscillation

       f = n / 2π (0.2335)² √ (210 10⁹/7800)

       f = n /0.34257 √ (26.923 10⁶)

       f = n /0.34257    5.1887 10³

       f = 15.1464 10³ n

a) We are asked for the uncertainty of the frequency (Df)

       Δf = | df / dL | ΔL + df /dE ΔE + df /dρ Δρ

in this case no  error is indicated in Young's modulus and density, so we will consider them exact

       ΔE = Δρ = 0

       Δf = df /dL  ΔL

       df = n / 2π   √E /ρ   | -2 / L³ | ΔL

       df = n / 2π 5.1887 10³ | 2 / 0.2335³) 0.005 10⁻³

       df = n 0.649

Absolute deviations must be given with a single significant figure

        Δf = 0.7 n

b, c) The uncertainty with the width and thickness of the canteliver is associated with the density

 

In your expression there is no specific dependency so the uncertainty should be zero

The exact equation for the natural nodes is

          f = n / 2π L² √ (E e /ρA)

where A is the area of ​​the cantilever and its thickness,

In this case, they must perform the derivatives, calculate and approximate a significant figure

        Δf = | df / dL | ΔL + df /de  Δe + df /dA  ΔA

        Δf = 0.7 n + n 2π L² √(E/ρ A) | ½  1/√e | Δe

               + n / 2π L² √(Ee /ρ) | 3/2 1√A23  |

the area is

        A = b h

        A = 24.9  3.3  10⁻⁶

        A = 82.17 10⁻⁶ m²

        DA = dA /db ΔB + dA /dh Δh

        dA = h Δb + b Δh

        dA = 3.3 10⁻³ 0.005 10⁻³ + 24.9 10⁻³ 0.005 10⁻³

        dA = (3.3 + 24.9) 0.005 10⁻⁶

        dA = 1.4 10⁻⁷ m²

let's calculate each term

         A ’= n / 2π L² √a (E/ρ A) | ½ 1 /√ e | Δe

         A ’= n/ 2π L² √ (E /ρ)      | ½ 1 / (√e/√ A) |Δe

        A ’= 15.1464 10³ n ½ 1 / [√ (24.9 10⁻³)/ √ (81.17 10⁻⁶)] 0.005 10⁻³

        A '= 0.0266  n

        A ’= 2.66 10⁻² n

       A ’’ = n / 2π L² √ (E e /ρ) | 3/2  1 /√A³ |

       A ’’ = n / 2π L² √(E /ρ) √ e | 3/2  1 /√ A³ | ΔA

       A ’’ = n 15.1464 10³ 3/2 √ (24.9 10⁻³) /√ (82.17 10⁻⁶) 3 1.4 10⁻⁷

       A ’’ = n 15.1464 1.5 1.5779 / 744.85 1.4 10⁴

       A ’’ = 6,738 10²

we write the equation of uncertainty

     Δf = n (0.649 + 2.66 10⁻² + 6.738 10²)

The uncertainty due to thickness is

    Δf = 3 10⁻² n

The uncertainty regarding the area, note that this magnitude should be measured with much greater precision, specifically the height since the errors of the width are very small

     Δf = 7 10² n

 d)    Δf = 7 10² n

e) the natural frequency n = 1

       f = (15.1 ± 0.7) 10³ Hz

7 0
3 years ago
Help me with physics​
Llana [10]

Answer:

z3

Explanation:

6 0
3 years ago
Two long, straight wires are parallel and 26 cm apart.
mezya [45]

Answer: 2.49×10^-3 N/m

Explanation: The force per unit length that two wires exerts on each other is defined by the formula below

F/L = (u×i1×i2) / (2πr)

Where F/L = force per meter

u = permeability of free space = 1.256×10^-6 mkg/s^2A^2

i1 = current on first wire = 57A

i2 = current on second wire = 57 A

r = distance between both wires = 26cm = 0.26m

By substituting the parameters, we have that

Force per meter = (1.256×10^-6×57×57)/ 2×3.142 ×0.26

= 4080.744×10^-6/ 1.634

= 4.080×10^-3 / 1.634

= 2.49×10^-3 N/m

5 0
3 years ago
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