The unit of the two quantities used in the Newton's third law is Newton, which is denoted by N.
Newton's third law states that there is equal and opposite reaction for every action, In other words, There is an equal reaction for every force, but in the direction opposite to direction of the force. These forces are the quantities which are used in this law. The SI unit of reaction and force is Newton, which is equal to kg-m/s². It is denoted by N. It is formulated as: Force = mass × Acceleration.
F = m × a
Hence the correct option is C.
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determine the mass of silver chloride precipitated when 2 grams of sodium chloride solition reacts with silver trioxonitrate (v) solution
The mass of the precipitate is 4.86 g.
What is the mass of the precipitate?We have to know that the equation of the reaction is the first way that we can be able to obtain the mass of the precipitate by the use of the stoichiometry of the reaction.
The equation of the reaction in this case is; AgNO3 + NaCl ---> AgCl + NaNO3. Then we are told that we have that.
Mass of the NaCl = 2 g
Number of moles = 2 g/58 g/mol
= 0.034 moles
Since the reaction is 1:1, we can see that the number of moles of the product is 2 g
Mass of the precipitate = 0.034 moles * 143 g/mol
= 4.86 g
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if you are not wearing a seat belt and the car you are driving hits a fixed barrier, you will hit the steering wheel with some force. this is because .
This is because S the force of the collision has thrown you forward you continue moving forward as the car slows down the steering wheel starts moving backward during the collision.
You will hit the steering wheel with some force. this is because ?
If you are not wearing a seat belt and the car you are driving hits a fixed barrier, you will hit the steering wheel with some force.There are two forces acting upon the book. One force - the Earth's gravitational pull - exerts a downward force. The other force - the push of the table on the book (sometimes referred to as a normal force) - pushes upward on the book.Positive g-force in an upward direction produces downward weight on an object. Negative g-force is an acceleration in a downward direction that produces weight force in an upward direction. Consider a person in an elevator accelerating downward.Fapp. An applied force is a force that is applied to an object by a person or another object. If a person is pushing a desk across the room, then there is an applied force acting upon the object.To learn more about force refers to:
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A colorimetric method calls for the use of 0.1 mL of serum, 5 mL of reagents, and 4.9 mL of water. What is the dilution of the serum in the final solution? A. 1 to 5 B. 1 to 10 C.Hto-50 How many milliliters of concentrated HSO4 (sp. gr assay 97 % ) are required to prepare 10 L of 0.1 N H2SO4? A. 1.84 B. 9.20 C. 27.5 D. 54.4 1.84 g/mL; D. 1 to 100
The dilution of the serum in the final solution is 0.01 when a colorimetric method calls for the use of 0.1 mL of serum, 5 mL of reagents, and 4.9 mL of water and 54.5mL of H2SO4 required.
Given for a dilution the amount of serum is = 0.1mL
The amount of reagents is = 5mL
The amount of water is = 4.9mL
By adding these all we get the total volume as: 0.1 + 5 + 0.4 =10mL
The dilution of serum will be the amount of serum in total volume of solution such that:
The dilution amount = 0.1/10 = 1/100 = 0.01
Given the volume of H2SO4 = 10L
The % of mass = 97%
The density of H2SO4 = 1.80g/mL
The normality of H2SO4 is = 0.1N
The concentration of H2SO4 = M
M = (97 x 1.8/ 97 x 100) x 1000 = 18M
So, we know that M1V1 = M2V2 such that:
18 x V = 0.1 x 10
V = 5.44 x 10^-2L = 54.4mL
Hence the volume of H2SO4 required is 54.4mL
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most galaxies including our own milky weay have supermassive black holes at their centers. what is the speed of s2
The speed of star S2 is 0.0063c.
The orbital radius is 1000 au.
1 au is 1.49 × 10^11 m.
The orbital radius in meters is 1.49 × 10^14 m.
The time period is given to be 15 years.
The speed can be calculated using the formula,
Speed = 2π × {Radius}/{Time}
Speed = 2π × {1.49 × 10^14}/{15×365×24×60×60} m/s
Speed = (9.36 × 10^14)/(4.73 × 10^8) m/s
Speed = 1.9 × 10^6 m/s
To have it in terms of c, divide and multiply the speed by speed of light c.
Speed = [(1.9 × 10^6) / (3 × 10^8)] c
Speed = 0.0063c.
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--The complete question is,
Most galaxies, including our own Milky Way, have supermassive black holes at their centers. Recently, astronomers were able to track one star, named S2, as it orbited the black hole at the center of our galaxy. The star's actual orbit is elliptical, but we'll model it as a circular orbit. S2 has a period of 15 years and an orbit radius of 1000 au, where 1 au = 1 astronomical unit is the distance of the earth from the sun.
What is the speed of S2? Give your answer as a fraction of c, where c is the speed of light, c=3.0×10^8m/s.--
the wave travels to the right. describe how the vertical displacement of the knot varies the next complete cycle
The wave resumes its former position at halftime or the halfway point. A point's movement within the half wavelength is essentially what constitutes a half period. It is most positive at 1/4 period since 1/2 period equals the same position. 1/2 divided by 2 is 1/4 = max position.
The wave moves vertically downward (1/4 cycle to maximum negative displacement), upward (1/4 cycle to maximum positive displacement), down (1/4 cycle) to equilibrium position/zero displacement, and correctly referencing either the maximum positive or negative displacement, or correctly referencing fractions of the cycle.
The figure given in the attachment below explains the continuous progressive wave with a knot in the rope, the direction of motion, equilibrium position.
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now, let's see what happens when the cannon is high above the ground. click on the cannon, and drag it upward as far as it goes (15 m m above the ground). set the initial velocity to 14 m/s m / s , and fire several pumpkins while varying the angle. for what angle is the range the greatest?
The angle that maximizes the range is 45 degrees.
The range of a projectile is the horizontal distance it covers from the starting point to the landing point. When a cannon is high above the ground, the projectile is subject to the effects of air resistance and gravity. The range is influenced by the initial velocity and launch angle.
The formula for the range of a projectile is given by R = (v^2 sin(2Θ))/g, where v is the initial velocity, Θ is the launch angle, and g is the acceleration due to gravity.
To find the launch angle that maximizes the range, we can take the derivative of R with respect to Θ and set it equal to zero. This will give us the maximum value of R, which corresponds to the angle that maximizes the range.
The result of this calculation shows that the launch angle that maximizes the range is 45 degrees. This means that if the cannon is fired at an angle of 45 degrees, the projectile will travel the greatest possible distance before hitting the ground.
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3. Two identical balls of mass 1.8 kg hit head on and rebounded from each other. One ball was traveling at 9.2 m/s and the other ball at 8.7 m/s in the opposite direction. After collision, the second ball rebounded at a speed of 9.1 m/s. What is the speed of the first ball?
The speed of the first ball is 8.6 m/s in opposite direction.
What is the speed of the first ball?
The final speed of the first ball is calculated by applying the principle of conservation of linear momentum as shown below.
mu₁ + mu₂ = mv₁ + mv₂
where;
m is the mass of the ballsu₁ is the initial velocity of firstu₂ is the initial velocity of secondv₁ is the final velocity of firstv₂ is the final velocity of second1.8 x 9.2 - 1.8 x 8.7 = 1.8v₁ + 1.8 x 9.1
0.9 - 16.38 = 1.8v₁
1.8v₁ = -15.48
v₁ = ( - 15.48 ) / 1.8
v₁ = -8.6 m/s
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body of 2.0 kg mass makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed. calculate the mass of the struck body.
The mass of the struck body is equal to 1/8th the mass of the first body.
In an elastic collision, both momentum and kinetic energy are conserved. The momentum of the system before the collision is equal to the momentum of the system after the collision.
Let's consider the mass of the struck body m2. The initial momentum of the first body before the collision is:
p1 = m1 * v1
where m1 is the mass of the first body (2.0 kg) and v1 is its initial velocity (v1 = v).
After the collision, the final momentum of the two bodies is:
p_total = m1 * v1 / 4 + m2 * v2
where v2 is the velocity of the second body after the collision.
Since the second body was initially at rest, its velocity before the collision was zero. The conservation of momentum equation can be written as:
m1 * v1 = m1 * v1 / 4 + m2 * v2
By solving m2, we get:
m2 = m1 * v1 / (4 + v1)
m2 = 2.0 kg * v1 / (4 + v1)
Since the velocity of the first body after the collision is 1/4th of its original speed, v1 / 4 = v1. Substituting this into the equation for m2, we get:
m2 = 2.0 kg * v1 / (4 + v1)
m2 = 2.0 kg * v1 / (4 + v1)
m2 = 2.0 kg * v1 / 8
m2 = 2.0 kg * v1 / 8
So, the mass of the struck body is equal to 1/8th the mass of the first body.
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the meridian is a north-south line on the sky that passes through the zenith. what is the right ascension of a star that is on the meridian at midnight on the 21st of september?
The right ascension of a star that is on the meridian at midnight on September 21st is 0 hours.
The right ascension of a star that is on the meridian at midnight on September 21st can be calculated based on the position of the sun on that day and time. On September 21st, the sun is at the vernal equinox, which marks the start of spring in the Northern Hemisphere and the beginning of the astronomical year. At the vernal equinox, the sun is at the celestial equator and its right ascension is zero hours.
As the meridian is a north-south line on the sky that passes through the zenith, a star that is on the meridian at midnight on September 21st will have the same right ascension as the sun at the vernal equinox, which is 0 hours.
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when a single point charge is located in a region of space where there is an electric field, it experiences a:
when a single point charge is located in a region of space where there is an electric field, it experiences a force.
The electric field exerts a force on the charged particle.
When the single point charge is locate in a space where there is an electric field, we tends to see that there is a force exerted on the particle by the field and the charge particle will experience that force in a particular direction as per the nature of charge on the particle.
If the charge if positive than the force will be in the parallel to the electric field lines but the force will be anti parallel if the charges particle is negative.
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The concentration of enzyme for each experiment was 5.0 uM. What is the kcat for the reaction at ph 4.5 when no chloride added when compound 3 is substrate?
DIDN'T GET WRONG BUT WANTED TO BETTER UNDERSTAND
2.5E-2 s^-1
Rate for compound 3 is 125 nM/s, and rate of product formation did not vary over time for first 5 minutes implies enzyme was saturated with substate.
Kcat=Vmax/[E]= 125 nM/s/(5.0 uM)= 2.5E-2 s^-1
The kcat for the reaction of the enzyme at pH 4.5 when no chloride is added and compound 3 is the substrate is 2.5E-2 s^-1. This value is calculated using the formula Kcat = Vmax / [E], where Vmax is the maximum velocity of the reaction, and [E] is the concentration of the enzyme.
In this experiment, the concentration of the enzyme is 5.0 uM, and the rate of product formation for compound 3 is 125 nM/s. As the rate of product formation did not vary over the first 5 minutes, it can be assumed that the enzyme was saturated with substrate. Using this information, the kcat can be calculated as:
Kcat = Vmax / [E]
= 125 nM/s / (5.0 uM)
= 2.5E-2 s^-1
Therefore, the kcat for the reaction at pH 4.5 when no chloride is added and compound 3 is the substrate is 2.5E-2 s^-1.
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during the collision, how significant was friction? consider that any impulse effects from friction would be based upon the magnitude of the friction and the time the friction has to act.
During the collision, friction was significant in the sense that it had an effect on the impulse of the collision.
The magnitude of the friction and the time it had to act are important factors in determining the effect of friction on the impulse of the collision.
In general, the greater the magnitude of the friction, the greater the effect on the impulse of the collision, and the longer the time that friction has to act, the greater the effect on the impulse of the collision.
Static friction occurs when the two objects in the collision are not moving relative to each other, and the magnitude of static friction depends on the surface roughness and the normal force between the two objects. The longer static friction has to act, the greater the total impulse of the collision.
Kinetic friction occurs when the two objects in the collision are moving relative to each other, and the magnitude of kinetic friction depends on the coefficient of kinetic friction between the two objects. The longer kinetic friction has to act, the greater the total impulse of the collision.
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if the same amount of sunlight is hitting an area of land and an are of water, which area would heat more rapidly
Water would heat more rapidly than land. The area of water on the land would heat more rapidly.
At the point when the sun is radiating on both an area of land and an area of water, the land will warm up quicker. This is on the grounds that land has a lower capacity to store heat contrasted with water. Consider it like this, when you empty high temp water into a container, it warms up quicker contrasted with when you empty it into a pot. The pot has a higher capacity to store heat, so it requires additional significant investment to warm up, yet it likewise remains warm for a more extended time frame.
The equivalent goes for the sun radiating ashore and water. The land warms up rapidly in light of the fact that it has a lower capacity to store heat, while the water takes more time to warm up however holds heat for a more extended timeframe. Thus, if you need to heat up rapidly on a bright day, it's ideal to track down a decent fix of grass or sand as opposed to hopping into a lake or sea.
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when converted to a voltage source, the equivalent voltage is a. 3.33 v b. 3000 v. c. 30 v. d. 0.3 v
When converted to a voltage source, the equivalent voltage is calculated to be 30 V.
From the figure, it is clear that, Current is 3 A and Resistance is 10 Ω.
As explained by Ohm's law, we know it is mathematically represented as, V = I R.
where, V is voltage
I is current
R is resistance
Let us follow these steps to solve this problem.
Convert the current source to equivalent voltage source between the load terminals.
Equivalent voltage source will have voltage source in series with resistance
Voltage source value = I × Rs
So, V = 3 × 10 = 30 V
Thus, the equivalent voltage is calculated to be 30 V.
The given question is incomplete. The complete question is 'See the circuit figure below.' The picture is given in the attachment below.
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A hurricane wind blows across a 7.70 m × 16.3 m flat roof at a speed of 160 km/hr . What is the pressure difference between inside the house and outside? How much force is exerted on the roof?
The pressure difference between inside the house and outside is 99701Pa and the force exerted on the roof is 12.5 x 10^6N.
The length of flat roof (L) = 7.70m
The width of roof (B) = 16.3m
The speed of hurricane (v) = 160km/h = 160 x 5/18 =
Let the pressure difference between inside the house and outside = ΔP
The atmospheric pressure = Pa = 101300Pa
Pressure inside the house = Pi
We know that Pi = pgd where g is gravitational constant, d is the area and p is the pressure.
Pi = 1.3 x 9.8 x (7.7 x 16.3) = 1598.99
(a) The pressure difference is = ΔP = Pa - Pi
ΔP = 101300Pa - 1598.99 = 99701Pa
(b) The force exerted on the roof = F
We know that pressure = force / area then F = P x A
Such that F = 99701Pa x (7.7 x 16.3)m^2 = 12.5 x 10^6N
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Find the net force acting on a box that has a force of 50 newtons pulling at an angle of 30 degrees and a force of 90 newtons pulling at a 160 degree angle
Answer:
Approximately [tex]69\; {\rm N}[/tex] at approximately [tex]126^{\circ}[/tex].
Explanation:
Assume that both angles in the question are relative to the positive [tex]x[/tex]-axis (towards the positive horizontal direction.)
Horizontal component ([tex]x[/tex]-component) of the two forces would be:
[tex](50\; {\rm N}) \, \cos(30^{\circ}) = \left(25\, \sqrt{3}\right)\; {\rm N} \approx 43.3\; {\rm N}[/tex].[tex](90\; {\rm N}) \, \cos(160^{\circ}) \approx (-84.6) \; {\rm N}[/tex].Note that the [tex]x[/tex]-component of the [tex]90\; {\rm N}[/tex] force is negative since this components points away from the positive [tex]x\![/tex]-direction.
Hence, the net force in the [tex]x[/tex]-component would be:
[tex](50\; {\rm N}) \, \cos(30^{\circ}) + (90\; {\rm N}) \, \cos(160^{\circ}) \approx (-41.3) \; {\rm N}[/tex].
(Again, this component is negative since it points away from the positive [tex]x[/tex]-axis.)
Similarly, the vertical component ([tex]y[/tex]-component) of the two forces would be:
[tex](50\; {\rm N}) \, \sin(30^{\circ}) = 25\; {\rm N}[/tex].[tex](90\; {\rm N}) \, \sin(160^{\circ}) \approx 30.8 \; {\rm N}[/tex].Hence, the net force in the [tex]y[/tex]-component would be:
[tex](50\; {\rm N}) \, \sin(30^{\circ}) + (90\; {\rm N}) \, \sin(160^{\circ}) \approx 55.8\; {\rm N}[/tex].
Refer to the diagram attached. The resultant net force is the vector sum of the components. Apply the Pythagorean Theorem to find this net force:
[tex]\begin{aligned}(\text{net force}) &= \sqrt{(\text{$x$-component})^{2} + (\text{$y$-component})^{2}} \\ &\approx \sqrt{(-41.3)^{2} + (55.8)^{2})}\; {\rm N} \\ &\approx 69\; {\rm N}\end{aligned}[/tex].
Find the angle of this net force relative to the positive [tex]x[/tex]-axis using the inverse cosine function [tex]\arccos[/tex]:
[tex]\begin{aligned}\arccos\left(\frac{(\text{$y$-component})}{(\text{net force})}\right) &\approx \arccos\left(\frac{55.8\; {\rm N}}{69\; {\rm N}}\right) \\ &\approx 126^{\circ}\end{aligned}[/tex].
(The units might need to be converted into degrees.)
A car slows down at -5. 00 m/s^2 until it comes to a complete stop after travelling 15. 0 m. What is the initial speed of the car?
The initial speed of the car is 12.25 m/s, if it slows down at an deceleration of -5 m/s², and completely stops after travelling 15 meter.
We can use the third equation of motion to solve for the initial speed:
v² = u² + 2as
Where, final velocity is v= 0 m/s,
u is the initial velocity,
a is the acceleration = -5.00 m/s², and
s is the distance travelled = 15.0 m
Rearranging the equation:
u² = v² - 2as
u = √(v² - 2as)
Substituting the values:
u = √(0² - 2 x (-5.00) x (15.0))
u = √150
u = 12.25 m/s
So, the initial speed of the car is 12.45 m/s.
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how do the speeds v0 , v1 , and v2 (at times t0 , t1 , and t2 ) compare?how do the speeds , , and (at times , , and ) compare? v0
The speeds v0, v1, and v2 (at times t0, t1, and t2) comparison is c: v0 = v2 > v1 > 0
ABOUT SPEEDSpeed is the amount of distance traveled by objects in each unit of time. Velocity is a type of quantity that depends on direction, so velocity is included in a vector quantity.
For one-dimensional motion, the direction of velocity can be expressed with a positive or negative sign. The concept of speed is divided into two, namely instantaneous speed and average speed.
Instantaneous speed is the speed of one object at a time, temporarily The average speed is the result of transfers at intervals each time. This speed can be measured with an instrument called a velocitometer.Your question is incomplete but most probably your full question was:
How do the speeds v0, v1, and v2 (at times t0, t1, and t2) compare?
a: v0 = v1 = v2 > 0
b: v0 = v2 > v1 = 0
c: v0 = v2 > v1 > 0
d: v0 > v1 > v2 > 0
e: v0 > v2 > v1 = 0
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2. in the ce amp, explain how re1 limits the gain? (hint, resistance reflection rule and the derivation for this circuit’s av)
Three purposes are served by the unbypassed emitter resistor RE1: 1. Set the required value for the voltage gain: The voltage gain of the amplifier reaches its maximum for the given circuit when RE1 = 0.
The gain can be reduced to a desirable value by setting RE1 to a nonzero value. By including a single resistor in the transistor's emitter circuit as shown, the bias voltage for the amplifier can be stabilized. The Emitter Resistance, or RE, is the name of this resistance. The potential divider network made consisting of the two resistors, R1, R2, and the power supply voltage Vcc, as depicted with the current flowing through both resistors, determines the quiescent Base voltage (Vb). The overall resistance RT will then be equal.
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an automobile traveling at 70.0 km/h has tires of 80.0 cm diameter. (a) what is the angular speed of the tires about their axles? (no response) 48.6 rad/s (b) if the car is brought to a stop uniformly in 35.0 complete turns of the tires (without skidding), what is the magnitude of the angular acceleration of the wheels? (no response) 5.37 rad/s2 (c) how far does the car move during the braking?
(a) The angular speed of the tires an automobile traveling at 70.0 km/h has tires of 80.0 cm diameter about their axles = 48.611 rad/s
(b) The magnitude of the angular acceleration of the wheels if the car is brought to a stop uniformly in 35.0 complete turns of the tires (without skidding) = 5.37 rad/s²
(c) The car moves during the braking = 87.92 m
The angular speed of the tire can be calculated using the linear velocity and radius.
We can find the remaining unknowns by using angular speed and rotational motion kinematic equations. We can calculate the distance traveled by using the formula for the length of the curve in terms of angle and radius.
The equation for angular speed:
ω = v/r
Where:
ω = angular speed
v = velocity
r = radius
70.0 km/h = 19.4 m/s
Diameter = 80 cm
Radius = 0.4 cm
Hence,
(a) the angular speed:
ω = (19.4 / 0.40)
= 48.611 rad/s
(b) the magnitude of the angular acceleration of the wheels:
θ = 35.0 x 2[tex]\pi[/tex] rad
= 219.8 rad
For rotational motion, use the third equation of motion.
ω² = ω₀² + 2[tex]\alpha[/tex]θ
Hence, the angular acceleration of the wheels:
[tex]\alpha[/tex] = (48.611)² / (2 x 219.8)
= 5.37 rad/s²
(c) the car move during the braking:
s = rθ
= 0.4 x 219.8
= 87.92 m
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On a car with disc/drum brakes, the front brakes grab quickly when light pedal pressure is applied. This problem could be caused by a bad:
A. proportioning valve
B. pressure safety switch
C. metering valve
D. residual check valve
On a car with disc/drum brakes, the front brakes grab quickly when light pedal pressure is applied. This problem could be caused by a bad (C). metering valve is correct option.
A brake drum is a revolving cylinder-shaped component that presses outwardly on a set of shoes or pads to create friction.
Drum brakes are brakes where the inside surface of the drum is pressed by the shoes. It is commonly referred to as a clasp brake when shoes press against the drum's exterior. Though such brakes are extremely uncommon, they are frequently referred to as pinch drum brakes when the drum is pinched between two shoes, much like a traditional disc brake. A band brake, a related design, wraps a flexible belt or "band" around the outside of a drum.
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two parallel-plate capacitors with different plate separation but with the same capacitance are connected in series to a battery. both capacitors are air-filled. when equilibrium is established, the quantity that is not the same for both capacitors is
The electric field between the plates will not be the same for both capacitors.
What is electric field?An electric field is a region of space which exerts an influence on the motion of charged particles, or the behavior of electric currents. It is generated by electric charges, either stationary or in motion, and can be either static, generated by a single charge or changing, generated by multiple charges. Electric fields are important because they are responsible for the attraction and repulsion of charged particles, such as electrons and protons, and can be used to create electric and magnetic fields. Electric fields can also be used to generate an electric current.
This is because the electric field is inversely proportional to the plate separation. Therefore, the electric field is stronger in the capacitor with the smaller plate separation and weaker in the capacitor with the larger plate separation.
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a wood block has a mass of 12.4 g and a density of 45.02 g/cm^3. find the volume of the block in cm^3.
Answer:
The volume of the block can be found by using the formula:
Density = Mass / Volume
Rearranging the formula to solve for volume:
Volume = Mass / Density
So,
Volume = 12.4 g / 45.02 g/cm^3
= 0.2747 cm^3
from its resting position, how long does it take the weight to bounce one direction, then the other, and then back to its resting position?
The weight takes 3 seconds to bounce one direction, then the other, and then back to its resting position.
From the table we can see that, initially at t = 0 s, the weight was at rest(h = 0), then in next 0.75 sec it moves upward up to a height of h = 15 cm, then at time, t = 1.5 sec its position was h = 0. But is didn't bounce back. So it moves further in downward direction and reach, h = -15 cm in 2.25 sec from the starting, then it bounced back and returns to its original position in 3 seconds, starting from the t = 0.
Hence the answer is 3 seconds.
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The given question is incomplete, the complete question is:
"The table shows the height in centimeters, that a weight bouncing from a spring would achieve if there were no friction, for a given number of seconds.
(TABLE SHOWN BELOW ON PICTURE)
From its resting position, how long does it take the weight to bounce one direction, then the other, and then back to its resting position?"
Prove that clockwise +/- anticlockwise = 0
If something is moving anticlockwise, it is moving in the opposite direction to the direction in which the hands of a clock move.
How do you determine the clockwise and counterclockwise directions?The terms clockwise and anti-clockwise are used to indicate the direction of a turn. So, which direction is clockwise is Clockwise is a turn to the right that follows the direction of a clock’s hands, while anti-clockwise is a turn to the left that opposes the direction of a clock’s hands.
We can detect which way the vector is heading by computing the dot product of (0,0,1). If (ab)(0,0,1) > 0, the points are anticlockwise. If (ab)(0,0,1)0, the points are arranged clockwise. If the vectors are same, they are parallel to one other. The loading is statically identical to the load FC applied at the beam’s center, plus a clockwise moment of magnitude FCt/2 and an anticlockwise moment of magnitude FCt/2.
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what is the repulsive force between two pith balls that are 7.2 cm apart and have equal charges of −25 nc? f: n
The repulsive force between two pith balls that having equal charges is calculated to be 108.51 × 10⁻⁵ N.
From the expression of Coulomb's law, the repulsive force between the two charges can be determined.
Mathematically, it is represented as, F = k q₁ q₂/r²
Where,
The value of k is 9.00×10⁹ Nm²/C²
Magnitude of equal charges is given as,
q₁ = q₂ = −25 nc = -25 × 10⁻⁹ c
Distance between pith balls r = 7.2 cm = 7.2 × 10⁻² m
Putting the values into the above equation, we have,
F = k q₁ q₂/r² = [(9 ×10⁹)(-25 × 10⁻⁹)(-25 × 10⁻⁹)]/(7.2 × 10⁻²)² = [25×25×9× 10⁻⁹]/(7.2 × 10⁻²)² = (5625 × 10⁻⁹)/(51.84 × 10⁻⁴) = 108.51 × 10⁻⁵ N
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what happens to h when path 2 is followed rather than path 1
When path 2 is followed, the value of h will depend on the actions taken along that path. It could remain the same, increase, decrease, or even become undefined.
A property whose value relies on the route used to get there is referred to by the thermodynamics phrase "path function." To put it another way, a path function is dependent on the route travelled to go from a starting state to a final one. path operation Physics uses mathematics to explain how a particle or object moves. In this method, an object's route is determined at certain periods using a set of equations. It is employed in a number of disciplines, including robotics, optics, acoustics, and mechanics.
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Consider a case where the potential energy of two interacting molecules is E=1.3 kJ/mol. Choose all options that apply:(A) The net force on either molecule is zero.(B) The molecules are an infinite distance apart.(C) The forces between molecules are attractive.(D) The forces between molecules are repulsive.
(C) The forces between the molecules are attractive.
The simplest form of matter, molecules are made up of two or more atoms that are chemically bound. They are a chemical substance's tiniest particle that nonetheless has its chemical properties. They can be straightforward or intricate, and they come in a range of sizes and shapes. Two or three atoms make up simple molecules like oxygen or water, but thousands of atoms make up complex compounds like proteins or DNA. The boiling point, melting point, and reactivity of an object are all determined by the interactions between its molecules. The type of bonding between the atoms and the overall electric charges of the molecule determine the attractiveness or repulsion of these interactions.
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an object on a rope is lowered at a steadily decreasing speed. is the magnitude of the tension force on the object greater than, less than, or equal to the magnitude of the weight force on the object?
The tension on the object on a rope that is lowered at a steadily decreasing speed must be greater than the magnitude of the weight force on the object.
Tension is the pulling force transmitted through a rope, string, wire, or other similar object when it is pulled. Weight, though, is known as the force of gravity.
Recall that force can cause acceleration to an object. Therefore, if the object is accelerating, there must be a net force acting on the object. In physics, acceleration is defined as the change in velocity/direction, which includes decreasing speed. When the pulling speed is steadily decreased, the tension in the rope will have to be greater than the force of gravity to cancel out the weight of the object.
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a satellite is orbiting a planet at distance r above its surface an has period of t. what would the distance above the surface have to be in order
The period of a satellite is proportional to the square root of its orbital radius, so if the period of a satellite is t at a distance of r, its period would be 8t at a distance of r × sqrt(8).
The formula that relates a satellite's period and its distance from the planet's surface is given by Kepler's Third Law of Planetary Motion, which states that:
T^2 = k × r^3
where T is the period, r is the distance from the planet's surface, and k is a constant that depends on the planet's mass. By rearranging this equation, we can find the new distance for an 8 times greater period:
r' = [tex](\frac{T}{8T} )^{\frac{2}{3} }[/tex] × r
r' = [tex](\frac{1}{8})^{\frac{2}{3} }[/tex] × r
r' = [tex]\frac{1}{2}[/tex] × r
So the new distance would have to be half the original distance in order for the period to become 8 times greater.
Complete question:
a satellite is orbiting a planet at a distance r above its surface and has a period of t. what would the distance above the character have to be in order for the period to become eight times greater?
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