A shot hitting the birdie high and deep to the backcourt is called what

Answers

Answer 1
I think it would be "Clear"
Answer 2

Final answer:

In badminton, a shot that sends the birdie high and deep to the backcourt is called a clear. It's a defensive move to create more playing space and opportunities.

Explanation:

A shot in badminton that is hit high and deep into the opponent’s backcourt is known as a clear. This type of shot is a defensive move, intended to give the player hitting the clear more time to return to a ready position and to push the opponent to the rear of their court, potentially creating opportunities for future aggressive plays. The clear can be performed as either a forehand or backhand stroke depending on the player's position and the shuttlecock's trajectory.


Related Questions

At the end of a factory production line, boxes start from rest and slide down a 30 ∘ ramp 5.8 m long. part a if the slide is to take no more than 3.9 s , what's the maximum allowed frictional coefficient

Answers

We can solve the problem by analyzing the forces acting along the direction of the inclined plane. There are two forces: the first one is the component of the weight parallel to the inclined plane, which is
[tex]mg \sin \alpha[/tex]
where [tex]g=9.81 m/s^2[/tex] and [tex]\alpha=30^{\circ}[/tex]. This force points along the direction of the motion (down). The second one is the frictional force, acting against the motion, as well along the inclined plane:
[tex]-m g \mu_D[/tex]
where [tex]\mu_D[/tex] is the dynamic frictional coefficient, and we have to find the value of this. The negative sign means the force is acting against the direction of the motion.

Newton's second law states that the resultant of the forces acting on the object is equal to the product between the mass m and the acceleration a:
[tex]mg \sin \alpha - mg \mu_D = ma[/tex]
So we can write the acceleration as
[tex]a=g \sin \alpha - g \mu_D[/tex]

Now, we can use the law of motion along the direction of the inclined plane. For an uniformly accelerated motion, we have
[tex]S= \frac{1}{2} a t^2 [/tex]
Assuming the object is initially at rest. But S, in our problem, is exactly the lenght of the plane:
[tex]S=L=5.8 m[/tex]
So we can substitute a inside the formula and we get
[tex]L= \frac{1}{2} (g \sin \alpha - g \mu_D) t^2 [/tex]
From which we get
[tex]\mu_D = \sin \alpha - \frac{2L}{gt^2} [/tex]
and using the time given by the problem, [tex]t=3.9 s[/tex] and [tex]L=5.8 m[/tex], we find
[tex]\mu_D = 0.422[/tex] 
The component of weight(mg) that is responsible for the motion of boxes on ramp is:
[tex]m*g*sin \alpha [/tex]

Where m = mass of the boxes.
g = Acceleration due to gravity = 9.8[tex]m/s ^{2} [/tex]
[tex] \alpha [/tex] = The angle the ramp makes with the ground. In this case it is 30°.

Since the frictional force is:
[tex]F_{f} = [/tex]μ*N.

Where,
μ = Frictional Coefficient
N = Normal to the ramp = [tex] m*g*cos\alpha [/tex]

Therefore, the frictional force becomes = [tex]F_{f}[/tex] = μ*[tex]m*g* cos \alpha [/tex]

Apply Newton's second law we would get:
[tex]m*g*sin \alpha [/tex] -  μ*[tex]m*g* cos \alpha [/tex] = [tex]m*a[/tex]

=> [tex]a[/tex]=[tex]g*(sin \alpha - [/tex]μ [tex]cos \alpha )[/tex] -- (A)

Now according to equation of motion:
[tex]x = x_{o} + v_{o}*t + 1/2 * a * t^{2} [/tex]

Where x = 5.8m
[tex]x_{o}[/tex] = 0
[tex]v_{o}[/tex] = 0
[tex]t^{2}[/tex] = 10.24

Plug in the value in the above equation you would get:
[tex]a[/tex] = 1.1328[tex]m/s ^{2} [/tex]


Plug in [tex]a[/tex] in equation (A) and solve for μ, you would get,
μ = 0.4439   

______________ is a measure of the force of gravity pulling on an object of a given mass.

Answers

I think the answer is weight.

A uniform bridge 20.0m long and weighing 4.00 x 10^5 N is supported by two pillars located 3.00m from each end. If a 1.96 x 10 ^4N car is parked 8.00m from one end of the bridge. How much force does each pillar exert?

Answers

Final answer:

The left support exerts a force of 2.0392 x 10⁵ N and the right support exerts a force of 2.1568 x 10⁵ N on the bridge. This is calculated by considering the total weight supported by the bridge and the location of the car.

Explanation:

The problem you are asking about relates to static equilibrium, a concept in Physics. To find the force exerted by each pillar, we must first find the total weight supported by the bridge. The total weight is the sum of the weight of the bridge itself (4.00 x 10⁵ N) and the weight of the car (1.96 x 10^4 N), which gives us a total of 4.196 x 10⁵ N.

The location of the car changes the balance of these forces. The first pillar is 8m from the car and the second pillar is 12m from the car. To find the force on each pillar, we apply the principle that the sum of the moments about any point in a system in equilibrium is zero. Using the left pillar as the point, the moments of the bridge's weight and the car's weight are counterbalanced by the force of the right pillar:

Force_right_pillar = (Bridge_weight * Bridge_mid_point + Car_weight * Distance_from_left_pillar) / Total_bridge_length = ((4.00 x 10⁵ N * 10m) + (1.96 x 10⁴ N * 8m)) / 20m = 2.1568 x 10⁵ N.

The force exerted by the left pillar is then the total weight minus the force on the right pillar: Force_left_pillar = Total_weight - Force_right_pillar = 4.196 x 10⁵ N - 2.1568 x 10⁵ N = 2.0392 x 10⁵ N. So, the left support exerts a force of 2.0392 x 10⁵ N, and the right one exerts a force of 2.1568 x 10⁵ N.

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The forces exerted by the pillars supporting a 20.0m long and [tex]4.00 * 10^5 N[/tex] bridge with a [tex]1.96 * 10^4 N[/tex] car parked 8.00m from one end are approximately [tex]2.13 * 10^5 N[/tex] and [tex]2.07 * 10^5 N[/tex].

To find the forces exerted by each pillar on the uniform bridge, we need to consider the torque (moment of force) and the equilibrium conditions of the bridge.

Here is the step-by-step solution:

Given Data:

Length of the bridge (L) = 20.0 mWeight of the bridge [tex](W_{bridge}) = 4.00 * 10^5 N[/tex]Weight of the car [tex](W_{car}) = 1.96 * 10^4 N[/tex]Distance of each pillar from the end = 3.00 mPosition of the car from one end = 8.00 m

The two pillars provide upward forces labeled as [tex]F_L[/tex] and [tex]F_R[/tex].

1. Identify the distances to center of mass:

Center of mass of bridge from left end = 10.0 mPosition of car from left end = 8.0 m

2. Set up equilibrium equations:

Since the bridge is in static equilibrium, the sum of the forces and the sum of the torques must be zero.

Sum of Vertical Forces:

[tex]F_L + F_R - W_{bridge} - W_{car} = 0\\F_L + F_R = 4.00 * 10^5 N + 1.96 * 10^4 N\\F_L + F_R = 4.196 * 10^5 N[/tex]

Sum of Torques about the left pillar (taking counterclockwise as positive):

[tex]-(W_{bridge} * 7.0 m) - (W_{car} * 5.0 m) + (F_R * 14.0 m) = 0\\-(4.00 * 10^5 N * 7.0 m) - (1.96 * 10^4 N * 5.0 m) + (F_R * 14.0 m) = 0\\-2.8 * 10^6 Nm - 9.8 * 10^4 Nm + 14 F_Rm = 0\\14 F_R = 2.898 **10^6 Nm\\F_R = 2.898 * 10^6 Nm / 14 m\\F_R = 2.07 * 10^5 N\\[/tex]

Substitute F_R back into the sum of the forces equation:

[tex]F_L + 2.07 * 10^5 N = 4.196 * 10^5 N\\F_L = 4.196 * 10^5 N - 2.07 * 10^5 N\\F_L = 2.126 * 10^5 N\\[/tex]

Result:

The forces exerted by the pillars are approximately:

[tex]F_L = 2.13 * 10^5 N\\F_R = 2.07 * 10^5 N[/tex]

Which of the following would most likely happen if water did not form hydrogen bonds? (2 points)

Answers

The answer choices for this question are:

A) Water would exist only as a solid.

B) Water would be able to dissolve all ionic compounds.

C) Water would not expand when it freezes.

D) Water would not exist on Earth.

The right answer is the option C) Water would not expand when if freezes.

Explanation:

Expansion of water when it freezes is an atypical behavior because substances contract when they freeze with the correpondant increase in density.

But when water forms the ice crystals, the strong hydrogen bonds do not permit an arrangement in which the molecules of water are closer and so they occupy more space (more volume). The length of the hydrogen bonds limits the distance between each water molecule. So, these bonds are the responsible of the expansion of water as it freezes.


Answer:

Water would not expand when it freezes

Explanation:

every 24 hours, about how many hours are spent socializing and watching tv?

Answers

My guess would be about roughly 15, give or take a few. Including school, and it really depends how much TV you watch. Me personally spend about 5 hours watching TV or on the computer/ phone. So it really depends but 15 is my guess.

Answer:

16+ hours  

Explanation:

because they have a job and jobs are long so they would take as much time as they can get

A sphere has surface area 1.25 m2, emissivity 1.0, and temperature 100.0°C. What is the rate at which it radiates heat into empty space? The Stefan-Boltzmann constant is 5.67 × 10-8 W/(m2 ∙ K4).

A.) 9.9 mW

B.) 3.7 W

C.) 7.1 W

D.) 1.4 kW

Answers

The total power emitted by an object via radiation is:
[tex]P=A\epsilon \sigma T^4[/tex]
where:
A is the surface of the object (in our problem, [tex]A=1.25 m^2[/tex]
[tex]\epsilon[/tex] is the emissivity of the object (in our problem, [tex]\epsilon=1[/tex])
[tex]\sigma = 5.67 \cdot 10^{-8} W/(m^2 K^4)[/tex] is the Stefan-Boltzmann constant
T is the absolute temperature of the object, which in our case is [tex]T=100^{\circ} C=373 K[/tex]

Substituting these values, we find the power emitted by radiation:
[tex]P=(1.25 m^2)(1.0)(5.67 \cdot 10^{-8}W/(m^2K^4)})(373 K)^4=1371 W = 1.4 kW[/tex]
So, the correct answer is D.

The 2779-m Brooklyn-Battery Tunnel, connecting Brooklyn and Manhattan, is one of the world's longest underwater vehicular tunnels. (a) Determine its fundamental frequency of vibration. (b) What harmonic must be excited so that it resonates in the audio region at 20 Hz or greater?

Answers

For a cylinder that has both ends open resonant frequency is given by the following formula:
[tex]f= \frac{nv}{2L} [/tex]
Where n is the resonance node, v is the speed of sound in air and L is the length of a cylinder.
The fundamental frequency is simply the lowest resonant frequency.
We find it by plugging in n=1:
[tex]f_0= \frac{v}{2L}=\frac{343}{2\cdot 2779}=0.062 Hz[/tex]
To find what harmonic has to be excited so that it resonates at f>20Hz we simply plug in f=20 Hz and find our n:
[tex]20= \frac{n343}{2\cdot 2779} =n\cdot f_0[/tex]
We can see that any resonant frequency is simply a multiple of a base frequency.
Let us find which harmonic resonates with the frequency 20 Hz:
[tex]20=n\cdot f_0\\ n=\frac{20}{0.062}=322.58[/tex]
Since n has to be an integer, final answer would be 323.

Binary ionic compounds are: Check all that apply.

A.Compounds with very different electronegativities
B.Made from two metals
C.Compounds that share electrons
D.Metals and nonmetal ions

Answers

I believe the answer is the last one; they are made of metals and non metal ions. Binary ionic compounds are ionic compounds that contain only two elements, one is the cation and the other is the anion. They are neutrally charged compounds made up of bonded ions, a cation an an anion. For example sodium chloride; contains Na+ (a cation with a +1) combines with Cl- (an anion with a -1 charge) to make sodium chloride.

Compounds with very different electronegativities AND metals and non metals.

on Earth, how many grams of mass does it take to produce a force of 1 newton?

Answers

The weight of an object of mass m is equal to:
[tex]W=mg[/tex]
where [tex]g=9.81 m/s^2 [/tex] is the gravitational acceleration.

If we require a mass m to produce a force of 1 N, it means its weight is 1 N. Therefore, the required mass is:
[tex]m= \frac{W}{g}= \frac{1N}{9.81 m/s^2}=0.10 kg [/tex]

HELP!! Physical or chemical property? For the questions..

Answers

1.Chemical
2.Physical
3.Physical
4.Chemical
5Physical
I am not sure if these are right, but I hope I helped (:

A bullet is fired at a speed of 2434 ft/s. what is this speed expressed in millimeters per minute?

Answers

44,512,992 mm/m. Hopefully...

Sam boards a ski lift, and rides up the mountain at 6 miles per hour. once at the top, sam immediately begins skiing down the mountain, averaging 54 miles per hour, and does not stop until reaching the entrance to the lift. the whole trip, up and down, takes 40 minutes. assuming the trips up and down cover the same distance, how many miles long is the trip down the

Answers

The general relationship between space S and time t for an uniform motion is
[tex]S=vt[/tex] 
with v being the velocity.
For the motion going up, we have [tex]v_{up}=6 mph[/tex], so we can write
[tex]S_{up} = 6 t_{up}[/tex]
While for the motion going down, we have [tex]v_{dn} = 54 mph[/tex], and so
[tex]S_{dn} = 54 t_{dn}[/tex]
The problem says that two distances covered up and down are the same, so we can write
[tex]6 t_{up} = 54 t_{dn}[/tex]
and so
[tex]t_{up} = 9 t_{dn}[/tex]

We also know that the total time of the motion (up+down) is 40 minutes, which corresponds to [tex] \frac{2}{3} [/tex] of hour. So we can write
[tex]t_{up} + t_{dn} = \frac{2}{3} [/tex]
Substituting [tex]t_{up} = 9 t_{dn}[/tex] as we found before, we can find the value of [tex]t_{dn}[/tex]:
[tex]9 t_{dn}+t_{dn} = \frac{2}{3} [/tex]
[tex]t_{dn} = \frac{1}{15} h [/tex]
And so we find also
[tex]t_{up}=9t_{dn}= \frac{3}{5}h [/tex]

And from [tex]t_{dn}[/tex], we can finally find how long is the trip going down:
[tex]S_{dn}=54 t_{dn}=54 \cdot \frac{1}{15} = 3.6 mil[/tex]
So, 3.6 miles.
Let us divide this problem into two parts:
1) Sam rides up the mountain.
2) Sam rides down the mountain.

1. 
Since speed is distance over time, as:
[tex]v = \frac{s}{t} [/tex]

Therefore, distance would be:
[tex]s_{up} = v_{up} * t_{up} [/tex]

Where s = distance,
v = speed,
t = time.

In the problem, Sam's speed while riding up is v = 6 miles/hour = (6 * 1609.34 / 60) = 160.934 meters/second(in SI Units). Plug this value in the above equation, you would get:

[tex]s_{up} = 160.934 * t_{up}[/tex] --- (A)

2. 
As Sam rides down the mountain, the speed given is:
[tex]v_{down} = 54 miles/h[/tex]

Convert it in SI units; the speed would be in SI unit:
v = 54 miles/hour = (54 * 1609.34 / 60) = 1448.406 meters/second(in SI Units). Plug this value in the distance equation, you would get:
[tex]s_{down} = 1448.406 * t_{down}[/tex]

Since the [tex]s_{up} = s_{down}[/tex], therefore,

[tex]160.934 * t_{up} = 1448.406 * t_{down}[/tex]

=> 
[tex] t_{up} = 9 t_{down}[/tex]

Now the condition is that the whole trip, up and down, takes 40 minutes(2400seconds), it means:
[tex]t_{up} + t_{down} = 2400[/tex]

Plug in the value of [tex]t_{up} [/tex] in the above equation, you would get:
[tex]t_{down} = 240[/tex]

Therefore,
[tex]s_{down} = 1448.406*240[/tex]
[tex]s_{down} = 347617.44[/tex] meters (in relation to seconds)

[tex]s_{down} = 5793.624[/tex] meters (in relation to hours)

Now the last step is to convert meters into miles, you would get:
[tex]s_{down} = 5793.624/1609.34 = 3.6miles[/tex]

So the answer is 3.6miles.

explain the role that the ant and the acacia tree play in their symbiotic relationships

Answers

The presence of mutualistic ants greatly reduces bacterial abundance on surfaces of acacia leaves and has a visibly positive effect on plant health

You are given two carts, a and
b. they look identical and you are told that they are made of the same material. you place a at rest on an air track and give b a constant velocity directed to the right so that it collides elastically with
a. after the collision, both carts move to the right, the velocity of b being smaller than what it was before the collision. what do you conclude?

Answers

The mass of a is smaller than the mass of b.
Final answer:

In the described collision, the momentum of cart B is partially transferred to cart A due to the conservation of momentum principle. Hence, cart A starts moving and cart B slows down but continues to move to the right.

Explanation:

The scenario you described relates to the physics principle of conservation of momentum. In the collision of the carts, the total momentum before the collision equals the total momentum after the collision. This is because in an elastic collision, both kinetic energy and momentum are conserved. Initially, only cart B had momentum (since it was moving), and cart A was at rest. After the collision, both carts moved to the right, indicating that some of cart B's momentum was transferred to cart A. The fact that cart B moved slower after the collision indicates that it transferred some of its initial momentum to cart A. Hence, we can conclude that cart A gained momentum while cart B lost some but kept enough to continue moving to the right.

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A long, straight, vertical wire carries a current upward. due east of this wire, in what direction does the magnetic field point? a long, straight, vertical wire carries a current upward. due east of this wire, in what direction does the magnetic field point? upward west north south downward east

Answers

Final answer:

The magnetic field is due east of a long, straight, vertical wire carrying a current upward points downward.

Explanation:

The magnetic field created by a long, straight, vertical wire carrying a current upward can be determined using the right-hand rule. According to the right-hand rule, if you point your right thumb in the direction of the current in the wire, your fingers will curl in the direction of the magnetic field.

Since the wire is carrying current upward, the magnetic field will point in the clockwise direction when viewed from above. Therefore, the magnetic field due east of the wire will point downward.

A body 'A' of mass 1.5kg travelling along the positive X-axis with speed of 4.5m/s collides with another body 'B' of mass 3.2kg, which initially is at rest as a result of the collision, 'A' is deflected and moves in a speed of 2.1m/s in a direction which is at an angle of 30 degree below the X-axis. 'B' is set in motion at an angle Φ above the X-axis. calculate the velocity of 'B' after the collision.

Answers

I already answered this question. 
Please refer to this link https://brainly.com/question/8743596.

What is the average speed of a human that runs a distance of 5 miles in 1 hour?

Answers

It would be 5 mph. Hope it helps! :)

What does the index of refraction directly measure? the angle between the incident ray and the normal line the angle between the refracted ray and the normal line the bending of light in a medium the reflection of light in a medium

Answers

The bending of light in a medium

Answer:

The answer is The bending of light in a medium.

Explanation:

The refractive index is the dimensionless number that expresses the relationship between the speed of light in the air and the speed of light in the densest medium, that is, the refractive index directly measures the ratio of the velocity of the light in a vacuum and the speed of light in a medium. This phenomenon occurs because the light rays follow a rectilinear path, but when they pass from one means of transport to another, it is refracted, because the light has a velocity distance according to the density of the material it passes through. The refractive index has variables that affect the measurement, which are temperature, wavelength and pressure.

find the mass of a flying discus that has a net force of 1.05 newtons and accelerates at 3.5 m/s^2

Answers

f=ma
m=1.05/3.5= 0.3kg

How far away from the sun would a planet mercury's size have to be before it would have an atmosphere?

Answers

it's 87.97 days of the orbital period going round the sun it's days of the shortest of all planets in the solar system.

a doghouse has a mass of 98 kg, and the coefficient of friction between it and the ground is 0.95. what is the maximum force of friction between the doghouse and the ground?

Answers

Answer:

Maximum force of friction, F = 912.38 N

Explanation:

It is given that,

Mass of doghouse, m = 98 kg

Coefficient of friction between doghouse and the ground, μ = 0.95

We have to find the maximum force of friction between the doghouse and the ground. It is given by :

F = μ N

N = normal force = mg

F = μmg

F = 0.95 × 98 kg × 9.8 m/s²

F = 912.38 N

Hence, this is the required solution.

Which ions are most abundant in an acid?

Answers

The hydrogen ion is the most abundant in an acid for it is a bare proton that associates with any water in the molecule, solution or reaction so the H+ ions produced by an acid exist as H3O+ ions.

Moreover, in the Arrhenius theory, an acid is a substance that dissociates to produce H+ ions in solution and a base as a substance that dissociates to produce OH- ions in solution. Meanwhile, in the Bronsted-Lowery theory, an acid is considered as a proton donor.

Animals, including humans, need shelter, food, water, air, and space in order to survive. How do animals satisfy their basic needs?

Answers

if you come from study island then its all of them

Food, shelter and clothes are basic need of the humans. Animals like tiger hunt the animals in the jungle, some animals leaves on green grass which are abundantly present in the jungle. for shelter they live in the group at suitable place.

What is Animal ?

Animals belong to the biological kingdom Animalia and are multicellular, eukaryotic creatures. Animals generally ingest organic matter, breathe oxygen, can reproduce sexually, and form from a hollow ball of cells called a blastula during early development. There are an estimated 7.77 million animal species in existence as of 2022, but only 2.16 million of them have been described, including 1.05 million insects, over 36,000 fish, over 11,700 reptiles, over 11,100 birds, and 6,596 mammals. Animals can be as little as 8.5 micrometres (0.00033 in) or as large as 33.6 metres (110 ft). They construct extensive food webs through their complex interactions with one another and their surroundings. Zoology is the study of animals from a scientific perspective.

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Describe the image of an object that is 38 cm from a concave mirror that has a focal length of 10 cm.

Answers

Let's start by using the mirror equation:
[tex] \frac{1}{f}= \frac{1}{d}+ \frac{1}{q} [/tex]
where f=10 cm is the focal lenght of the mirror, d=38 cm is the distance of the object from the mirror, while q is the distance of the image from the mirror.
For the sign convention, f is taken as positive for a concave mirror. Therefore, we can solve the equation for q:
[tex] \frac{1}{q}= \frac{1}{f}- \frac{1}{d} = \frac{1}{10 cm}- \frac{1}{38 cm} [/tex]
from which we find q=13.6 cm.

The fact that q is positive means that the image is real, so it is on the same side of the object, with respect to the mirror.

Then we can also find the size of the image with respect to the original object. The magnification is given by
[tex]M=- \frac{q}{d} =- \frac{13.6}{38}=-0.36 [/tex]
The negative sign means that the image is inverted, and the size of the image is 0.36 times the size of the object.

Final answer:

A concave mirror with a focal length of 10 cm will form a virtual, upright image that is 11.43 cm away from the mirror. The magnification of the image will be -0.301.

Explanation:

A concave mirror with a focal length of 10 cm will form an image of an object that is 38 cm away. To determine the characteristics of the image, we can use the mirror equation: 1/f = 1/di + 1/do, where f is the focal length, di is the image distance, and do is the object distance. Rearranging the equation, we get di = 1 / (1/f - 1/do). Plugging in the given values, di = 1 / (1/10 - 1/38) = 11.43 cm.

The negative sign indicates that the image is virtual and upright. Furthermore, to determine the magnification of the image, we can use the formula: magnification = -di / do. Plugging in the values, magnification = -11.43 / 38 = -0.301.

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A 92-kg astronaut and a 2000-kg satellite are at rest relative to a space station. the astronaut pushes on the satellite, giving it a speed of 0.14 m/s directly away from the station. seven and a half seconds later the astronaut comes into contact with the station. part a what was the initial distance from the station to the astronaut?

Answers

Momentum is conserved. 

Momentum of the satellite is 1600*0.14=224 kg-m/sec 
Resulting momentum of the astronaut will be equal -- hence 
86*v=224 or the velocity of the astronaut is 2.60 m/sec 

In 7.5 seconds the astronaut would travel 19.53 meters

Answer:

22.8 s

Explanation:

First we calculate the change in momentum of the satellite.

ΔP = 2000 kg (0.14 m/s 0 m/s) = 280 kg·m/s

The change in momentum of the satellite is equal to the change in momentum of the astronaut as a result of Newton's Third law. Therefore:

280 kg·m/s = 92kg * v

                 v = 3.04 m/s

d = speed * time = 3.04 m/s * 7.5 s = 22.8 m

Planet B has a tilt of 45 degrees. What seasonal changes would be expected on this planet?
A. Extreme temperature changes between seasons 
B. Little to no change in temperatures between seasons
 
C. Seasonal changes along the equator only
 
D. Seasonal changes at the poles only

Answers

... Extreme temperature changes between seasons.

... Extreme changes in length of daylight and darkness.

... Larger areas around both poles where daylight/darkness
can last more than a whole day. 
Altogether, these areas cover half of the planet. 

Answer:

Extreme temperature changes between seasons

Explanation:

If our Earth is at a rotational degree of 23 degrees, then the planet is going to experience bigger changes between how hot and how cold it is going to be.

Calculate the kinetic energy of an 86-kg scooter moving at 16 m/s

Answers

The kinetic energy of an 86 - kilogram scooter moving at 16 m / s would be  11008 Joules.

What is mechanical energy?

Mechanical energy is the combination of all the energy in motion represented by total kinetic energy and the total stored energy in the system which is represented by total potential energy.

As given in the problem, we have to calculate the kinetic energy of an 86 - kilogram scooter moving at 16 m / s ,

The mass of the scooter = 86 kilogram

The velocity fo the scooter = 16 meters per second

The kinetic energy of the scooter = 0.5 × m × v²

                                                        = 0.5 × 86 × 16²

                                                        = 11008 Joules

Thus , the kinetic energy of an 86 - kilogram scooter moving at 16 m / s would be  11008 Joules .

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The kinetic energy of an 86-kg scooter moving at 16 m/s is calculated using the formula KE = ½ mv², resulting in 10,976 Joules or 10.976 kJ of kinetic energy.

To calculate the kinetic energy of an 86-kg scooter moving at 16 m/s, you can use the kinetic energy formula KE = ½ mv², where m is mass and v is velocity. Substituting the given values, we get KE = ½ (86 kg) (16 m/s)². After performing the calculation, the kinetic energy of the scooter is found to be 10,976 Joules or 10.976 kJ.

Why can an object still be seen when it is at absolute zero?

Answers

As the temperature goes down, the chaotic motion (velocity) of atoms start decreasing. If the temperature hits the absolute zero (which, in reality, is impossible to achieve), the atoms of the body would freeze, making the body still and stiff. One thing to note here is that the atoms do not get destroyed when the temperature reaches the absolute zero. That is the reason why the object can still be seen when it is at absolute zero.

A weightlifter raises a 200-kg barbell through a height of 2 m in 2.2 s. the average power he develops during the lift is

Answers

Power is calculated as work per unit time, and work in turn is calculated as force multiplied by distance. In this case, the force required is equivalent to the weight of the barbell multiplied by acceleration due to gravity.
P = W/t = Fd/t = mgd/t = (200 kg)(9.81 m/s^2)(2 m)/2.2 s = 1783.64 Watts.

Answer:

Power, P = 1781.81 watts

Explanation:

It is given that,

Mass of the barbell, m = 200 kg

It is lifted to a height of, h = 2 m

Time taken, t = 2.2 s

We need to find the develops during the lift. We know that the pwer developed by an object is equal to the rate of doing work. Mathematically, it is given by :

[tex]P=\dfrac{W}{t}[/tex]

[tex]P=\dfrac{mgh}{t}[/tex]

[tex]P=\dfrac{200\times 9.8\times 2}{2.2}[/tex]

P = 1781.81 watts

So, the average power developed during the lift is 1781.81 watts. Hence, this is the required solution.

The work input = _____. Fi × di Fi ÷ di Fo × do Fo ÷ do

Answers


I believe the work input= Fi × di. Work input is work done on a machine equal to the effort force times the distance through which the force is applied. On the other hand the work output is the work that is done by a machine equals resistance force times the distance through which the force is applied.
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