Answer:
Rubber
Glass
Pure water
Oil
Air
Diamond
Dry wood
Dry cotton
Plastic
Asphalt
Other strong insulators include:
Fiberglass
Dry paper
Porcelain
Ceramics
Quartz
Explanation:
I hope this helps
How much net force is needed to accelerate a 200 kg satellite 9.8 m/s2 ?
Answer:
1960 NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 200 × 9.8
We have the final answer as
1960 NHope this helps you
Which force results from charged particles
Answer:
electromagnetic force
Explanation:
high heels shoes are more likely to damage floor than flat shoes . do you agree? explain.
yesssss I am agree ......
Answer:
Yes, I agree. The pressure exerted upon the floor is larger for high heel shoes than flat shoes, hence causing more damage to the floor.
Explanation:
Pressure is force/area. High heel shoes have a sharp point at which they touch the floor, while flat shoes have a flatter surface at which they touch the floor. Due to you exerting the same force anyway, the area is smaller for high heel shoes.
As you might know, the smaller a denominator is, the larger the number is. As an example, 3/1 is larger than 3/2. 3/1 is 3, and 3/2 is 1.5. Hence, the force is larger for high heel shoes, and it will cause more damage.
Hope this helped!
What is the difference
between interplanting and intercropping?
Answer:
Interplanting is the practice of planting a fast-growing crop between a slower-growing one to make the most of your garden space. ... Intercropping enables you to boost the health of all plants because it can enhance soil fertility and cooperation among different plants.
Intercropping refers to large, often commercial, farming operations and is the procedure of planting alternating crop rows. Growing two or more crops in close proximity is known as intercropping.
What is interplanting?Starting to grow two or more crops in close vicinity is known as intercropping.
The most common goal of intercropping is to increase yield on a given plot of land by utilizing resources that would otherwise go unused by a single crop.
Intercropping is the cultivation of more than one crop in the same space at the same time. Intercropping is the simultaneous planting of two or more species in a mixture, or the interplanting of one species during the growth of another.
The terms are frequently and regionally used interchangeably. Intercropping is the practice of planting alternating crop rows in large, often advertising, farming operations.
Thus, this is the difference between intercropping and interplanting.
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During a tennis match, a player serves the ball at 23.6 m/s, with the center of the ball leaving the racquet horizontally 2.37 m above the court surface. The net is 12 m away and 0.90 m high. When the ball reaches the net,
(a) does the ball clear it and
(b) what is the distance between the center of the ball and the top of the net? Suppose that, instead, the ball is served as before but now it leaves the racquet at 5.00° below the horizontal. When the ball reaches the net,
(c) does the ball clear it and
(d) what now is the distance between the center of the ball and the top of the net?
Answer:
a
Yes it clears
b
[tex]b= 0.19 \ m[/tex]
c
No it does not clear
d
[tex]z= 0.86 \ m[/tex]
Explanation:
From the question we are told that
The speed at which the player serves the ball is [tex]v = 23.6 \ m/s[/tex]
The height of the ball above the ground is [tex]h = 2.3 7 \ m[/tex]
The distance of the net is [tex]d = 12 \ m[/tex]
The height of the net is [tex]H = 0.9 \ m[/tex]
Generally the time taken for the ball to reach the net is mathematically represented as
[tex]t = \frac{d}{v}[/tex]
=> [tex]t = \frac{12}{23.6}[/tex]
=> [tex]t = 0.508 \ s[/tex]
Generally the change in height of the ball after t is mathematically represented as
[tex]\Delta h = ut + \frac{1}{2} gt^2[/tex]
Here u is the initial velocity which is zero given that the ball was at rest initially
So
[tex]\Delta h = 0* t + \frac{1}{2} * 9.8 * 0.50 8^2[/tex]
=> [tex]\Delta h =1.28 \ m[/tex]
Generally the new height of the ball is mathematically evaluated as
[tex]s= h-\Delta h[/tex]
=> [tex]s = 2.37 - 1.28[/tex]
=> [tex]s = 1.09 \ m[/tex]
From the value obtained we see that [tex]s > H[/tex] hence the ball clears the net
Generally the distance between the center of the ball and the top of the net is mathematically represented as
[tex]b = s - H[/tex]
=> [tex]b = 1.09 - 0.90[/tex]
=> [tex]b= 0.19 \ m[/tex]
Given that the ball makes an angle of [tex]5^o[/tex] with the horizontal , the velocity along the x-axis is
[tex]v_x = v cos(5)[/tex]
=> [tex]v_x = 23.6 cos(5)[/tex]
=> [tex]v_x = 23.5 \ m/s[/tex]
The velocity along the y-axis is
[tex]v_y = v sin(5)[/tex]
=> [tex]v_y = 23.6 sin(5)[/tex]
=> [tex]v_y = 2.06 \ m/s[/tex]
Generally the time taken for the ball to reach the net is
[tex]t = \frac{d}{v_x}[/tex]
=> [tex]t = \frac{12}{23.5}[/tex]
=> [tex]t =0.508 \ s[/tex]
Generally the change in height of the ball after t seconds is
[tex]c = v_yt + \frac{1}{2}gt^2[/tex]
=> [tex]c = 2.06 * 0.508 + \frac{1}{2}* 9.8 * 0.508 ^2[/tex]
=> [tex]c = 2.33[/tex]
Generally the new height of the ball after time t seconds is
[tex]e = h - c[/tex]
=> [tex]e = 2.37 - 2.33[/tex]
=> [tex]e = 0.04 \ m[/tex]
From the value obtained we see that [tex]e < H[/tex] hence the ball does not clear the net
Generally the distance between the center of the ball and the top of the net is mathematically represented as
[tex]z = H-e[/tex]
=> [tex]z = 0.90 - 0.04[/tex]
=> [tex]z= 0.86 \ m[/tex]
(a) Yes, the ball clears the net.
(b) The distance between the center of the ball and the top of the net is 0.203 m.
(c) No, the ball does not clear the net.
(d) Now, the distance between the center of the ball and the top of the net is -0.85 m.
What is a Projectile motion?When any object or body is launched with some initial velocity and making some angle with the horizontal, the body travels in a parabolic path. It is known as the projectile motion.
Given,
The horizontal distance traveled by the ball is 12 m.
The height of the top of the net is 0.90 m.
The height of the horizontal launch of the ball is 2.37 m.
The time for the horizontal motion of the projectile that is the ball is,
[tex]\begin{aligned} {{v}_{0x}}&={{S}_{x}}t \\ t&=\dfrac{{{S}_{x}}}{{{v}_{0x}}} \\ &=\dfrac{{{S}_{x}}}{{{v}_{0}}\cos 0{}^\circ } \\ &=\dfrac{{{S}_{x}}}{{{v}_{0}}} \end{aligned}[/tex]
The equation for the vertical motion of the projectile can be solved by substituting the above result.
[tex]\begin{aligned} y&={{y}_{0}}+{{v}_{0y}}t-\frac{1}{2}g{{t}^{2}} \\ y&={{y}_{0}}+\left( {{v}_{0}}\sin 0{}^\circ \right)\left( \frac{{{S}_{x}}}{{{v}_{0x}}} \right)-\frac{1}{2}g{{\left( \frac{{{S}_{x}}}{{{v}_{0x}}} \right)}^{2}} \\ \left( 0.90\text{ m}+h \right)&=\left( 2.37\text{ m} \right)+0-\frac{1}{2}\left( 9.8\text{ m/}{{\text{s}}^{2}} \right){{\left( \frac{12\text{ m}}{23.6\text{ m/s}} \right)}^{2}} \\ h&=2.37\text{ m}-\text{1}\text{.267 m}-0.90\text{ m} \\ &=0.203\text{ m} \end{aligned}[/tex]
Now, consider the case when the ball is thrown with an angle [tex]\bold{5^{\circ}}[/tex] with the horizontal.
The horizontal component of the initial velocity is [tex]\bold{v_0 \cos 5{}^\circ.}[/tex]
The vertical component of the initial velocity is [tex]\bold{v_0 \sin 5{}^\circ.}[/tex]
For the horizontal distance traveled by the ball in this case, the time taken can be calculated as below,
[tex]\begin{aligned} {t}'&=\frac{{{{{S}'}}_{x}}}{{{{{v}'}}_{0x}}} \\ &=\frac{12\text{ m}}{\left( 23.6\text{ m/s} \right)\cos 5{}^\circ } \\ &=0.51\text{ s} \end{aligned}[/tex]
Now, the vertical distance above the ground, y’, traveled by the projectile till reaching the net can be determined as,
[tex]\begin{aligned} {y}'&={{{{y}'}}_{0}}+{{{{v}'}}_{0y}}{t}'-\frac{1}{2}g{{{{t}'}}^{2}} \\ &=0\text{ m}-\left( \left( 23.6\text{ m/s} \right)\sin 5{}^\circ \right)\left( 0.51\text{ s} \right)-\frac{1}{2}\left( 9.8\text{ m/}{{\text{s}}^{2}} \right){{\left( 0.51\text{ s} \right)}^{2}} \\ &=-1.05\text{ m}-1.28\text{ m} \\ &=-2.32\text{ m} \end{aligned}[/tex]
The height above the top of the net can be determined by adding the above result with (2.37 m - 0.90 m) which is the height of the net’s top relative to the launch position.
[tex]\begin{aligned} {h}'&=\left( 2.37\text{ m}-0.90\text{ m} \right)+{y}' \\ &=\left( 2.37\text{ m}-0.90\text{ m} \right)-2.32\text{ m} \\ &=-0.85\text{ m} \end{aligned}[/tex]
Thus, the distance between the center of the ball and the top of the net is -2,32 m.
When the ball leaves the racquet at 5.00° below the horizontal, the distance between the center of the ball and the top of the net is -0.85 m.
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PLEASE HELP ASAP WILL GIVE BRAINLIEST!!!!
compare and contrast the strength of the forces between two objects with a mass of 1 kg each, a charge of 1 C, and at a distance of 1 m from each other.
Explanation:
Please mark me as the brainliest answer
Answer:
The electrostatic force is larger by a factor of 8.988E9 / 6.674E-11 = 1.35E20
Explanation:
William gave you all the ingredients, but not the answer. Being both 1/r^2 forces (resulting from massless mediators in QFT), and with unit charges in the problem (1 coulomb, 1 Kg), separated by unit distance (1m), the only nontrivial numerical values in the problem are the constants of proportionality: Coulomb's constant (k) and Newton's gravitational constant (G). So to "compare and constrast": the ratio of the forces is simply the ratio of these constants. The electromagnatic force is 1.35 X 10^20 times stronger than the gravitational force. Assuming positive charge on both objects (the problem is ambiguous on this), they are repelled, whereas the much weaker gravitational force is attractive. (Gravity only has one kind of "charge" - it's unsigned - and the force is always attractive).
The problem doesn't say the objects are pointlike: if they have some extent and are either conductive or made of some dielectric, then things get messy because the charge distribution on the object won't be uniform then, but save that for grad school. :-)
What will happen if the atom rearrange?
1. a new substance will form
2. start to boil
3. nothing will happen
choose the correct answer
Answer:
In a chemical reaction, only the atoms present in the reactants can end up in the products. No new atoms are created, and no atoms are destroyed. In a chemical reaction, reactants contact each other, bonds between atoms in the reactants are broken, and atoms rearrange amd form new bonds to make the products.
A student is measuring the volumes of nectar produced by a flowering plant for an experiment. He measures nectar from 50 flowers using a graduated cylinder that measures to the nearest milliliter(mL). Which statement describes a change that could help improve the results of his experiment
The question is incomplete, the complete question is;
A student is measuring the volumes of nectar produced by a flowering plant for an experiment. He measures nectar from 50 flowers using a graduated cylinder that measures to the nearest millilitre (mL). Which statement describes a change that can help improve the results of his experiment?
A.) His measurements will be more precise if he takes measurements from an additional 100 flowers. B.) His measurements will be more accurate if he uses a graduated cylinder that measures to the nearest tenth of a mL. C.) His measurements will be more precise if he uses a graduated cylinder that measures to the nearest tenth of a mL. D.) His measurements will be more accurate if he takes measurements from an additional 100 flowers.
Answer:
His measurements will be more accurate if he uses a graduated cylinder that measures to the nearest tenth of a mL.
Explanation:
In the measurements of volume using most graduated cylinders, the cylinders are calibrated to the nearest tenth owing to the uncertainty in the measurement of volume.
Hence if a cylinder has measures to the nearest milliliter(mL), then he can improve his experiment by using a graduated cylinder that measures to the nearest tenth of a mL
please help me.
During a football game, one of the players on the home team kicks the football that has a
mass of 0.6 kg so that the ball accelerates toward the opposing team at
23 m/s2. If no other forces act on the ball, how much force did the kicker apply to the
football?
Answer:
[tex]Force = 13.8N[/tex]
Explanation:
Given
[tex]Mass = 0.6kg[/tex]
[tex]Acceleration = 23m/s^2[/tex]
Required
Determine the applied force
From the question, we understand that no other force acts on the ball.
i.e. the only applied force on the ball is the force applied by the striker.
So, we apply Newton's second law to solve this question.
And this implies that:
[tex]Force = Mass * Acceleration[/tex]
[tex]Force = 0.6kg * 23m/s^2[/tex]
[tex]Force = 13.8N[/tex]
Hence, the applied force by the striker on the ball is 13.8N
Suppose a 125 N force is applied to a lawnmower handle at an angle of 35° with the ground and the lawnmower moves along the surface of the ground. If the lawnmower moves 56 m, how much work was done? (hint: use cos to find the x of force vector)
Answer:
Workdone is 5734.06Nm.
Explanation:
Given the following data;
Force applied = 125N
Angle = 35°
Distance = 56m
To find the workdone by the lawnmower, we would first of all find the horizontal component of the force applied.
[tex] Horizontal force, Fx = mgCosd[/tex]
Where;
Fx represents the horizontal force. m is the mass of an object. g is the acceleration due to gravity. d is the angle of inclination (theta).mg = weight = 125N
Substituting into the equation, we have;
[tex] Fx = 125 * Cos35[/tex]
[tex] Fx = 125 * 0.8192[/tex]
Fx = 102.39N
Workdone is given by the formula;
[tex] Workdone = force * distance[/tex]
[tex] Workdone = 102.39 * 56[/tex]
Workdone = 5734.06Nm
Therefore, the work done by the lawnmower is 5734.06Nm.
Someone please do this ! ASAP I’ll give brainliest!
Answer:
The first law, an object will not change its motion unless a force acts on it.
The second law, the force on an object is equal to its mass times its acceleration.
The third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.
HELP
which two changes to a metal wire both increases resistance? the answer is B but why ?
Answer:
option C decreasing its thickness and increasing its temperature.Explanation:
Resistance is directly proportional to length and temperature of the wire and inversely to area.if you increase the temperature the resistance will increase.(resistance is directly proportional to temperature)if you decrease its thickness (area) then the resistance will increase ( resistance is inversely proportional to area)hope it helps:)
Earthquakes are essentially sound waves—called seismic waves—traveling through the earth. Because the earth is solid, it can support both longitudinal and transverse seismic waves. The speed of longitudinal waves, called PP waves, is 8000 m/sm/s. Transverse waves, called SS waves, travel at a slower 4500 m/sm/s. A seismograph records the two waves from a distant earthquake. The SS wave arrives 2.0 minmin after the PP wave. Assume that the waves travel in straight lines, although actual seismic waves follow more complex routes.
Question
Earthquakes are essentially sound waves—called seismic waves—traveling through the earth. Because the earth is solid, it can support both longitudinal and transverse seismic waves. The speed of longitudinal waves, called P waves, is 8000 m/s Transverse waves, called S waves, travel at a slower 4500 m/s. A seismograph records the two waves from a distant earthquake. The S wave arrives 2.0 min after the PP wave.How far away is the Earthquake. Assume that the waves travel in straight lines, although actual seismic waves follow more complex routes.
Answer:
The distance is [tex]d = 1.23 *10^{6} \ m[/tex]
Explanation:
From the question we are told that
The speed of longitudinal seismic wave is [tex]v_p = 8000 \ m/s[/tex]
The speed of Transverse seismic wave is [tex]v_s = 4500 \ m/s[/tex]
The time difference between the arrival of longitudinal seismic with respect to Transverse waves is [tex]\Delta t = 2.0\ min = 120\ seconds[/tex]
Generally the time difference between the arrival of longitudinal seismic with respect to Transverse waves is mathematically represented as
[tex]\Delta t = t_p - t_s[/tex]
=> [tex]\Delta t =\frac{d}{v_p} -\frac{d}{v_s}[/tex]
=> [tex]d = \frac{\Delta t}{ \frac{1}{v_p} - \frac{1}{v_s} }[/tex]
=> [tex]d = \frac{120 }{ \frac{1}{8000} - \frac{1}{4500} }[/tex]
=> [tex]d = 1.23 *10^{6} \ m[/tex]
Which could most likely describe the three surfaces?
Surface 1 is ice, Surface 2 is gravel, and Surface 3
is blacktop.
Surface 1 is gravel, Surface 2 is ice, and Surface 3
is blacktop.
Surface 1 is blacktop, Surface 2 is gravel, and
Surface 3 is ice.
Surface 1 is blacktop, Surface 2 is ice, and Surface
3 is gravel.
Answer:
Surface 1 is blacktop, Surface 2 is gravel, and Surface 3 is ice.
Explanation:
Hope this helps! :]
Answer:
C. Surface 1 is blacktop, Surface 2 is gravel, and
Surface 3 is ice.
Explanation:
Jorge conducted an experiment,and included the graph shown below as part of his lab report.
Jorges experiment involved which if the following?
-A Chemical change.
-A change in the chemical properties of a substance.
-A physical change.
-the formation of a new substance.
The correct answer is A physical change
Explanation:
Jorge's experiment shows water at different temperatures; in this experiment, it is expected at low temperatures such as -20°C water is in solid-state (ice), at medium temperatures such as 40°C water is in a liquid state (liquid water), and at high temperatures such as 120°C water is in gaseous state (water vapor). This implies during this experiment the changing factor is the physical state (solid, gas, or liquid), and this is a physical change because only the physical properties of water change but not its composition or identity. According to this, the correct answer is physical change.
A slab of glass 8.0 cm thick is placed upon a printed page. If the refractive index of the glass is 1.5, how far from the surface would the letters appear to be when viewed from directly above?
Answer:
5.3 cm
Explanation:
This question is an illustration of real and apparent distance.
From the question, we have the following given parameters
Real Distance, R = 8.0cm
Refractive Index, μ = 1.5
Required
Determine the apparent distance (A)
The relationship between R, A and μ is:
μ = R/A
i.e.
Refractive Index = Real Distance ÷ Apparent Distance
Substitute values in the above formula
1.5 = 8/A
Multiply both sides by A
1.5 * A = A * 8/A
1.5A = 8
Divide both side by 1.5
1.5A/1.5 = 8/1.5
A = 8/1.5
A = 5.3cm
Hence, the letters would appear at a distance of 5.3cm
Katie and Mark sit next to one another in class. She has a mass of 40 kg and his mass is 65 kg.
If they are 0.5 meters apart, what is the gravitational force between Mark and Katie?
1. An electric lamp a marked 240V, 6A. What
its resistance when it is operated at the correct
voltage?
Answer:
The resistance of the lamp is 4Ω.
Explanation:
You have to apply voltage formula :
V = I × R
R = V ÷ I
R = 240 ÷ 60
R = 4 Ω
a sheet of metal is 2mm wide 10cm tall and 15cm long. it was 4g. what is the density?
Answer:
Ro = 133 [kg/m³]
Explanation:
In order to solve this problem, we must apply the definition of density, which is defined as the relationship between mass and volume.
[tex]Ro = m/V[/tex]
where:
m = mass [kg]
V = volume [m³]
We will convert the units of length to meters and the mass to kilograms.
L = 15 [cm] = 0.15 [m]
t = 2 [mm] = 0.002 [m]
w = 10 [cm] = 0.1 [m]
Now we can find the volume.
[tex]V = 0.15*0.002*0.1\\V = 0.00003 [m^{3} ][/tex]
And the mass m = 4 [gramm] = 0.004 [kg]
[tex]Ro = 0.004/0.00003\\Ro = 133 [kg/m^{3}][/tex]
You kick a soccer ball across a field the ball travels across the field and slowly comes to a stop. Since the ball slowly comes to a stop what does this mean about the net force on the ball as it rolls across the field?
Answer:
It slowly decreases and the friction acting on it slowing it down becomes the bigger net force, if that makes sense :)
Explanation:
If you kick a soccer ball across a field the ball travels across the field and slowly comes to a stop. Since the ball slowly comes to a stop, this means about the net force on the ball as it rolls across the field would decrease and finally would be less than the friction force.
What is Newton's second law?Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum. The mathematical expression for Newton's second law is as follows
As given in the problem if you kick a soccer ball across a field the ball travels across the field and slowly comes to a stop. Since the ball slowly comes to a stop.
Since the ball stops gradually, this implies that the net force exerted on it as it rolls over the field will eventually diminish until it is less than the friction force.
Learn more about Newton's second law here, refer to the link given below ;
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A space vehicle is coasting at a constant velocity of 22.4 m/s in the y direction relative to a space station. The pilot of the vehicle fires a RCS (reaction control system) thruster, which causes it to accelerate at 0.206 m/s2 in the x direction. After 40.5 s, the pilot shuts off the RCS thruster. After the RCS thruster is turned off, find (a) the magnitude and (b) the direction of the vehicle's velocity relative to the space station. Express the direction as an angle (in degrees) measured from the y direction.
Answer:
a. 23.9 m/s b. 69.58°
Explanation:
a. Since the space vehicle is moving at a constant velocity of 22.4 m/s in the y direction relative to the space station, its initial vertical velocity v = 22.4 m/s.
Also, since the space vehicle moves in the y direction, its initial horizontal velocity u = 0 m/s.
Since its acceleration a = 0.206 m/s² for time, t = 40.5 s, we find the final horizontal velocity u' from u' = u + at.
Substituting the values of the variables into the equation, we have
u' = u + at
u' = 0 m/s + 0.206 m/s² × 40.5 s
= 8.343 m/s
≅ 8.34 m/s
The resultant velocity relative to the space station V = √(v² + u'²)
= √[(22.4m/s)² + (8.34 m/s)²]
= √[501.76 m²/s² + (69.56 m²/s²]
= √[571.32 m²/s²]
= 23.9 m/s
b. The direction of the vehicle's velocity relative to the space station is thus θ = tan⁻¹(v/u')
= tan⁻¹(22.4 m/s/8.34 m/s)
= tan⁻¹(2.6959)
= 69.58°
A crossbow is fired horizontally off a cliff with an initial velocity of 15 m/s. If the arrow takes 4s to hit the ground, what is the range of the projectile?
Answer:
The range of the projectile is 60 meters
Explanation:
To determine the range/distance of the projectile, the formula for velocity is used;
velocity = distance/time
where velocity is 15 m/s
time is 4 seconds
distance is unknown
From the formula above, distance is made the subject and thus
distance = velocity × time
distance = 15 × 4
distance = 60 m
The range of the projectile is 60 meters
Explain why when a firefighter rescues a dog that has fallen through ice on a
lake, they put their ladder on the ice first and then crawl out to the dog on the
ladder.
Answer:
This can spread their weight.
Explanation:
The ladder has a much larger surface area than the firefighter. Hence, his weight is spread out much more than usual, decreasing the pressure on the ice, preventing the ice from breaking/cracking.
Hope this helped!
Read the elapsed time on the stopwatch and answer the questions.
In which digit is there the least amount of confidence?
How many significant figures does this measurement have? I need help quick
Answer:
1.7
2.4
Explanation:
Answer:
1.7
2.4
Explanation:
Select all the correct answers.
Which situations describe an elastic collision?
(A) Two glass marbles bounce off each other.
(B) Rodrick flops onto his sofa and sinks into the cushion.
(C) A tossed water balloon flattens when it lands on the grass.
D) A bowling ball knocks over five pins.
During combustion reactions, explain why the energy of the reactants must exceed the total energy of the products
Answer:
In these reactions the products are higher in energy than the reactants. ... This barrier is due to the fact that to make CO2 and H2O we have to break 4 carbon-hydrogen bonds and some ...
Explanation:
can y’all please help me with this 3 part question?
Answer:
Vf = 210 [m/s]
Av = 105 [m/s]
y = 2205 [m]
Explanation:
To solve this problem we must use the following formula of kinematics.
[tex]v_{f} =v_{o} +g*t[/tex]
where:
Vf = final velocity [m/s]
Vo = initial velocity = 0 (released from the rest)
g = gravity acceleration = 10 [m/s²]
t = time = 21 [s]
Vf = 0 + (10*21)
Vf = 210 [m/s]
Note: The positive sign for the gravity acceleration means that the object is falling in the same direction of the gravity acceleration (downwards)
The average speed is defined as the sum of the final speed plus the initial speed divided by two. (the initial velocity is zero)
Av = (210 + 0)/2
Av = 105 [m/s]
To calculate the distance we must use the following equation of kinematics
[tex]v_{f} ^{2} =v_{o} ^{2} +2*g*y\\\\(210)^{2} = 0 + (2*10*y)[/tex]
44100 = 20*y
y = 2205 [m]
A grapefruit falls from a tree and hits the ground 0.72 s later.
How far did the grapefruit drop?
What was its speed when it hit the ground?
Answer:
The grapefruit dropped 2.54 m and hit the ground at 7.06 m/s
Explanation:
Free Fall Motion
A free-falling object falls under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. Free-falling objects do not encounter air resistance.
If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is [tex]g = 9.8 m/s^2.[/tex]
The final velocity of a free-falling object after a time t is given by:
vf=g.t
The distance traveled by a dropped object is:
[tex]\displaystyle y=\frac{gt^2}{2}[/tex]
Given a grapefruit free falls from a tree and hits the ground t=0.72 s later, we can calculate the height it fell from:
[tex]\displaystyle y=\frac{9.8\cdot 0.72^2}{2}[/tex]
y = 2.54 m
The final speed is computed below:
[tex]vf=9.8\cdot 0.72[/tex]
vf = 7.06 m/s
The grapefruit dropped 2.54 m and hit the ground at 7.06 m/s
What is the mass of 2.5 mol of Ca, which has a molar mass of 40 g/mol?
Answer:100 g of ca
Explanation:
A tortoise can run with a speed of 0.10 m/s, and a hare can run 20 times as fast. In a race, they both start at the same time, but the hare stops to rest for 2.0 minutes. The tortoise wins by a shell (20 cm)!How long does the race take?What is the length of the race?
Answer: a. 126.21secs
b. 12.621 meters.
Explanation:
Given data:
Speed of tortoise = 0.1m/s.
Speed of hare = 2m/s.
Solution:
a. Distance traveled = Speed* Time
Speed of tortoise = 0.1 m/s
Speed of hare = 20*0.1 m/s = 2 m/s
2 minutes = 2* 60 s = 120 s
Let the time taken for the race be t seconds.
• Distance moved by tortoise
= (0.1 /s)* (t s)
= 0.1*t meter
•Hare has run for a time of (t - 120)s.
distance moved by hare
= Speed * Time
= (2 m/s)*(t- 120)s
= (2t - 240) meter.
Since hare is 20 cm (0.2 m) behind the tortoise, therefore
(0.1*t - 0.2) meter
= (2t - 240) meter
0.1*t - 0.2 = 2t - 240
Collect like terms
239.8 = 1.9t
Divide both sides by 1.9
t = 126.21secs
The race lasted for 126.21secs
b. Length of race
= Distance moved by tortoise
= 0.1*126.21 meter
= 12.621 meter
The length of the race is 12.621 meters.