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BaLLatris [955]
2 years ago
11

What would the charge be on an ion with 8 protons 7 neutrons and 6 elements?

Physics
1 answer:
andre [41]2 years ago
8 0

Answer:

-2

Explanation:

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Q2. You push a crate up a ramp with a force of 10 N. Despite your pushing, the crate slides down the ramp 4 m. How much work did
Ksivusya [100]

Answer:

40 J

Explanation:

From the question given above, the following data were obtained:

Force (F) = 10 N

Distance (s) = 4 m

Workdone (Wd) =?

Work done is simply defined as the product of force and distance moved in the direction of the force. Mathematically, we can express the Workdone as:

Workdone = force × distance

Wd = F × s

With the above formula, we can obtain the workdone as follow:

Force (F) = 10 N

Distance (s) = 4 m

Workdone (Wd) =?

Wd = F × s

Wd = 10 × 4

Wd = 40 J

Thus, 40 J of work was done.

5 0
2 years ago
An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get
Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

mV_i = (m-m_O)V_f - m_OV_O

where,

m=Total mass

m_O = Mass of Object

V_i = Velocity before throwing

V_f = Final Velocity

V_O = Velocity of Object

Our values are:

m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg

Solving to find the final speed, after throwing the object we have

V_f=\frac{mV_0+m_TV_O}{m-m_O}

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) 5.3Kg\rightarrow 15m/s

V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}

V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}

V_{f1}=0.8511m/s

B) 7.9Kg\rightarrow 11.2m/s

V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}

V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}

V_{f2} = 2.0173m/s

C) 10.5Kg\rightarrow 7m/s

V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}

V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}

V_{f3} = 3.63478m/s

Therefore the final velocity of astronaut is 3.63m/s

7 0
3 years ago
two test cars of equal mass moving towards each other collide on a horiontal frictionless surface. Before the collision, car A h
LiRa [457]

Answer:

4 m/s

Explanation:

m1 = m2 = m

u1 = 20 m/s, u2 = - 12 m/s

Let the speed of composite body is v after the collision.

Use the conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

m x 20 - m x 12 = (m + m) x v

20 - 12 = 2 v

8 = 2 v

v = 4 m/s

Thus, the speed of teh composite body is 4 m/s.

4 0
2 years ago
After observing the electric field in your trials above, where was the electric field the strongest? what was the direction of t
earnstyle [38]
<span>Answer: Answer is The direction of the electric field is always directed in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding the source charge.</span>
8 0
3 years ago
Two descriptions about physical quantities are given below:
Semenov [28]

Answer:

quantity A is mass and quantity B is wright

5 0
2 years ago
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