What force is needed to accelerate an object 5 m/s if the object has a mass of 10kg?

Answers

Answer 1

Answer: The force that is needed to accelerate an object 5 m/s if the object has a mass of 10kg 50N because you multiply 5 and 10

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Which change of state do particles in a material become farther apart?

Answers

Particles become farther apart from each other in the change of state between liquid and gas.
Hope that helped =)

Hello Please Help With A couple Science questions

Answers

#1 A) Long wavelength and high frequency.

#2) A) Brightness.

#3) D) They are used to examine things.

#4) B) Compression and visible light waves.

For the first one is A.
Second one is A.
Third one is D.
Forth one is B.

Hope this helps:)

Does a force cause motion or a change in motion

Answers

it changes the motion as Newton's second law of motion states that a force, acting on an object, will change its velocity by changing either its speed or its direction or both. If your basketball goes rolling into the street and is hit by a bike, either the ball will change direction or its speed or both.

A graduated cylinder is filled to an initial volume of 12.7 ml. A rock Is dropped into the graduated cylinder. The final volume of the graduated cylinder is 18.2ml. What is the Rocks volume in both ml and cm3? What method was used to determine this?

Answers

Answer:

V = 5.5 mL

Explanation:

The volume filled in the graduated cylinder is 12.7 mL

now when a stone is dropped into the cylinder then the volume of liquid is raised to final level of 18.2 mL

so as per the theory of given by Archimidies we can say that the volume of the object is exactly same as the volume displaced by the object

So here the volume displaced by the object is given as

$V_(displaced) = V_f - V_i$

$V_(displaced) = 18.2 mL - 12.7 mL$

$V_(displaced) = 5.5 mL$

so the volume of the object is given as

$V = V_(displaced)$

$V = 5.5 mL$

First of all, ml and cm^3 are equal. The volume of the rock is 5.5 ml. This is found by subtracting the volume of just the water from the combined volume of the rock and the water leaving you with just the volume of the rock. This is method is using water displacement.

Three resistors (R1 = 120 Ohms, R2 = 330 Ohms, and R3 = 240 Ohms) and an ideal inductor (L = 1.6 mH) are connected to a battery (V = 9 V) through a switch as shown in the figure below.The switch has been open for a long time before it is closed at t = 0. At what time t0, does the current through the inductor (I3) reach a value that is 63% of its maximum value?

Answers

The time at which the current through the inductor reaches 63% of the maximum current is 4.85 $\mu$ s

What is current?

The current is defined as the flow of the charge in the circuit is is the rate of flow of the charge.

At t=0 s there is no current in the circuit because the switch is not closed and the circuit is not complete. The current across the LR circuit increases exponentially, when the switch is closed, and becomes steady after a certain time.

Given that

The value of resistor is .120 ohm

The value of resistor is .330 ohm

The value of resistor is .240ohm

The value of the inductor is .1.6 mh

The voltage applied across the circuit is .9 V

To determine the value of effective resistance of this circuit we need to look at the circuit from inductor’s side i.e., from inductor’s side the resistors is connected in series with the parallel combination of resistors

The effective resistance of the circuit is:

$R_(eff)=R_a+(R_1* R_2)/(R_1+R_2)$ …… (1)

Here, $R{eff$ is the effective resistance of the circuit. Now substituting the values.

$R_(eff)=240+(120* 330)/(120+330)=328\ ohm$

The current through the inductor is:

$i=i_o(1-e^{(tR_(eff))/(L)})$ ...... (2)

Here, is the current across the inductor, io is the maximum current in the circuit and L is the inductance across the inductor.

The current across the inductor is equal to the 63% of the maximum current in the circuit.

The current across the inductor is:

i=0.63io

Substitute 0.63io for 328 ohm , for 1.6 mH and for L in equation (2).

$0.63 i_o=i_o(1-e^(-t(328))/(1.6)})$

Simplify the above expression.

$e^((-2.05*10^6))=0.37$

Taking natural log on both sides and simplify.

$t=4.85* 10^(-6)\ s$

$t=4.85 \mu s$

Thus, the time at which the current through the inductor reaches 63% of the maximum current is $t=4.85 \mu s$

To know more about current follow

brainly.com/question/24858512

The time at which the current through the inductor reaches 63% of the maximum current is $\fbox{\begin\n4.85 \mu s\end{minispace}}$ or $\fbox{\begin\n4.85 * {10^( - 6)}\,{\text{s}}\end{minispace}}$ .

Further Explanation:

At $t = 0\,{\text{s}}$ there is no current in the circuit because the switch is not closed and the circuit is not complete. The current across the LR circuit increases exponentially, when switch is closed, and becomes steady after certain time.

Given:

The value of resistor is $120\,\Omega$ .

The value of resistor is $330\,\Omega$ .

The value of resistor is $240\,\Omega$ .

The value of the inductor is $1.6\,{\text{mH}}$ .

The voltage applied across the circuit is $9\,{\text{V}}$ .

Concept:

To determine the value of effective resistance of this circuit we need to look at the circuit from inductor’s side i.e., from inductor’s side the resistors ${R_3}$ is connected in series with the parallel combination of resistors ${R_1}$ and ${R_2}$ .

The effective resistance of the circuit is:

$\fbox{\begin\n{R_(eff)} = {R_3} + \frac{{{R_1} * {R_2}}}{{{R_1} + {R_2}}}\end{minispace}}$ …… (1)

Here, ${R_(eff)}$ is the effective resistance of the circuit.

Substitute the $120\,\Omega$ for ${R_1}$ , $330\,\Omega$ for ${R_2}$ and $240\,\Omega$ for ${R_3}$ in equation (1).

$\begin{aligned}{R_(eff)}&=240\,\Omega+\frac{{\left( {120\,\Omega } \right) * 330\,\Omega }}{{120\,\Omega +330\,\Omega }} \n&=328\,\Omega\n \end{aligned}$

The current through the inductor is:

$\fbox{\begin\ni = {i_0}\left( {1 - {e^{ - \frac{{t{R_(eff)}}}{L}}}} \right)\end{minispace}}$ ...... (2)

Here, $i$ is the current across the inductor, ${i_0}$ is the maximum current in the circuit and $L$ is the inductance across the inductor.

The current across the inductor is equal to the 63% or times of the maximum current in the circuit.

The current across the inductor is:

$i = 0.63{i_0}$

Substitute $0.63{i_0}$ for $i$ , $328 \Omega$ for ${R_(eff)}$ and $1.6\,{\text{mH}}$ for $L$ in equation (2).

$0.63{i_0} = {i_0}\left( {1 - {e^{ - \frac{{t\left( {328\Omega } \right)}}{{\left( {1.6\,{\text{mH}}} \right)}}}}} \right)$

Simplify the above expression.

${e^{ - \left( {2.05 * {{10}^6}} \right)t}}= 0.37$

Taking natural log on both sides and simplify.

$\begin{aligned}t&=4.85\, * {10^( - 6\,)}\,{\text{s}} \n&=4.85\mu \text{s}}\n\end{aligned}$

Thus, the time at which the current through the inductor reaches 63% of the maximum current is $\fbox{\begin\n4.85 \mu s\end{minispace}}$ or $\fbox{\begin\n4.85 * {10^( - 6)}\,{\text{s}}\end{minispace}}$ .

Learn more:

1. Conservation of energy brainly.com/question/3943029

2. Average translational energy brainly.com/question/9078768

3. The motion of a body under friction brainly.com/question/4033012

Answer Details:

Grade: Middle School

Subject: Physics

Chapter: Current Electricity

Keywords:

Resistor circuit, LR circuit, current, current across inductor, time constant, 4.85 microsecond, 4.85 microsec, 4.85 micros, 4.85*10-6 s, 4.85*10^6 s, 4.85*10-6 sec, 4.85*10^6 sec.

SQUA is a square. If side SQ has endpoints S(-2,8) and Q(4,10), find the perimeter to two decimal places?HURRY!!

Answers

Answer : The perimeter of square is, 25.28 unit.

Step-by-step explanation :

First we have to calculate the distance of SQ.

Using distance formula:

$d=√((x_2-x_1)^2+(y_2-y_1)^2)$

where,

d = distance between the two coordinates

x and y are the coordinates.

To calculate the distance of SQ:

$d=√((x_2-x_1)^2+(y_2-y_1)^2)$

$d=√((4-(-2))^2+(10-8)^2)$

$d=√((36)^2+(4)^2)$

$d=√(40)$

$d=6.32$

As we know that SQUA is a square that means all sides are equal.

So, Side SQ = Side QU = Side UA = Side SA = 6.32

Now we have to calculate the perimeter of the square.

Perimeter of square = Side SQ + Side QU + Side UA + Side SA

Perimeter of square = 6.32 + 6.32 + 6.32 + 6.32

Perimeter of square = 25.28 unit.

Therefore, the perimeter of square is, 25.28 unit.

Final answer:

To find the perimeter of the square, calculate the length of one side using the distance formula and multiply it by 4.

Explanation:

To find the perimeter of a square, we need to know the length of one side and multiply it by the number of sides. In this case, the length of one side can be found using the distance formula, which is the square root of the sum of the squared differences in the x-coordinates and y-coordinates of the endpoints of the side. So, the length of side SQ is:

Length of SQ = √((-2 - 4)2 + (8 - 10)2) = √((-6)2 + (-2)2) = √(36 + 4) = √40 ≈ 6.32

The perimeter of the square is found by multiplying the length of one side by 4, since a square has 4 equal sides:

Perimeter = 4 * Length of SQ = 4 * 6.32 = 25.28

Therefore, the perimeter of the square to two decimal places is 25.28 units.

Learn more about Perimeter of a Square here:

brainly.com/question/29192128

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