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stira [4]
3 years ago
9

A partical of an atom that carries a negative charge is called a a/n

Physics
2 answers:
OlgaM077 [116]3 years ago
8 0
Negative neutron. You’re welcome.
jeka57 [31]3 years ago
7 0
It is called a negative nuetron/atom
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Two ropes are attached to either side of a 100.0 kg wagon as shown below. The rope on the right is pulled at an angle 40.0° rela
NikAS [45]

The acceleration of the wagon is found by applying Newton's Second Law of motion.

1. The responses for question 1 are;

  • x-component of the tension in the rope on the right is approximately <u>91.93 N</u>
  • y-component of the tension in the rope on the right is approximately <u>71.135 N</u>
  • x-component of the tension in the rope on the left is -80.0 N
  • y-component of the tension in the rope on the left is 0

2. The net force in the x-direction is approximately <u>11.93 N</u>

3. The net acceleration of the wagon in the horizontal direction is approximately <u>0.1193 m/s²</u>.

Reasons:

The given parameters are;

Mass of the wagon, m = 100.0 kg

Angle of inclination to the horizontal of the rope to the right, θ = 40.0°

Tension in the rope on the right = 120.0 N

Direction in which the rope on the left is pulled = To the west

Tension in the rope on the left = 80.0 N

1. The <em>x</em> and <em>y</em> component of the tension in the rope on the right are;

x-component = 120.0 N × cos(40.0°) ≈ <u>91.93 N</u>

y-component = 120.0 N × sin(40.0°) ≈ <u>77.135 N</u>

The <em>x</em> and <em>y</em> component of the tension in the rope on the left are;

x-component = 80.0 N × cos(180°) = <u>-80.0 N</u>

y-component = 80.0 N × sin(180°) = <u>0.0 N</u>

2. The net force in the horizontal direction, Fₓ, is found as follows;

Fₓ = The x-component of the rope on the left + The x-component of the rope on the right

Which gives;

Fₓ = 91.93 N - 80.0 N = <u>11.93 N</u>

3. The net acceleration of the block is given as follows;

According to Newton's Second Law of motion, we have;

Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ

Fₓ = m × aₓ

Therefore;

\displaystyle a_x = \frac{F_x}{m}  \approx \frac{11.93 \, N}{100.0 \, kg} = \mathbf{0.1193 \ m/s^2}

  • The acceleration of the wagon in the horizontal direction, aₓ ≈ <u>0.1193 m/s²</u>.

Learn more here:

brainly.com/question/20357188

8 0
2 years ago
A 545-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli
Art [367]

Answer

given,

mass of satellite = 545 Kg

R = 6.4 x 10⁶ m

H = 2 x 6.4 x 10⁶ m

Mass of earth = 5.972 x 10²⁴ Kg

height above earth is equal to earth's mean radius

a) satellite's orbital velocity

   centripetal force acting on satellite = \dfrac{mv^2}{r}

     gravitational force = \dfrac{GMm}{r^2}

    equating both the above equation

    \dfrac{mv^2}{r} = \dfrac{GMm}{r^2}

      v = \sqrt{\dfrac{GM}{r}}

      v = \sqrt{\dfrac{6.67 \times 10^{-11}\times 5.972 \times 10^{24}}{2 \times 6.4 \times 10^6}}

          v = 5578.5 m/s

b) T= \dfrac{2\pi\ r}{v}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

          T = 14416.92 s

          T = \dfrac{14416.92}{3600}\ hr

          T = 4 hr

c) gravitational force acting

  F = \dfrac{GMm}{r^2}

  F = \dfrac{6.67 \times 10^{-11}\times 545 \times 5.972 \times 10^{24} }{(6.46 \times 10^6)^2}

     F = 5202 N

4 0
3 years ago
a 230 kg roller coaster reaches the top of the steepest hill with a speed of 6.2 km/h. It then descends the hill, which is at an
igor_vitrenko [27]

Answer: 81.619 kJ

Explanation:

Given

Mass of roller coaster is m=230\ kg

It reaches the steepest hill with speed of u=6.2\ km/h\ or \ 1.72\ m/s

Hill to bottom is 51 m long with inclination of 45^{\circ}

Height of the hill is h=51\sin 45^{\circ}=36.06\ m

Conserving energy to get kinetic energy at bottom

Energy at top=Energy at bottom

\Rightarrow K_t+U_t=K_b+U_b\\\Rightarrow \dfrac{1}{2}mu^2+mgh=K_b+0\\\\\Rightarrow K_b=0.5\times 230\times 1.72^2+230\times 9.8\times 36.06\\\Rightarrow K_b=340.216+81,279.24\\\Rightarrow K_b=81,619.456\ J\\\Rightarrow K_b=81.619\ kJ

8 0
2 years ago
Which of the following would increase​
bazaltina [42]

Answer:

1 and 3

Explanation:

<u>1 and 3  </u>

Increasing coils increases strength

   COOLING the wire would increase current flow and strength of magnet

Adding an iron core will definitely increase the strength of the electromagnet

7 0
1 year ago
Read 2 more answers
You are moving a desk that has a mass of 36 kg; its acceleration is 0.5 m / s 2. What is the force being applied
valentina_108 [34]

Answer:

18 N

Explanation:

Force can be found using the following formula.

f= m*a

where m is the mass and a is the acceleration.

We know the desk has a mass of 36 kilograms. We also know that its acceleration is 0.5 m/s^2.

m= 36 kg

a= 0.5 m/s^2

Substitute these values into the formula.

f= 36 kg * 0.5 m/s^2

Multiply 36 and 0.5

f=18 kg m/s^2

1 kg m/s^2 is equivalent to 1 Newton, or N.

f= 18 Newtons

The force being applied is 18 kg m/s^2, Newtons, or N

6 0
3 years ago
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