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The size of the gravitational force depends on the mass of the objects involved. The greater the mass, the greater the gravitational force.
The normal force of weight will not differ for different placements but the force of pressure applied to the surface of your hand is per unit area.
Since, Pressure is equal to the ratio of Force per unit area.
Thus, the side with the smallest area i.e. 3 cm x 6 cm will exert the most pressure on your hand.
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Related Questions
Myth: An organism's kingdom only describes physical characteristics.
Fact:
Evidence:
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Organisms are classified into the different kingdoms based on physical appearance, their function, how they reproduce and how they obtain their food.
EVIDENCE:
Kingdom Plantae: They obtain their food through photosynthesis
Reproduce asexually.
Kingdom Animalia : They obtain their food through hunting or consuming other organisms.
Reproduce sexually.
The kingdom of an organism does not only describe physical characteristics. It is evident by that plant and animal kingdom are classified on the basis of photosynthetic ability.
Kingdom:
In classification kingdom is the highest in the hierarchy. A kingdom includes the organism of similar classes.
For Example: plant and animal kingdom
Plant Kingdom:
It includes the multicellular organisms who have cell wall and produce glucose using Photosynthesis. Such as Bryophytes, Pteridophytes, and gymnosperms.Animal Kingdom:
This includes the the organisms with cell membrane only. They rely o others for their nutrition. Such as mammals and fishesTherefore, the fact is that the organism's kingdom does not only describes physical characteristics.
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A Geiger counter is like an electroscope that discharges whenever ions formed by a radioactive particle produce a conducting path. A typical Geiger counter consists of a thin conducting wire of radius 0.002 cm stretched along the axis of a conducting cylinder of radius 2.0 cm. The wire and the cylinder carry equal and opposites charges of 8.0 x 10-10 C all along their length of 10.0 cm. What is the magnitude of the electric field at the surface of the wire
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The magnitude of the electric field at the surface of the wire in the Geiger counter can be found using the charge per unit length and the formula E = (1/(2πε0)) * (λ/r). Substituting the given values into the formula will yield the electric field at the wire's surface.
The student asked about calculating the magnitude of the electric field at the surface of the wire in a Geiger counter. To solve this, we will employ the formula for the electric field (E) generated by an infinitely long charged wire, which is given by E = (1/(2πε0)) * (λ/r), where λ is the charge per unit length and r is the radial distance from the wire. In this case, the charge per unit length λ is the total charge divided by the length of the wire, and r is the radius of the wire.
Given: Charge (Q) = 8.0 x 10-10 C, Length (L) = 10.0 cm = 0.1 m, and Wire radius (a) = 0.002 cm = 0.00002 m.
First, calculate the charge per unit length: λ = Q/L = (8.0 x 10-10 C) / (0.1 m) = 8.0 x 10-9 C/m.
Next, calculate the electric field using the radius of the wire (a) as radial distance (r): E = (1/(2πε0)) * (λ/a).
Using the value for the vacuum permittivity ε0 (approximately 8.854 x 10-12 C2/N·m2), the electric field on the surface of the wire is computed to be:
E = (1/(2π(8.854 x 10-12))) * (8.0 x 10-9 / 0.00002) N/C.
Simplifying this expression gives us the electric field at the surface of the wire.
Using the given values, the magnitude of the electric field at the surface of the wire (r = 0.002 cm or 0.00002 m) and L = 10 cm or 0.1 m is calculated as follows:
Convert the radius to meters: r = 0.002 cm = 0.00002 m.
Plugging in the values: E = (8.0 x 10-10 C) / (2π * 8.854 x 10-12 C2/Nm2 * 0.00002 m * 0.1 m).
Calculate the electric field: E = 2.87 x 106 N/C.
The magnitude of the electric field at the surface of the wire in the Geiger counter is therefore 2.87 x 106 Newtons per Coulomb (N/C).
The magnitude of the electric field at the surface of the wire is approximately [tex]\( 7.19 \times 10^{6} \, \text{N/C} \)[/tex].
The magnitude of the electric field at the surface of the wire is given by the formula:
[tex]\[ E = \frac{\sigma}{\varepsilon_0} \][/tex]
where [tex]\( \sigma \)[/tex] is the surface charge density of the wire, and [tex]\( \varepsilon_0 \)[/tex]is the vacuum permittivity [tex](\( 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \)).[/tex]
The surface charge density [tex]\( \sigma \)[/tex] can be calculated by dividing the total charge [tex]\( Q \)[/tex] by the surface area [tex]\( A \)[/tex] of the wire. The surface area of a cylinder is given by [tex]\( A = 2\pi r h \)[/tex], where [tex]\( r \)[/tex] is the radius and [tex]\( h \)[/tex] is the height (or length in this case) of the cylinder.
Given:
- Radius of the wire, [tex]\( r = 0.002 \, \text{cm} = 2 \times 10^{-5} \, \text{m} \) (since 1 cm = 0.01 m)[/tex]
- Length of the wire, [tex]\( h = 10.0 \, \text{cm} = 0.1 \, \text{m} \)[/tex]
- Total charge on the wire, [tex]\( Q = 8.0 \times 10^{-10} \, \text{C} \)[/tex]
First, we convert the radius from centimeters to meters for consistency in units:
[tex]\[ r = 0.002 \, \text{cm} \times \frac{1 \, \text{m}}{100 \, \text{cm}} = 2 \times 10^{-5} \, \text{m} \][/tex]
Now, we calculate the surface area of the wire:
[tex]\[ A = 2\pi r h = 2\pi (2 \times 10^{-5} \, \text{m})(0.1 \, \text{m}) \][/tex]
[tex]\[ A = 4\pi \times 10^{-6} \, \text{m}^2 \][/tex]
Next, we calculate the surface charge density [tex]\( \sigma \)[/tex]:
[tex]\[ \sigma = \frac{Q}{A} = \frac{8.0 \times 10^{-10} \, \text{C}}{4\pi \times 10^{-6} \, \text{m}^2} \][/tex]
[tex]\[ \sigma = \frac{8.0 \times 10^{-10} \, \text{C}}{4\pi \times 10^{-6} \, \text{m}^2} \approx \frac{8.0 \times 10^{-10}}{12.56637 \times 10^{-6} \, \text{m}^2} \][/tex]
[tex]\[ \sigma \approx 6.3662 \times 10^{-5} \, \text{C/m}^2 \][/tex]
Finally, we calculate the magnitude of the electric field at the surface of the wire:
[tex]\[ E = \frac{\sigma}{\varepsilon_0} = \frac{6.3662 \times 10^{-5} \, \text{C/m}^2}{8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2} \][/tex]
[tex]\[ E \approx 7.19 \times 10^{6} \, \text{N/C} \][/tex]
Therefore, the magnitude of the electric field at the surface of the wire is approximately [tex]\( 7.19 \times 10^{6} \, \text{N/C} \)[/tex].
What is the frequency of a wave that has a wave speed of 20 m/s and a wavelength of 0.20 m
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I hope it help you
A rock of mass m is thrown horizontally off a building from a height h. the speed of the rock as it leaves the thrower's hand at the edge of the building is v0, as shown. m h v0 m what is the kinetic energy of the rock just before it hits the ground? 1. kf = 1 2 m v2 0 2. kf = 1 2 m v2 0 − m g h 3. kf = 1 2 m v2 0 + m g h 4. kf = m g h − 1 2 m v2 0 5. kf = m g h
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In fact, the total energy of the rock when it leaves the thrower's hand is the sum of the gravitational potential energy U and of the initial kinetic energy K:
[tex]E=U_i+K_i=mgh + \frac{1}{2}mv_0^2 [/tex]
As the rock falls down, its height h from the ground decreases, eventually reaching zero just before hitting the ground. This means that U, the potential energy just before hitting the ground, is zero, and the total final energy is just kinetic energy:
[tex]E=K_f[/tex]
But for the law of conservation of energy, the total final energy must be equal to the tinitial energy, so E is always the same. Therefore, the final kinetic energy must be
[tex]K_f = mgh + \frac{1}{2}mv_0^2 [/tex]
[tex]K_f=mgh+\frac{1}{2}mv_o^2[/tex]
Given information:
A rock of mass m is thrown horizontally off a building from a height h
As, the total energy of rock at the time of leaving the thrower's hand is the sum of gravitational potential energy and the initial kinetic energy.
[tex]E=U_i+K_i\\E=mgh+(1/2)mv_0^2[/tex]
As, the height from the ground decreases, the potential energy before hitting the ground is zero and the total final energy is just kinetic energy:
[tex]E=K_f[/tex]
As, the law of conservation of energy states the total final energy must be equal to the initial energy.
Hence, the final kinetic energy will be
[tex]K_f=mgh+\frac{1}{2}mv_o^2[/tex]
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A 0.55-kg ball, attached to the end of a horizontal cord, is revolved in a circle of radius 1.3 m on a frictionless horizontal surface. if the cord will break when the tension in it exceeds 75 n, what is the maximum speed the ball can have?
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[tex]T=F_c = m \frac{v^2}{r} [/tex]
where T is the tension of the cord, [tex]F_c[/tex] is the centripetal force, [tex]m=0.55 Kg[/tex] is the mass of the ball, v its speed and [tex]r=1.3 m[/tex] is the radius of the circle.
The maximum allowed tension is T=75 N, before the cord breaks. Using this value inside the formula, we can find which is the maximum allowed value fot the speed v:
[tex]v= \sqrt{ \frac{T r}{m} }= \sqrt{ \frac{(75 N)(1.3 m)}{(0.55 kg)} }=13.3 m/s [/tex]
The force act on the body tries to attract the body inward towards the circle when the body is executing circular motion. The maximum value of speed will be 13.3 m/sec.
What is centripetal force?The force operating on an object in curvilinear motion directed toward the axis of rotation or center of curvature is known as centripetal force.
Newton is the unit of centripetal force.
The centripetal force is always perpendicular to the direction of displacement of the item. The centripetal force of an item traveling on a circular route always works towards the center of the circle.
Due to rotation, the tension is act in the chord which is equal to the centripetal force act the ball.
[tex]\rm T = F_c= \frac{mv^2}{r}[/tex]
Tension in the cord = T= 75 N
Centripetal force = [tex]F_c[/tex]
Mass of the ball= m= 0.55 Kg
Linear speed =v= ?
The radius of the circle.= r = 1.3 m
T is the maximum tension before the cord breaks. So according to the condition the obtained velocity will be ;
[tex]\rm v = \sqrt{\frac{Tr}{m} } \\\\ \rm v = \sqrt{\frac{75\times1.3}{0.55} }\\\\ \rm v = 13.3 \;m/sec[/tex]
Hence the maximum value of speed will be 13.3 m/sec.
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the record for a ski jump is 180 m set in 1989. Assume the jumper comes off the end of the ski jump horizontally and falls 90 m vertically before contacting the ground . what was the initial horizontal speed of the jumper?
Answers
[tex]S_x(t)=v_x t[/tex]
[tex]S_y(t)=h- \frac{1}{2} gt^2[/tex]
First, let's find the time at which the jumper reaches the ground. This happens when Sy(t)=0:
[tex]0=h- \frac{1}{2} g t^2[/tex]
and so
[tex]t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2\cdot 90m}{9.81 m/s^2} }=4.28 s [/tex]
Then, we can find the horizontal speed. In fact, we know that at the time t=4.28 s, when the jumper reached the ground, he covered exactly 180 m, so Sx=180 m. Using this into the law of motion in x, we find
[tex]v_x= \frac{S_x}{t} = \frac{180 m}{4.28 s} =42 m/s[/tex]
A bag of sports equipment has a mass of 10.0 kilograms and a weight of
Answers
The weight of the bag of sports equipment is 98N
From the question,
We are to determine the weight of a bag of sports equipment that has a mass of 10.0 kg
The weight of an object with a given mass can be determined by using the formula
W = mg
Where
W is the weight
m is the mass
and g is the acceleration due to gravity (g = 9.8 m/s²)
From the given information,
m = 10.0 kg
∴ Weight of the bag of sports equipment,
W = 10.0 × 9.8
W = 98 N
Hence, the weight of the bag of sports equipment is 98N
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a boat travels in a straight path that is 25 . west of north which describes the values of the west and north components of the boat's displacement ?
Both components are positive numbers.
Both components are negative numbers.
The west component is a negative number, and the north component is a positive number.
The west component is a positive number, and the north component is a negative number.
Answers
the values of the west and north components of the boat's displacement belongs to the second quadrant, therefore the component X is negative and the component Y is positive
Answer:
C on edge
Explanation:
If you mix equal amounts of cyan pigments and magenta pigments on a sheet of white paper, what color will you see on the paper?
A. Red
B. Blue
C. Black
D. Yellow
E. Cyan
F. Magenta
Answers
Hope this helps! :)
One mole of ideal gas is slowly compressed to one-third of its original volume. in this compression, the work done on the gas has magnitude 672 j . for the gas, cp=7r/2.
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An engine performs 2700 J of work on a scooter. The scooter and rider have a combined mass of 150 kg. If the scooter started at rest, what is the speed of the bike after the work is performed?
Answers
[tex]W=K= \frac{1}{2} M v^2 [/tex]
where M is the combined mass of scooter and rider, and v is the velocity of the scooter. Therefore, we can find the velocity as:
[tex]v= \sqrt{ \frac{2W}{M} } = \sqrt{ \frac{2\cdot 2700 J}{150 Kg} }=6 m/s[/tex]
What happens to light when it changes speed?
A)It reflects
B)It polarizes
C)It refracts.
Answers
When light changes speed, it REFRACTS.
Your answer is going to be option C.
Hope this helps you.
Have a great day!
Answer:
The correct answer is C) Refracts
Referring to the map of the temperature field in the room, where was the warmest field found?
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The warmest field on a temperature map is indicated by warmer colors such as red or orange. On maps like those from the WMAP spacecraft, red represents higher temperatures. In thermographic maps of buildings, warmer areas show where heat transfer is most severe.
Explanation:The warmest field on a map showing temperature variations can typically be found in regions marked with a warmer color, such as red or orange, indicating higher temperatures. In the context of the map used by the WMAP spacecraft to show fluctuations in the cosmic microwave background, red represents areas of higher temperature and higher density. Similarly, thermographic maps of buildings, like Figure 1.32, use color to indicate variations in temperature, with warmer areas potentially signifying regions where heat transfer is most severe, such as through windows.
It's important to carefully analyze the colors and legends provided on the map to determine the precise locations of the warmest fields. When maps show global temperature changes, such as Figure 24.9.5, the land areas and the Arctic are noted for having experienced the greatest increases in temperature relative to mid-20th-century baseline values.
Almost all of the electricity that people use is produced by
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What is a property of a transparent object? A. Almost all of the light rays that reach it are scattered. B. Almost all of the light rays that reach it are absorbed. C. Almost all of the light rays that reach it are reflected. D. Almost all of the light rays that reach it pass through it.
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A bolt is dropped from a bridge under construction, falling 80 m to the valley below the bridge. (a) how much time does it take to pass through the last 17 % of its fall? what is its speed (b) when it begins that last 17 % of its fall and (c) just before it reaches the ground?
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A particular battery lasts 14.5 hours. what percentile is its lifetime on?
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The battery with a lifetime of 14.5 hours is at the 40.13th percentile in terms of battery lifetimes, assuming a normal distribution with a mean of 15 hours and a standard deviation of 2 hours.
Certainly, to find the percentile of the battery's lifetime, we'll use the Z-score formula and then refer to the standard normal distribution table. Let's assume a normal distribution with a mean (μ) of 15 hours and a standard deviation (σ) of 2 hours. The battery's lifetime is X = 14.5 hours.
The Z-score is calculated as:
Z = (X - μ) / σ
Substitute the values:
Z = (14.5 - 15) / 2 = -0.25
Now, we need to find the percentile associated with this Z-score. Looking up -0.25 in the standard normal distribution table, we find that the corresponding percentile is approximately 40.13%.
The question probable may be:
What is the percentile of a battery's lifetime with a mean (μ) of 15 hours and a standard deviation (σ) of 2 hours when its actual lifetime is X = 14.5 hours? Use the Z-score formula and the standard normal distribution table to determine the percentile associated with the Z-score of -0.25.
This battery's lifetime falls in the 22.66th percentile.
The lifetime of a battery is normally distributed with a mean (μ) of 16 hours and a standard deviation (σ) of 2 hours. To find what percentile a particular battery lasting 14.5 hours falls into, we need to calculate its Z-score.
The Z-score is given by:
[tex]Z = \frac{X- \mu}{\sigma}[/tex]
where X is the observed value.
The Z-score of the given battery is:
Z = [tex]\frac{14.5 - 16}{2}[/tex] = [tex]\frac{-1.5}{2}[/tex] = - 0.75
According to the Z-table, a Z-score of - 0.75 corresponds to a cumulative probability of approximately 0.2266.
This means that 14.5 hours is at the 22.66th percentile.
Thus, the lifetime of a particular battery lasting 14.5 hours falls at the 22.66th percentile of the lifetime distribution.
The complete question is:
The lifetime of a battery in a certain application is normally distributed with mean μ = 16 hours and standard deviation σ = 2 hours.
Activities:
What is the probability that a battery will last more than 19 hours?Find the 10th percentile of the lifetimes.A particular battery lasts 14.5 hours. What percentile is its lifetime on?What is the probability that the lifetime of a battery is between 14.5 and 17 hours?A 0.200 kg block of a substance requires 3.59 kJ of heat to raise its temperature by 20.0 K. What is the specific heat of the substance? A. 2,020 J/(kg * K) B. 383 J/(kg*K) C. 130 J/(kg * K) D. 897 J/(kg*K)
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A water balloon weighing 19.6 N rests on a table. The balloon has an area of 0.015 m^2 in contact with the table. What pressure does the balloon exert on the table?
Answers
Substituting the known values, we get that p=1306.67 P= 1306.67 [tex] \frac{N}{m^2} [/tex]. That is regarding the measure of P. The direction of the pressure is always the same as that of the force, in this case downwards (towards the center of the Earth).
A train travels 67 kilometers in 1 hours, and then 81 kilometers in 5 hours. What is its average speed?
Answers
Explanation:
1) Average speed definition and formula
The average speed is the total distance run divided by the time elapsed:
S = distance / time
2) Distance 1 = 67 km
3) Distance 2 = 81 km
4) Total distance traveled = 67 km + 81 km = 148 km
5) time 1 = 1 hour
6) time 2 = 5 hours
7) total time = time 1 + time 2 = 1 h + 5 h = 6 h
8) Average speed:
S = 148 km / 6 h = 24.7 km/h
Three people are pushing a 500 kg of box in the same direction. applied forces are 30 n, 20 n, and 10 n respectively. if the acceleration of the box is 0.02 m/s2, what is the magnitude of a force created by friction?
Answers
[tex]F=30 N+20 N+10 N=60 N[/tex]
This force is pushing toward the direction of the motion, while the frictional force [tex]F_f[/tex] points in the opposite direction.
We can write Newton's second law applied to the block: the resultant of the two forces must be equal to the product between the block's mass and its acceleration
[tex]F-F_f = ma[/tex]
We know the mass of the block, m=500 kg, and the acceleration, [tex]a=0.02 m/s^2[/tex], so we can find the friction:
[tex]F_f = F-ma=60 N - (500 kg)(0.02 m/s^2)=50 N[/tex]
when you squeeze the rubber bulb of a perfume atomizer how do you change the air pressure at the top of the tube
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During one year, eight moose in a population died and two moose were born. Three moose immigrated from another population and five emigrated to find mates.
What was the population growth during this year?
Answers
-8 hope it helps i just took the test
HURRY The compressed spring of a dart gun has potential energy of 50 J. If the spring constant is 200 N/m, what is the displacement of the spring?
A. 0.5 m
B. 0.2 m
C. 0.7 m
D. 0.4 m
Answers
E = 0.5 * k * x²
k spring constant
x spring compression
x = √(2 * E / k) = 0.7
Answer: its C on edge 2023. =0.7
Explanation: He's right just putting it in a quicker format.
If the above two waveforms were sound waves, we would hear the ___________ wave louder. If the above two waveforms were light waves, we would see the ________ wave dimmer.
Answers
We can rewrite the sentence as follows:
"If the above two waveforms were sound waves, we would hear the superposed wave louder. If the above two waveforms were light waves, we would see the superposed wave dimmer."
Answer:
the answer is green, red
The charge per unit length on a long, straight filament is -94.5 µc/m. (a) find the electric field 10.0 cm from the filament, where distances are measured perpendicular to the length of the filament. (take radially inward toward the filament as the positive direction.)
Answers
[tex]E(r)= \frac{\lambda}{2 \pi \epsilon _0 r} [/tex]
where [tex]\lambda[/tex] is the linear density of charge and [tex]\epsilon _0 =8.85 \cdot 10^{-12} F/m[/tex] is the electric permittivity.
In our problem, the charge density is [tex]\lambda = -94.5 \mu C/m= -94.5 \cdot 10^{-6} C/m[/tex]. We want to calculate the electric field at [tex]r=10.0 cm=0.1 m[/tex], which is
[tex]E(0.1 m)= \frac{94.5 \cdot 10^{-6} C/m}{2 \pi (8.85 \cdot 10^{-12} F/m) (0.1 m)}=1.7 \cdot 10^7 V/m [/tex]
and since the charge on the wire is negative, the field points toward the wire.
The electric field at 10.0 cm from a long, straight filament with a charge per unit length of -94.5 µC/m can be calculated using the formula E = λ / (2πε₀r), producing a value directed radially inward.
Explanation:Finding the Electric Field from a Charged Filament
The problem is asking us to calculate the electric field generated by a long, straight filament with a given charge per unit length. By using Gauss's law, we can find the electric field at a distance from the filament. Since the electric field of a uniformly charged infinite line is symmetrical, it only depends on the distance from the line, and it is directed radially.
We can use the formula for the electric field created by an infinitely long straight filament, which is:
E = λ / (2πε0r)
Where:
ε0 is the permittivity of free space (approximately 8.85 x 10-12 C2/N·m2).
r is the distance from the filament, which is 10.0 cm (or 0.10 meters).
By substituting the known values into the equation:
E = (-94.5 x 10-6 C/m) / (2π x 8.85 x 10-12 C2/N·m2 x 0.10 m)
We perform the calculation to find the magnitude and direction of the electric field. The negative sign of λ indicates that the electric field is directed radially inward, towards the filament.
A ball is thrown at a 60.0° angle above the horizontal across level ground. it is thrown from a height of 2.00 m above the ground with a speed of 23.7 m/s and experiences no appreciable air resistance. the time the ball remains in the air before striking the ground is closest to
Answers
The time the ball remains in the air before striking the ground is closest to 4.20 seconds.
To find the time the ball remains in the air before striking the ground, we can use the equations of motion. Since the ball is thrown at an angle above the horizontal, we'll need to analyze both the horizontal and vertical components of its motion separately.
Given:
- Initial velocity [tex](\(v_0\))[/tex] = 23.7 m/s
- Launch angle[tex](\(\theta\))[/tex] = 60.0°
- Initial height (h) = 2.00 m
First, we'll find the horizontal and vertical components of the initial velocity:
Horizontal component: [tex]\(v_{0x} = v_0 \cdot \cos(\theta)\)[/tex]
Vertical component: [tex]\(v_{0y} = v_0 \cdot \sin(\theta)\)[/tex]
[tex]\[v_{0x} = 23.7 \, \text{m/s} \cdot \cos(60.0^\circ)\]\\v_{0x} \approx 23.7 \, \text{m/s} \cdot \frac{1}{2}\]\\v_{0x} \approx 11.85 \, \text{m/s}\]\\v_{0y} = 23.7 \, \text{m/s} \cdot \sin(60.0^\circ)\]\\v_{0y} \approx 23.7 \, \text{m/s} \cdot \frac{\sqrt{3}}{2}\]\\v_{0y} \approx 20.57 \, \text{m/s}\][/tex]
Now, let's find the time it takes for the ball to reach the maximum height using the vertical motion equation:
[tex]\[v_y = v_{0y} - g \cdot t\][/tex]
At the maximum height, the vertical velocity[tex](\(v_y\))[/tex] becomes zero. So:
[tex]\[0 = v_{0y} - g \cdot t_{\text{max}}\][/tex]
Solving for [tex]\(t_{\text{max}}\):[/tex]
[tex]\[t_{\text{max}} = \frac{v_{0y}}{g}\]\\t_{\text{max}} = \frac{20.57 \, \text{m/s}}{9.8 \, \text{m/s}^2}\]\\t_{\text{max}} \approx 2.10 \, \text{s}\][/tex]
Now, let's find the total time the ball is in the air. We know that the time to reach maximum height and the time to fall from maximum height to the ground are equal due to symmetry (neglecting air resistance). Therefore, the total time in the air is twice the time to reach maximum height:
Total time in the air [tex]\( = 2 \times t_{\text{max}}\)[/tex]
[tex]\[ \text{Total time in the air} \approx 2 \times 2.10 \, \text{s} \]\\ \text{Total time in the air} \approx 4.20 \, \text{s} \][/tex]
So, the time the ball remains in the air before striking the ground is closest to 4.20 seconds.
Graph the velocity of a car accelerating at a uniform rate from 7.0 m/s to 12.0 m/s in 2.0 s. Calculate the accerleration
Answers
a = (v - v₀) / t
a = (12 - 7) / 2
a = 2.5 m/s²
Final answer:
Velocity vs. time graph for uniform acceleration is a straight line, acceleration calculated using the formula a = (Vf - Vi) / t. The acceleration during the first 8 seconds is 1 m/s², and during the last 6 seconds it is -1 m/s².
Explanation:
To graph the velocity of a car accelerating at a uniform rate from 7.0 m/s to 12.0 m/s in 2.0 s, plot a velocity vs. time graph where the time axis (x-axis) spans at least 2 seconds and the velocity axis (y-axis) spans from 7.0 m/s to 12.0 m/s. The graph will be a straight line starting at the point (0, 7.0) and ending at the point (2.0, 12.0) since the acceleration is uniform.
To calculate the acceleration (a), use the formula a = (Vf - Vi) / t, where Vf is the final velocity, Vi is the initial velocity, and t is the time taken for the change in velocity. Substituting the given values, we get a = (12.0 m/s - 7.0 m/s) / 2.0 s = 2.5 m/s².
The acceleration of the car during the first 8 seconds (from 2 m/s to 10 m/s) is calculated by a = (Vf - Vi) / t = (10 m/s - 2 m/s) / 8 s = 1 m/s². During the last 6 seconds, when the car slows down from 10 m/s to 4 m/s, the acceleration is a = (4 m/s - 10 m/s) / 6 s = -1 m/s². Note the negative acceleration indicates a deceleration or slowing down.
A box is initially sliding across a frictionless floor toward a spring which is attached to a wall. the box hits the end of the spring and compresses it, eventually coming to rest for an instant before bouncing back the way it came. the work done by the spring on the box as the spring compresses is:
Answers
[tex]U= \frac{1}{2}kx^2 [/tex]
where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.
The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:
[tex]W=-\Delta U = \frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2 [/tex]
In our problem, initially the spring is uncompressed, so [tex]x_i=0[/tex]. Therefore, the work done by the spring when it is compressed until [tex]x_f[/tex] is
[tex]W=- \frac{1}{2}kx_f^2 [/tex]
And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.
How many electrons are in an electrically neutral atom of boron? A) 3 B) 5 C) 6 D) 8
Answers
If you want an explanation, here it is.
The atomic number gives the number of protons. Protons which have a positive charge are balanced by an equal number of electrons in a neutral atom. Boron number 5 has five protons and therefore as a neutral atom also has five electrons.
Answer:
B. five electrons
Explanation: