50
1. Force = ? N mass = 65 kg acceleration = 25 m/s2 (Record your answer
by number only and don t.include units...)

Answer: i can help you

Explanation:

[tex]2.5[/tex]×[tex]10^{11}[/tex] electrons are needed to make an object neutral which is carrying a charge of [tex]q=+4[/tex]×[tex]10^{-8}[/tex]

To learn more about charge, refer to

https://brainly.com/question/18102056

#SPJ2

Answer:

the answer is B

Explanation:

materialism is just wanting to always buy new things B does the opposite

Answer:

The final velocity of the first ball (0.3 kg) is 0.4 m/s in negative x direction.

Explanation:

Given;

mass of the first object, m₁ = 0.3 kg

initial velocity of the first ball, u₁ = 2.8 m/s

mass of the second ball, m₂ = 0.4 kg

initial velocity of the second ball, u₂ = 0

let the final velocity of the first ball, = v₁

let the final velocity of the second ball, = v₂

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.3 x 2.8) + (0.4 x 0) = 0.3v₁ + 0.4v₂

0.84 = 0.3v₁ + 0.4v₂

2.8 = v₁ + 1.333v₂ -------equation (1)

Apply one-direction velocity;

u₁ + v₁ = u₂ + v₂

2.8 + v₁ = 0 + v₂

v₂ = 2.8 + v₁

substitute the value of v₂ into equation (1)

2.8 = v₁ + 1.333v₂

2.8 = v₁ + 1.333(2.8 + v₁)

2.8 = v₁ + 3.732 + 1.333v₁

2.8 - 3.732 = v₁ + 1.333v₁

-0.932 = 2.333v₁

v₁ = -0.932 / 2.333

v₁ = -0.4 m/s

Therefore, the final velocity of the first ball (0.3 kg) is 0.4 m/s in negative x direction.

Answer:

The bicyclist's acceleration is 2.2m/s^2

Explanation:

Given

[tex]u = 2.5m/s[/tex] ---- Initial Velocity

[tex]v = 12.5m/s[/tex] ---- Final Velocity

[tex]t = 4.5s[/tex] ---- Time

Required

Determine the acceleration

This will be solved using the first equation of motion

[tex]v = u + at[/tex]

Substitute values for v, u and a

[tex]12.5 = 2.5 + a * 4.5[/tex]

[tex]12.5 = 2.5 + 4.5a[/tex]

Collect Like Terms

[tex]4.5a = 12.5 - 2.5[/tex]

[tex]4.5a = 10.0[/tex]

Solve for a

[tex]a = 10.0/4.5[/tex]

[tex]a = 2.2m/s^2[/tex] ---- (approximated)

Hence, the bicyclist's acceleration is 2.2m/s^2

Complete Question

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.

Answer:

The torque is  [tex]\tau = 34.3 \ N\cdot m[/tex]

Explanation:

From the question we are told that

   The mass of the steel ball is  [tex]m = 3.0 \ kg[/tex]

    The length of arm is  [tex]l = 70 \ cm = 0.7 \ m[/tex]

    The mass of the arm is [tex]m_a = 4.0 \ kg[/tex]

Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

       [tex]r = \frac{l}{2}[/tex]

=>    [tex]r = \frac{ 0.7}{2}[/tex]  

=>    [tex]r = 0.35 \ m[/tex]  

Generally the magnitude of torque about the athlete shoulder is mathematically represented as

      [tex]\tau = m_a * g * r + m * g * L[/tex]

=>    [tex]\tau = 4 * 9.8 * 0.35 + 3 * 9.8 * 0.70[/tex]

=>    [tex]\tau = 34.3 \ N\cdot m[/tex]

Answer:

In the clarification portion elsewhere here, the definition of the concern is mentioned.

Explanation:

So like optical telescopes capture light waves, introduce it to concentrate, enhance it, as well as make it usable through different instruments via study, so radio telescopes accumulate weak signal light waves, introduce that one to focus, enhance it, as well as make this information available during research. To research naturally produced radio illumination from stars, galaxies, dark matter, as well as other natural phenomena, we utilize telescopes.

Optical telescopes detect space-borne visible light. There are some drawbacks of optical telescopes mostly on the surface:

Although radio telescopes monitor space-coming radio waves. Those other telescopes, when they are already typically very massive as well as costly, have such an improvement surrounded by optical telescopes. They should be included in poor weather and, when they travel through the surrounding air, the radio waves aren't obscured by clouds. Throughout the afternoon and also some at night, radio telescopes are sometimes used.

Answer:

[tex]a=200\ m/s^2[/tex] and t = 0.5 s

Explanation:

Given that,

Mass of the object, m = 90 gram = 0.09 kg

Diameter of the circle, d = 2.56 m

Radius, r = 1.28 m

Speed, v = 16 m/s

We need to find the object acceleration. On the circular path, acceleration is given by :

[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{(16)^2}{1.28}\\\\a=200\ m/s^2[/tex]

So, the object's acceleration is [tex]200\ m/s^2[/tex].

Let t is time it will take to complete the around. Speed is given by :

[tex]v=\dfrac{2\pi r}{t}\\\\t=\dfrac{2\pi r}{v}\\\\t=\dfrac{2\pi \times 1.28}{16}\\\\t=0.5\ s[/tex]

So, time to cover one complete round is 0.5 s.

Answer:

Changes in the object's momentum (answer D)

Explanation:

A net force will cause an object to change its velocity, and that will affect the object's momentum, which is defined by the product of the object's mass times its velocity.

So, select the last option (D) in the given list.