1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
mixas84 [53]
2 years ago
10

In an object's net charge positive or negative if it loses electrons?

Physics
1 answer:
Greeley [361]2 years ago
7 0
<span>Answer: If a neutral atom gains electrons, then it will become negatively charged. If a neutral atom loses electrons, then it become positively charged. 6. Determine the quantity and type of charge on an object that has 3.62 x 1012 more protons than electrons.</span>
You might be interested in
The unit for measuring electric power is the <br> A. ampere.<br> B. volt.<br> C. ohm.<br> D. watt.
Drupady [299]
The correct answer is D: Watt. This unit was named after James Watt, and is used to express the equivalent of one joule per second in energy. In experiments and on the packaging for electrical products such as light-bulbs, the measurement will usually be written in its abbreviated format: W.
<span />
6 0
3 years ago
How do you know when water is boiling?
vladimir1956 [14]

Answer:

when the steam starts coming out

Explanation:

5 0
3 years ago
Scenario
Anvisha [2.4K]

Answer:

1) t = 23.26 s,  x = 8527 m, 2)   t = 97.145 s,  v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

         v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

         y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

         y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

         0 = y₀ + 0 - ½ g t2

          t = \sqrt{ \frac{2y_o}{g} }

          t = √(2 2650/9.8)

          t = 23.26 s

this is the horizontal scrolling time

          x = v₀ t

          x = 366.67  23.26

          x = 8527 m

the speed at the point of arrival is

         v_y = v_{oy} - g t = 0 - gt

         v_y = - 9.8 23.26

         v_y = -227.95 m / s

Module and angle form

        v = \sqrt{v_x^2 + v_y^2}

         v = √(366.67² + 227.95²)

        v = 431.75 m / s

         θ = tan⁻¹ (v_y / vₓ)

         θ = tan⁻¹ (227.95 / 366.67)

         θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

           cos θ = v₀ₓ / v₀

           sin θ = v_{oy} / v₀

            v₀ₓ = v₀ cos θ

            v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

            y = y₀ + v_{oy} t - ½ g t²

            0 = y₀ + vo sin θ t - ½ g t²

            0 = -50 + vo sin 50 t - ½ 9.8 t²

            x = x₀ + v₀ₓ t

            0 = x₀ + vo cos θ t

            0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

          50 = 0,766 vo t – 4,9 t²

          400 =   0,643 vo t

resolved

          50 = 0,766 ( \frac{400}{0.643 \ t}) t – 4,9 t²

          50 = 476,52 t – 4,9 t²

          t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

         t = [97.25 ± \sqrt{97.25^2 - 4 \ 10.2}] / 2

         t = 97.25 ±97.04] 2

         t₁ = 97.145 s

         t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

         t = 97.145 s

let's find the initial velocity

         x = x₀ + v₀ cos 50 t

         0 = -400 + v₀ cos 50 97.145

         v₀ = 400 / 62.44

         v₀ = 6.4 m / s

5 0
2 years ago
On a highway curve with a radius of 46 meters, the maximum force of static friction that can act on a 1,200 kg car going around
Mekhanik [1.2K]

Answer:

v\approx 16.956\,\frac{m}{s}

Explanation:

The motion of the vehicule on a highway curve can be modelled by the following equation of equilibrium:

\Sigma F = f = m\cdot \frac{v^{2}}{R}

The maximum speed is:

v = \sqrt{\frac{f\cdot R}{m} }

v = \sqrt{\frac{(7500\,N)\cdot (46\,m)}{1200\,kg} }

v\approx 16.956\,\frac{m}{s}

7 0
3 years ago
A projectile is fired with an initial velocity of 120.0 meters per second at an angle, θ above the horizontal. If the projectile
k0ka [10]

Answer:

θ = 62.72°

Explanation:

The projectile describes a parabolic path:

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = x₀+ vx*t   Formula (1)

vx = v₀x

Where:  

x: horizontal position in meters (m)

x₀: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m/s

v₀x: Initial speed in x  in m/s

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

y= y₀+(v₀y)*t - (1/2)*g*t² Formula (2)

vfy= v₀y -gt Formula (3)

Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Data

v₀ = 120 m/s  , at an angle  θ above the horizontal

v₀x= 55 m/s

x-y components of the initial  velocity ( v₀)

v₀x = v₀*cosθ Equation (1)

v₀y = v₀*sinθ   Equation (2)

Calculating of the angle θ

We replace data in the  Equation (1)

55 =  120*cosθ

cosθ = 55 / 120

\theta = cos^{-1}(  \frac{55}{120} )

θ = 62.72°

3 0
2 years ago
Other questions:
  • Dexter uses a compass to walk 140m due east. What is his distance and displacement?
    12·1 answer
  • A capacitor with plates separated by a distance d is charged to a potential difference ΔVC. All wires and batteries are disconne
    13·1 answer
  • We will look at an example of a banked curve. In this case the normal force provides the needed radial acceleration. This allows
    7·1 answer
  • Atmospheric pressure and density at the top of Mount Everest ​
    6·1 answer
  • How is sound created (at the particle level)?
    8·1 answer
  • Which of the following is contact force
    8·1 answer
  • 6. Which does more wor:
    5·1 answer
  • 1
    8·1 answer
  • A 1kg mass is thrown to a height of 2cm. what is the potential energy​
    9·2 answers
  • 20. A 0.450-kg baseball comes off a bat and goes straight up. At a height of 10.0 m, the
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!