I would like to know if it would be m/s² or just m/s and explain why. thanks​

Answer:

Explanation:

The gravitational force = G * m1 * m2 / r^2

That means as the distance increases, r get's bigger and m1 and m2 don't change, the force decreases. Read that sentence carefully. In physics, it is very important to know what goes up and goes down means and especially why.

Answer: HP = Horse Power.

Explanation: it is the unit given to tell the motor's particular power and 1hp = 746 watts.

Answer:

D is the correct answer for this question.

Explanation:

I had done this already

Answer:

speed and acceleration

Explanation:

speed is a scalar quantity

acceleration is a vector quantity

Answer:14

Explanation:cuz

The average distance between the two stars is 792 light years

Let the mass of the first star be [tex]m_1[/tex]

Let the mass of the second star be [tex]m_2[/tex]

The combined mass of the two stars, [tex]m_1+m_2=5.5[/tex] solar masses

The orbital period of the stars, P = 12 years

Average distance between the two stars, D = ?

The average distance between the two stars can be calculated using Kepler's equation

[tex]D=(m_1+m_2)P^2[/tex]

Substitute [tex]m_1+m_2=5.5[/tex] and P = 12 into the formula [tex]D=(m_1+m_2)P^2[/tex]

[tex]D=5.5(12^2)[/tex]

D = 5.5(144)

D = 792 light years

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The characteristics of the thermistors allow to design an experiment to analyze the dependence of resistance with temperature;

      1) We set up the circuit

      2) We look for the current of the circuit and we keep it fixed.

      3) We look for the thermistor resistances for each temperature.

      4) Plot Resistance versus Temperature.

A Thermitor is an element whose resistance changes with temperature, for example a platinum resistance.

An experiment that we can carry out to measure is to set up the attached circuit where we have:

Let's start by looking for the current in the circuit.

In thermistors, the resistance for room temperature is given, so if we measure the voltage across the fixed resistor, we can shrink the voltage across the thermistor.

           [tex]V_{source} = V_R + V_{therm} \\V_{therm} = V_{source} - V_R \\ i = \frac{V_{therm} }{R_{therm}}[/tex]

This current remains fixed, since the fixed resistance is greater than the resistance of the thermistor throughout the range.

Second step.

Now we can change the temperature that is measured in the thermometer, for each temperature the voltage in the fixed resistance is measured, the resistance in the thermistor is calculated.

        [tex]R_{therm} = \frac{V_{source} - V_R}{i}[/tex]  

Third step.

We end with a graph of resistance versus temperature of the curve obtained, it is possible to understand the behavior of the resistance of the thermistor with temperature.

In conclusion with the characteristics of thermistors we can design an experiment to analyze the dependence of resistance with temperature:

      1) We set up the circuit

      2) We look for the current of the circuit and we keep it fixed.

      3) We look for the thermistor resistances for each temperature.

      4) Plot Resistance versus Temperature.

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Answer:

Increase 9 times

Explanation:

Answer:

T = C / R      t = time, C =constant of proportionality, R = rate

180 min = C / 400 gal/min      

C = 72000 gal      to empty tank

T = 72000 gal / 600 gal/min = 120 min = 2 hrs

Answer:

Explanation:

The current is the same through components connected in series. If you want to measure the current in a component, the ammeter must be placed in series.

__

The voltage is the same at the terminals of components connected in parallel. If you want to measure the voltage across a component, the voltmeter must be connected across the component, that is, in parallel with it.

[tex]\huge \mathbb \pink{ANSWER:}[/tex]

To measure the current through and the voltage across a resistor in a circuit, you should place the ammeter in series with the resistor and the voltmeter in parallel with the resistor.

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Answer:

Your answer would be D :)

Explanation:

An object with a higher mass will always fall faster due to gravity acting upon it. 15 N is more massive than the 10 N object, so statement D is correct.

Answer:

100.22

Explanation:

Have a great day!

Hi there!

We know that:

∑F = ma

If you tripled the net force and halved the mass:

3F = 1/2ma

Rearrange for 'a':

3F × 2 = ma

6F = ma

Thus, the new acceleration would be 6 times the original, or 36 m/s².

Answer:

No work is done on the weight. :3

To solve this we must be knowing each and every concept related to work and its calculation. Therefore, 250J is the work done in joule by the rope on the weight.

Work in physics is the energy delivered to or out of an item by applying force across a displacement. It is frequently expressed in its most basic form as the combination of displacement and force.

When a force is applied, it is said to produce positive work if it has a proportion in the orientation of the movement of a site of application. Work done is positive when the direction of force acting on the object and displacement of the object both are in the same direction.  

Mathematically,    

Work= force ×displacement

force =20 N

displacement=2.00 m

displacement=2×3.14×2=12.5m

substituting all the given values in the above equation, we get

Work= 20 ×12.5

Work =250J

Therefore, 250J is the work done in joule by the rope on the weight.

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#SPJ5

Answer:

It is the same

Explanation:

I Jsut know

From Rutherford  and Marsden scattering experiment, we can conclude that the particle shown in red will come straight back from the foil because it is deflected by the charge in the gold nucleus.

Rutherford proposed Planetary model atom, which visualized an atom to consists of a positively charged heavy core called the nucleus around which negatively charged electrons circle in orbits much as planets move round the sun

Thus, from Rutherford  and Marsden scattering experiment, we can conclude that the particle shown in red will come straight back from the foil because it is deflected by the charge in the gold nucleus.

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the answer is reflected or u can say repelled

Answer:

p=I×B

Explanation:

If a current 'I' flows through a given element in your circuit, losing voltage 'V' in the process, then the power 'P' dissipated by that circuit element is the product of that current and voltage.

Answer:

combine them all

6 m

Explanation:

[tex] \huge \bf༆ Answer ༄[/tex]

The given terms in the question are : -

We have been given the task to find out time (t)

The formula that can be used to find time is : -

[tex] \sf \: I = \dfrac{ Q}{t}{} [/tex]

Rearrange the formula,

[tex] \sf \: t = \dfrac { Q}{I }[/tex]

Solve for time (t)

[tex]꧁ \: \large \frak{Eternal \: Being } \: ꧂[/tex]

Hi there!

We can begin by using the derived kinematic equation:

[tex]t = \sqrt{\frac{2h}{g}[/tex]

Plug in the given values and let g = 9.8 m/s²:

[tex]t = \sqrt{\frac{2(20)}{g}} = \boxed{2.02 s}[/tex]

Now, we can solve for its final velocity using the equation:

[tex]v_f = v_i + at[/tex]

It is dropped from rest, so vi = 0 m/s.

[tex]v_f = at[/tex]

[tex]v_f = 9.8(2.02) = \boxed{19.796 m/s}[/tex]

Answer:

a. The ball would reach the ground in approximately [tex]2.02\; \rm s[/tex].

b. The velocity of the ball right before landing would be approximately [tex]19.8\; \rm m\cdot s^{-1}[/tex].

(Assumptions: the ball was dropped with no initial velocity; air resistance on the ball is negligible; [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex].)

Explanation:

Under these assumptions, the acceleration of this ball would be constantly [tex]a = g = 9.81\; \rm m\cdot s^{-2}[/tex] (same as the gravitational field strength) during the descent.

Displacement of the ball: [tex]x = 20.0\; \rm m[/tex].

Initial velocity of the ball: [tex]v_{0} = 0\; \rm m\cdot s^{-1}[/tex].

Let [tex]t[/tex] denote the duration of this descent.

The SUVAT equation [tex]x = (1/2)\, a\, t^{2} + v_{0}\, t[/tex] relates the known quantities [tex]a[/tex], [tex]x[/tex], and [tex]v_{0}[/tex] to the unknown [tex]t[/tex].

Substitute the known quantities into this equation and solve to find the value of [tex]t\![/tex]:

[tex]\displaystyle 20.0\; {\rm m} = \frac{1}{2}\times 9.81\; {\rm m\cdot s^{-2}} \times t^{2} + 0\; {\rm m \cdot s^{-1}} \times t[/tex].

[tex]\displaystyle 20.0\; {\rm m} = \frac{1}{2}\times 9.81\; {\rm m\cdot s^{-2}} \times t^{2}[/tex].

[tex]\displaystyle t^{2} = \frac{20.0\; \rm m}{(1/2) \times 9.81\; \rm m\cdot s^{-2}}[/tex].

Since [tex]t > 0[/tex]:

[tex]\begin{aligned}t &= \sqrt{\frac{20.0\; \rm m}{(1/2) \times 9.81\; \rm m\cdot s^{-2}}} \\ &\approx 2.01928\; \rm s\\ &\approx 2.02\; \rm s \\ & (\text{Rounded to 2 sig. fig.})\end{aligned}[/tex].

Since the acceleration of this ball is constant, the velocity of the ball right before landing would be:

[tex]\begin{aligned}v_{1} &= a\, t \\ &\approx 9.81\; \rm m\cdot s^{-2} \times 2.01928\; \rm s \\ &\approx 19.8\; \rm m \cdot s^{-1} \end{aligned}[/tex].

Hi there!

There are two torques acting on the system:

τ (Arthur) = RW = R(790N)

τ (Violet) = RW = 4(500N) = 2000N

∑τ = 0 = τ(Violet) - τ(Arthur)

Thus:

τ(Violet) = τ(Arthur)

2000 = 790R

R ≈ 2.53m

Answer:

c.

Explanation:

Answer:

here is your answer

1+1=2

2+2=4

4+4=8

Answer:

Solar eclipses occur when the Moon passes between Earth and the Sun, leaving a moving region of shadow on Earth's surface. Lunar eclipses occur when Earth passes between the Sun and the Moon, casting a shadow on the Moon. This is an annular eclipse.

(credits to "britannica")

Explanation:

Answer:

The time to reach max height is    h = 1/2 g t^2  

The time to fall is the same as that for an equivalent rise

m g h = 1/2 m v^2       KE vs PE

Here:   (Vf - VI) / g = t = -9  / -9.8 = .92 s

Answer:

facilitated diffusion

The energy of characteristic x rays tend to increase for heavier elements

because the charge of the nucleus increases.

X ray is a form of an electromagnetic spectrum and the energy is directly

proportional to the charge of the nucleus. This implies that an increase in

the nucleus charge will lead to a corresponding increase in x ray energy

and vice versa.

X rays energy is also usually more in heavier elements than when

compared to lighter elements

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