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laila [671]
3 years ago
11

What two things does force depend on​

Physics
1 answer:
faust18 [17]3 years ago
8 0

Answer:

When dealing with the force of gravity between two objects, there are only two things that are important – mass, and distance or (acceleration)

Explanation:

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PLS HELP WITH THIS FOR BRAINLIST IF ITS RIGHT
Zolol [24]

Answer:

the first one is Primary

the second one I think it's Mature but I don't know

5 0
3 years ago
When the speed of the bottle is 2 m/s, the KE is kg m2/s2. When the speed of the bottle is 3 m/s, the KE is kg m2/s2. When the s
d1i1m1o1n [39]

mass of the bottle in each case is M = 0.250 kg

now as per given speeds we can use the formula of kinetic energy to find it

1) when speed is 2 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(2)^2 = 0.5 J

2) when speed is 3 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(3)^2 = 1.125 J

3) when speed is 4 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(4)^2 = 2 J

4) when speed is 5 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(5)^2 = 3.125 J

5) when speed is 6 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(6)^2 = 4.5 J

3 0
3 years ago
Read 2 more answers
The spring in the muzzle of a child's spring gun has a spring constant of 730 N/m. To shoot a ball from the gun, first the sprin
Korolek [52]

Answer:

a. V=11.84 m/s

b.x=0.052m

Explanation:

a).

Given

K=730 N/m,m=0.053kg, h=1.90m.

v_f^2=v_i^2+2*g*h

v_i^2=2*g*h=2*9.8m/s^2*1.9m

v_i=\sqrt{2*9.8m/s^2*1.9m}=\sqrt{37.24 m^2/s^2}

v_i=6.1 m/s

v_i=V*sin(31)

V=\frac{v_i}{sin(31)}=\frac{6.1m/s}{sin(31)}

V=11.84 m/s

b).

K_k=\frac{1}{2}*K*x^2

No friction on the ball so:

x^2=\frac{2*K_k}{K}

x=\sqrt{\frac{2*0.053kg*9.8m/s^2*1.9m}{730N/m}}

x=\sqrt{2.7x10^{-3}m^2}=0.052m

5 0
3 years ago
Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere A h
vodka [1.7K]

Answer:

Explanation:

Given

Radius of A is twice of B i.e.

R_A=2R_B

Also Potential of both sphere is same

V_A=V_B

V=\frac{kQ}{R}

thus

k\frac{Q_A}{R_A}=K\frac{Q_B}{R_B}

\frac{Q_A}{Q_B}=\frac{R_A}{R_B}

\frac{Q_A}{Q_B}=\frac{2}{1}=2

\frac{Q_B}{Q_A}=\frac{1}{2}

(b)Ratio of \frac{E_B}{E_A}

Electric Field is given by E=\frac{kQ}{R^2}

thus E_A=\frac{kQ_A}{R_A^2}----1

E_B=\frac{kQ_B}{R_B^2}----2

Divide 2 by 1

\frac{E_B}{E_A}=\frac{Q_B}{R_B^2}\times \frac{R_A^2}{Q_A}

\frac{E_B}{E_A}=\frac{1}{2}\times 4=2

8 0
3 years ago
Two motorcyclists are riding side-by-side at night and the distance between their center-mounted headlights is 1.40 m. (a) If th
dezoksy [38]

Answer:

θ = 1.591 10⁻² rad

Explanation:

For this exercise we must suppose a criterion when two light sources are considered separated, we use the most common criterion the Rayleigh criterion that establishes that two light sources are separated census the central maximum of one of them coincides with the first minimum of the other source

         

Let's write the diffraction equation for a slit

       a sin θ = m λ

The first minimum occurs for m = 1, also field in these we experience the angles are very small, we can approximate the sin θ = θ

             θ = λ / a

In our case, the pupil is circular, so the system must be solved in polar coordinates, so a numerical constant is introduced.

           θ = 1.22 λ / D

Where D is the diameter of the pupil

 Let's apply this equation to our case

        θ = 1.22 600 10⁻⁹ / 0.460 10⁻²

        θ = 1.591 10⁻² rad

This is the angle separation to solve the two light sources

6 0
3 years ago
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