Without using the Gizmo, use your equation to calculate the acceleration of a puck that is in uniform circular motion with a radius of 3.0 m and a velocity of 9.0 m/s:___________.

Answer:

d = 8.06 (units)

Explanation:

In this problem, we are given two points with X & y coordinates. therefore we must use the Pythagorean theorem with the respective coordinates of each point to determine the required distance.

[tex]A = (-6, 1)\\B = (2, 2)[/tex]

Now using the Pythagorean theorem.

[tex]d=\sqrt{(x_{1}-x_{0})^{2} +(y_{1}-y_{o})^{2} }\\d=\sqrt{(2-(-6))^{2} +(2-1)^{2} }\\d=\sqrt{64+1} \\d=8.06[/tex]

I suppose you mean to say that the wall is frictionless in the first scenario? Also, I assume the block is to held in place. Construct a free body diagram for the block. There are 3 (in part A) or 4 (in part B) forces acting on it.

• its weight w, pulling it downward

• the normal force (magnitude n), pushing outward from the wall to the left

• the push as described, with magnitude p

• static friction (mag. f ), opposing the upward net force and thus pointing downward.

The static friction force has a magnitude proportional to that of the normal force. If the coefficient of static friction is µ, then

f = µ n

so if the wall is frictionless with µ = 0, then f = 0 and does not need to be considered.

(A) If µ = 0, then by Newton's second law we have

• net vertical force:

∑ F = p sin(50°) - w = 0

and we don't need to consider the net horizontal force to determine p from here. We get

p = w / sin(50°) = (4 kg) (9.8 m/s²) / sin(50°) ≈ 51 N

(B) If µ = 0.25, then Newton's second law gives

• net vertical force:

∑ F = p sin(50°) - w - f = 0

p sin(50°) - f = (4 kg) (9.8 m/s²)

p sin(50°) - f = 39.2 N

• net horizontal force:

∑ F = p cos(50°) - n = 0

p cos(50°) - f /0.25 = 0

[since f = 0.25 n]

p cos(50°) - 4f = 0

Multiply the first equation by -4, then add it to the second equation to eliminate f and solve for p :

-4(p sin(50°) - f ) + (p cos(50°) - 4f ) = -4 (39.2 N) + 0

p (cos(50°) - 4 sin(50°)) = -156.8 N

p = (156.8 N) / (4 sin(50°) - cos(50°)) ≈ 65 N

Answer:

R = 1.699 × [tex]10^{4}[/tex]

Explanation:

given data

height H = 1.70 m

time is t = 11.1 s

to find out

radius r of Earth

solution

we get here cosθ so

cos(θ) = R ÷ (R+1.70)          ................1

here θ = 360 × 11.1 ÷ (24×60×60)

θ = 0.04625

so put here value in eq 1

cos(0.04625) = R ÷ (R+1.70)

so radius R will be

0.9999 = R ÷ (R+1.70)

1.6999 + 0.999 R = R

R = 1.699 × [tex]10^{4}[/tex]

Answer:

Chemical

Explanation:

Answer:

As the service marks the start of every rally and subsequently dictates its flow, it is a crucial aspect of the game to get right in badminton.

These are the four main types of services in badminton and most can be executed with either your forehand or backhand.

1. Low serve

This low serve is almost a gentle tap over the net with the shuttle, with the aim of flying just over the net, yet falling just over the front line of his service court. It must not be too high or predictable, otherwise it would be easy for your opponent to do an outright smash or net kill.

2. High serve

The high serve is a powerful strike upwards with the shuttle, that aims to travel a great distance upwards and fall deep at the rear end of the court.

Although it is a strong serve and the popular choice of beginner players, its a serve that isn't so easy to disguise especially since you're using a forehand grip. Your opponent will already expect the shuttlecock to land at the back of the court.

Do remember that shuttlecocks have to fall within the corresponding service areas and this is different in singles and doubles.

3. Flick serve

This flick serve is also played upwards but at a much lesser altitude. It is most common for players to use their backhand to execute the flick serve and the trajectory is lower as this grip has less power.

The whole point of the backhand flick serve is deception, by mixing your serves up and making it look like you're doing a low serve. For this reason, serving with your backhand is thus very popular with competitive players.

It becomes hard for your opponent to predict if you are going to do a flick or a low serve as your stroke will look exactly the same until the point of contact.

4. Drive Serve

This is an attacking serve that is used by top badminton players like Lin Dan. The idea is to hit the shuttle directly at your opponent, limiting their return options and catching them off guard, winning you easy points. It's a good change of pace but it is also risky as if your opponent is prepared, he could just smash the shuttlecock back at you.

This serve is executed with your forehand through underarm action and following through. The shuttle should be dropped a bit sideways rather than in front of your body and hit flatter.

Now that you've determined the type of serve you want to make, here are a four tips on how to execute these serves well.

1. Keep your feet still

During the service, some part of both your feet must be in contact with the ground for it to be a legal serve.

2. Disguise your shots

Make sure your stroke is the same up to the point of contact with the shuttle. This will make your serve possible to predict only at the last possible second. Advanced players can try to trick their opponent by making it deliberately look like you're leaning back and about to do a high serve when you're really going to do a low serve.

3. Observe your opponents position

Is your opponent leaning towards the back already anticipating a high serve to the rear-court? In that case, you may want to execute a low serve to catch him off-guard. Always be aware of the position of your opponent. Try to imagine what he's expecting and do the opposite to gain an advantage.

4. Mix up your serves

Using just one type of service will make you too easy to predict. Make sure you incorporate at least two types of serves into your play. Once you've mastered the basic high and low serves, you can learn the flick and drive serves to add more dimension to your play.

In a nutshell, executing a service well allows you to start the rally strong and dictate its flow.

Explanation:

I hope it's help

Answer:

b

Explanation:

doing same amount of work while increase in amount of heat released

Explanation:

We need to use the conservation of energy to help us solve this one.

The energy imparted onto the bullet projectile is equal to the energy imparted into the recoiling gun. We can ignore potential energy and only consider the kinetic energy of the two masses (bullet and gun).

Remember that kinetic energy is expressed as [tex]K_E = \frac{1}{2}mv^2[/tex]

Conservation of energy means that both the gun and the bullet have the same amount of energy.

[tex]\frac{1}{2}m_gv_g^2 = \frac{1}{2}m_b v_b^2[/tex]

We'll want to put the mass terms into the same units. So the gun has a mass of 2340 grams.

Answer:

40m to the East

Explanation:

Displacement is the distance moved in a specific direction. When writing the displacement value of a moving body, the direction must be put in the description.

Displacement takes into account the start and finish position of a body.

                                                 100m

Start         --------------------------------------------------------------  →

                                                               60m

                                       Final ←----------------------------------

Displacement  = 100m  - 60m  = 40m

 Therefore, the displacement is 40m due east

Complete question is;

A 5 × 10^(5) kg subway train is brought to a stop from a speed of 0.5 m/s in 0.4 m by a large spring bumper at the end of its track. what is the force constant k of the spring

Answer:

781250 N/m

Explanation:

From conservation of energy, potential energy is equal to kinetic energy.

Thus;

½mv² = ½kx²

where;

m = mass of train

v = velocity of train

k = force constant of spring

x = the distance the train went while being stopped

We are given;

Mass; m = 5 × 10^(5) kg

Velocity; v = 0.5 m/s

Distance; x = 0.4 m

Thus, from ½mv² = ½kx²

Divide both sides by ½ to get;

mv² = kx²

k = mv²/x²

k = [(5 × 10^(5)) × 0.5²]/0.4²

k = 781250 N/m

Answer: it will take 1000 guitar players for sound intensity level to be 100 dB

Explanation:

Given that;

noise level of a single player = 70 dB

Intensity of one guitar will be;

SL= 10 × log(I/I₀)

where I₀ is hum threshold intensity = 10⁻¹² W/m²

I is intensity of sound produced by 1 guitar

SL is intensity level of one guitar

so we substitute

70 = 10 × log(I/10⁻¹²)

70/10 = log (I/10⁻¹²)

I = 10⁻¹² × 10⁷

I = 10⁻⁵ W/m²  

now Suppose n guitars produce sound intensity of 100 dB, then intensity of n guitars (In) will be;

100 = 10 × log(In/10⁻¹²)

log(In/10⁻¹²)  = 100/10

In = 10⁻¹² × 10¹⁰ =  10⁻² Wm²

we know that Now intensity of 1 guitar I = 10⁻⁵ W/m²

so Intensity of n guitars will be;

n × I = In

n = In / I

we substitute

n = 10⁻² Wm² /  10⁻⁵ W/m²

n = 1000

Therefore, it will take 1000 guitar players for sound intensity level to be 100 dB

Question:

A spherical balloon with a 36 cm diameter is being deflated. Its volume V is a function of its radius r according to [tex]V(r) = \frac{4}{3}\pi r^3.[/tex]

As it's deflating, it is much easier to measure its radius than its volume. Suppose the radius of the balloon is [tex]r(t) = 18 - 18e^{-0.24t}[/tex] cm at t seconds.

Answer:

[tex]V'(4) = 201.143[/tex]

[tex]r(4) = 11.106[/tex]

Explanation:

Given

[tex]V(r) = \frac{4}{3}\pi r^3.[/tex]

[tex]r(t) = 18 - 18e^{-0.24}[/tex]

Solving (a):  V'(4)

First, we determine V('r)

[tex]V(r) = \frac{4}{3}\pi r^3.[/tex]

Differentiate w.r.t r

[tex]V'(r) = 3 * \frac{4}{3}\pi r^{3-1}[/tex]

[tex]V'(r) = 3 * \frac{4}{3}\pi r^2[/tex]

[tex]V'(r) = 4\pi r^2[/tex]

Substitute 4 for r and take [tex]\pi = \frac{22}{7}[/tex]

[tex]V'(4) = 4 * \frac{22}{7} * 4^2[/tex]

[tex]V'(4) = 4 * \frac{22}{7} * 16[/tex]

[tex]V'(4) = \frac{4 * 22* 16}{7}[/tex]

[tex]V'(4) = \frac{1408}{7}[/tex]

[tex]V'(4) = 201.143[/tex]

This means that the volume of the balloon when the radius is deflated to 4 seconds is 201.143 cm^3

Solving (b): r(4)

Substitute 4 for t in [tex]r(t) = 18 - 18e^{-0.24t}[/tex]

[tex]r(4) = 18 - 18e^{-0.24*4}[/tex]

[tex]r(4) = 18 - 18e^{-0.96}[/tex]

[tex]r(4) = 18 - 18*0.383[/tex]

[tex]r(4) = 11.106[/tex]

This means that the radius of the balloon when at 4 seconds of deflation is 11.106 cm

Answer:

energy of position

Explanation:

I think that is the answer

Answer:

Weight of the runner is 60 kilograms

Explanation:

Momentum is the product of weight and velocity.

The formula is : p=mv   where m is mass in kg and v is velocity in m/s

Given in the question ;

v= 20 m/s  and p=1200 kg*m/s

Applying the formula as;

p= mv

1200 = m * 20

1200/20 = m

60 kg = m

Answer:

Explanation:

The best terms that represent vector quantity are magnitude and direction. Magnitude and direction are both qualities in vector quantity. The basic measurement of vector quantity is the meter all others would best be described as scalars. So to best describe vector quantity use the meter uni

This question is incomplete, the complete question is;

1.80 kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a physical pendulum. the period for small angle oscillations is 0.940 s.

a) what is the moment of inertia of the wrench about an axis through the pivot?

b) If the wrench is initially displaced 0.400 rad from its equilibrium position, what is the angular speed of the wrench as it passes through the equilibrium position

Answer:

a) the moment of inertia of the wrench about an axis through the pivot is 0.0987 Kg.m²

b) the angular speed of the wrench as it passes through the equilibrium position is 2.6559 rad/s

Explanation:

Given that

mass m = 1.80 kg

L or d = 0.250 m

period T = 0.940 sec

g = 9.8

a)

we take a look at the expression for the moment of inertia

I = T²mgL / 4π²

we substitute

I = ((0.940)² × 1.80 × 9.8 × 0.250) / 4 × π²

I = 3.896676 / 4 × π²

I = 0.0987 Kg.m²

Therefore the moment of inertia of the wrench about an axis through the pivot is 0.0987 Kg.m²

b)

If the wrench is initially displaced 0.400 rad from its equilibrium position,

the angular speed of the wrench as it passes through the equilibrium position = ?

Using conservation of energy;

mg × d × ( 1 - cos∅) = 1/2×I×w²

we substitute

1.80 × 9.8 × 0.250 × ( 1 - cos(0.400 rad)) =  1/2 × 0.0987  × w²

4.41 × ( 1 - 0.921060994 ) = 0.04935 × w²

0.000441 = 0.04935w²

w² = 0.3481 / 0.04935

w² = 7.0536

w = √7.0536

w = 2.6559 rad/s

Therefore, the angular speed of the wrench as it passes through the equilibrium position is 2.6559 rad/s

Answer:

The new gravitational force will be of 4.5 units

Explanation:

Recall that the formula for the gravitational force between two objects of masses m1 and m2 separated by a distance d is given by:

[tex]F_g=G\,\frac{m1*m2}{d^2}[/tex]

in our case, we are told that such gives 72 units of force:

[tex]F_g=G\,\frac{m1*m2}{d^2} =72[/tex]

Then we change the distance between the objects to 4 times the original (4 * d), such will produce a new gravitational force Fg':

[tex]F_g'=G\,\frac{m1*m2}{(4*d)^2} =G\,\frac{m1*m2}{16*d^2} = \frac{1}{16} *G\,\frac{m1*m2}{d^2}=\frac{1}{16} *\,72=4.5[/tex]

Therefore the new gravitational force would be of 4.5 units

Answer:

The distance of the object from the concave mirror determines if it is a virtual or real image

Explanation:

Concave mirrors form both real and virtual images. When the concave mirror is placed very close to the object, a virtual and magnified image is obtained and if we increase the distance between the object and the mirror, the size of the image reduces and real images are formed.

Answer:

Work = 250 Joules.

Explanation:

Given the following data;

Potential difference, V = 25V

Charge, C = 10 Coulombs

To find the work required;

Work = Charge * potential difference

Substituting into the equation, we have

Work = 10*25

Work = 250 Joules.

Therefore, the amount of work required to move the charge from A to B is 250 Joules.

Answer:

average

Explanation:

Answer:

f = 0.0833 Hz

Explanation:

Frequency is defined as the no. of cycles per unit time. The frequency of water waves can be given by the following formula:

[tex]Frequency\ of\ waves = f = \frac{No.\ of\ Waves}{time}\\[/tex]

here,

No. of Waves = No. of times seagull moves up and down = 5

Time = 1 minute = 60 sec

Therefore, using these values in the formula, we get:

[tex]f = \frac{5}{60\ sec}[/tex]

f = 0.0833 Hz

Answer:

A medium in which speed of light is more is known as optically rarer medium and a medium in which speed of light is less is said to be optically denser medium. For example in air and water, air is raer and water is a denser medium.

Explanation:

Answer:Rarer medium is less dense

Explanation:

Rarer medium is less dense than dense medium because rarer medium doesn't have a very high amount of particles packing it together. It's easier to see through a rarer medium, so you're able to get a clearer picture of what's on the other side of it. Denser medium is, simply put, more dense.

Answer:

d. 268 s

Explanation:

Constant Speed Motion

An object is said to travel at constant speed if the ratio of the distance traveled by the time taken is constant.

Expressed in a simple equation, we have:

[tex]\displaystyle v=\frac{d}{t}[/tex]

Where  

v = Speed of the object

d = Distance traveled

t  = Time taken to travel d.

From the equation above, we can solve for d:

d = v . t

And we can also solve it for t:

[tex]\displaystyle t=\frac{d}{v}[/tex]

Two cars are initially separated by 5 km are approaching each other at relative speeds of 55 km/h and 12 km/h respectively. The total speed at which they are approaching is 55+12 = 67 km/h.

The time it will take for them to meet is:

[tex]\displaystyle t=\frac{5}{67}[/tex]

t = 0.0746 hours

Converting to seconds: 0.0746*3600 = 268.56

The closest answer is d. 268 s

Answer:

Yes

Explanation:

Answer:

Periodic Motion :–

This is a type of motion where the object repeats its motion after a fixed interval of time.

Examples: pendulum of a clock, motion of child on a swing etc.

Answer:

Periodic motion is a type of motion that is being performed. You can take a rocking chair, bouncing ball, and a swing as an example of periodic motion.

Periodic motion is just another word for the motion because it describes any type of object performing movements even an object vibrating is another great example of periodic motion. If you want to talk about the motion of the universe take Earth and the moon orbiting around the sun it is an object performing.

Hope this helped :)

Explanation:

Since Pressure = Force / Area,

We have 550N / (0.003m²) = 180k Pa. (2s.f.)

Answer:70 watts

Explanation: I took the test and got it correct

Answer:

a) v = 32.8 m/s

b) v= 15.4 m/s

Explanation:

a)

        [tex]v = v_{o} + a*t (1)[/tex]

        where v₀ is the initial velocity and a is the acceleration, being t the

       time elapsed.

       [tex]v = 23.6 m/s + (1.30 m/s2 * 7.10 s) = 32. 8 m/s (2)[/tex]

b)  

       [tex]v = 23.6 m/s + ((-1.15 m/s2) * 7.10 s)) = 15. 4 m/s (3)[/tex]

Answer:

Alpha radiation is a kinda massive type of radiation, this would mean that alpha radiation is weakly penetrating. Because of this:

a) When we have particles of smoke in the path of the alpha radiation, the smoke will "stop" a big part of the radiation, and when the detector stops detecting the radiation, it will let the alarm go off.

b) The radiation of alpha particles can be harmful, yes, but the smoke detector is shielded (Again, alpha radiation is weakly penetrating, then shielding the smoke detector is an easy task), this means that the radiation can not get out of the smoke detector, then it is not a problem for the people living near it.

Explanation:

Q is the charge, 9×10^9×3×10^4 is the magnitude ,d = 5 cm therefore q=3×10^4÷5^2=2.7×10^14÷25=1.08×10^13 c