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Balloons and static electricity
Answer the following questions:

Answer:

2.is the most massive object in the system, and less massive objects orbit more massive objects

is considered as a single focus by the objects describing elliptical paths about the sun.

The force applies an acceleration in the reverse direction of

29 N = (0.145 kg) a   →   a = 200 m/s²

and in order to hit the ball back with speed 40.0 m/s, this acceleration is applied over time t such that

200 m/s² = (40.0 m/s - (-40.0 m/s)) / t   →   t = (80.0 m/s) / (200 m/s²) = 0.4 s

Answer:

See the explanation below.

Explanation:

First, we must determine the mass of the baseball, we know that the weight of a body is defined as the product of mass by gravity.

[tex]w=m*g[/tex]

where:

w = weight = 1.47[N]

m = mass [kg]

g = gravity acceleration = 9.81 [m/s²]

Now replacing:

[tex]m=w/g\\m = 1.47/9.81\\m = 0.149[kg][/tex]

We now know that kinetic energy is converted to potential energy as the ball descends. By means of the following equation, we can determine the potential energy when the baseball is 10 meters high.

[tex]E_{pot}=m*g*h\\[/tex]

where:

Epot = potential energy [J]

m = mass = 0.149[kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = 10 [m]

Now replacing:

[tex]E_{pot}=0.149*9.81*10\\E_{pot}=14.7[J][/tex]

In theory, this same energy must be converted to kinetic energy just before the ball hits the floor. But we see that we only have 12 [J] of kinetic energy.

That is to say, that of the 14 [J] that were had as potential energy (mechanical energy) 2 [J] was converted to internal energy, and the rest was converted to kinetic energy (mechanical energy)

[tex]E_{int}=14-12\\E_{int}=2[J][/tex]

Note: Potential, kinetic and elastic energies are forms of mechanical energy.

Given :

A race car moves along a circular track of radius 100m at a velocity of 25m/s.

To Find :

(a) What is  the time taken to complete one lap of the circular track.

(b) What is the time taken for 10  laps.

Solution :

Circumference of circular track,

[tex]C = 2\pi r\\\\C = 2\times \pi \times 100\ m\\\\C = 628.32 \ m[/tex]

a) Time taken to complete one lap is :

[tex]t= \dfrac{628.32}{25}\ s\\\\t =25.13 \ s[/tex]

b) Time taken to complete 10 laps is :

[tex]t_{10} = 10\times t\\\\t_{10} = 10\times 25.13\ s\\\\t_{10} = 251.3\ s[/tex]

Hence, this is the required solution.

Answer:

[tex]\boxed {\boxed {\sf \frac{14}{3} \ or \ 4.6667 \ m/s^2}}[/tex]

Explanation:

Acceleration can be found using the following formula.

[tex]a=\frac{v_f-v_1}{t}[/tex]

where [tex]v_f[/tex] is the final velocity, [tex]v_i[/tex] is the initial velocity and t is the time.

The lizard started at 12.0 m/s and accelerated up to its final velocity of 40.0 m/s in 6 seconds.

Therefore:

[tex]v_f= 40.0 \ m/s \\v_i= 12.0 \ m/s \\t= 6 \ s[/tex]

Substitute the variables into the formula.

[tex]a=\frac{40.0 \ m/s - 12.0 \ m/s}{6 \ s}[/tex]

Solve the numerator first and subtract.

[tex]a=\frac{ 28 \ m/s}{6 \ s}[/tex]

Divide.

[tex]a= \frac{14}{3} \ m/s/s= \frac{14}{3} \ m/s^2[/tex]

[tex]a=4.66667 \ m/s^2[/tex]

The lizard's average acceleration is 14/3 or 4.66667 m/s²

Answer:

4 2/3 m/s

Explanation:

first thing to find the average acceleration is to figure out what the increase in speed was, we can do that by subtracting the original speed from the speed after accelerating, that looks like:

40 - 12 = 28

so the lizard accelerated 28 m/s in 6 seconds, to find the average increase in m/s every second, we divide the m/s by the seconds, which gives us:

28 / 6 = 4.66 = 4 2/3 m/s

Answer:

7.5 × 10¹⁵ Hz

Explanation:

CHECK THE COMPLETE QUESTION BELOW;

Amateur radio operators in the United States can transmit on several bands. One of those bands consists of radio waves with a wavelength near 40nm.

Calculate the frequency of these radio waves.

The frequency of these radio waves can be calculated using the expression below

ν = c/λ

Where

v = Frequency of the radio waves

c= Speed of light= 3.00 × 10⁸ m/s

λ= Wavelength= 40 nm

= 40 × 10⁻⁹ m

ν = 3× 10⁸/40 × 10⁻⁹

= 7.5 × 10¹⁵ Hz

Hence, the frequency of these radio waves is 7.5 × 10¹⁵ Hz

Answer:

6.3Ns

Explanation:

Impulse can be calculated by using the formula as follows:

I = F × ∆t

Where;

I = impulse (Ns)

F = Force exerted (N)

∆t = change in time (s)

According to the information provided in the question, I = ?, F = 15N, ∆t (time interval) = 0.42s

I = 15 × 0.42

I = 6.3Ns

Answer:

The tangential force will act as a torque on the body, increasing its angular velocity and thus also increasing its kinetic energy. By the work-kinetic-energy theorem, work has been done on the body. Yes, in non-uniform circular motion the work done on the object is non-zero, for the reason you stated.

Explanation:

Answer:

R = 1.8 m

Explanation:

This is a simple harmonic movement exercise, at the bottom of the swing the acceleration is vertical upwards and the speed is tangential to the trajectory, that is horizontal; the expression for the centralized acceleration is

              [tex]a_{c}[/tex] = v² / R

              R = v² /a_{c}

where the radius is equal to the length of the swing

let's calculate

            R = 8.1 / 4.5

            R = 1.8 m

Answer:

\alpha  = \frac{2F}{3m} \ \frac{1}{r}

maximmun x =r

Explanation:

In this exercise we are asked for the maximum angular acceleration, let's start by writing the second law of / newton for rotational motion. Let's fix our reference system at the midpoint of the bar that has a length 2r

       Σ τ  = (I₁ + I₂) α

where I₁ and I₂ moment of inertia of the capsule with masses m and 2m, respectively. Let's treat these capsules as point particles

       I₁ = m r²

       I₂ = 2m r²

the troque of a pair of force is the force times the distance perpendicular to the point of application of the force which is the same for both forces, we will assume that the counterclockwise rotation is positive

      Στ  = F x + F x

the angular acceleration is the same because they are joined by the bar of negligible mass, let us substitute

     2 F x = (m r² + 2m r²) α

     α = [tex]\frac{2F x}{3m r^{2} }[/tex]

     α = [tex]\frac{2F }{3m } \ \frac{x}{r^{2} }[/tex]

let's analyze this expression

* for the application point in the center (x = 0) at acceleration is zero

* for the point of application of the torque at the ends the acceleration is

              [tex]\alpha = \frac{2F}{3m} \ \frac{1}{r}[/tex]

this being its maximum value

Answer:

0.02 HZ

Explanation:

There is no way to explain it.

Answer:

C.0.02 Hz

Explanation:

Answer:

The mountain is 403 m tall

Explanation:

Horizontal Launch

When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled by gravity until it hits the ground.

The range or maximum horizontal distance traveled by the object can be calculated as follows:

[tex]\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}[/tex]

It's given that if a ball is kicked from the top of the Sugarloaf Mountain, it will land near its base without hitting the side. The initial speed is v=9.37 m/s and it reaches a horizontal distance of d=85 m. Solving the equation for h:

[tex]\displaystyle h=\frac{d^2g}{2v^2}[/tex]

Substituting:

[tex]\displaystyle h=\frac{85^2*9.8}{2*9.37^2}[/tex]

[tex]\displaystyle h=403 \ m[/tex]

The mountain is 403 m tall

Answer:

60%

Explanation:

Heya!

For this problem, use the formula:

s = Vo * t + (at^2) / 2

Since the initial velocity is zero, the formula simplifies like this:

s = (at^2) / 2

Clear a:

2s = at^2

(2s) / t^2 = a

a = (2s) / t^2

Data:

s = Distance = 518 m

t = Time = 7,48 s

a = Aceleration = ¿?

Replace according formula:

a = (2*518 m) / (7,48 s)^2

Resolving:

a = 1036 m / 55,95 s^2

a = 23,34 m/s^2

The aceleration must be 23,34 meters per second squared

Answer:

xxxxx

Explanation:

Density is the mass of an object divided by its volume. Density often has units of grams per cubic centimeter (g/cm3). Remember, grams is a mass and cubic centimeters is a Volume (the same volume as 1 milliliter).

Answer:

The pressure exerted by the brick on the table is 18,933.3 N/m².

Explanation:

Given;

height of the brick, h = 0.1 m

density of the brick, ρ = 19,300 kg/m³

acceleration due to gravity, g = 9.81 m/s²

The pressure exerted by the brick on the table is calculated as;

P = ρgh

P = (19,300)(9.81)(0.1)

P = 18,933.3 N/m²

Therefore, the pressure exerted by the brick on the table is 18,933.3 N/m².

Answer:

book resting on a shelf has no potential energy relative to the shelf since it has a height of zero meters relative to the shelf. However, if the book is elevated to some height above the shelf, then it has potential energy proportional to the height at which it resides above the shelf.

An object can have both kinetic and potential energy at the same time. For example, an object which is falling, but has not yet reached the ground has kinetic energy because it is moving downwards, and potential energy because it is able to move downwards even further than it already has. The sum of an object's potential and kinetic energies is called the object's mechanical energy.

As an object falls its potential energy decreases, while its kinetic energy increases. The decrease in potential energy is exactly equal to the increase in kinetic energy.

Another important concept is work.

The string moves to the right, as it restores its original position with the median plane of the bow. As a result, the string "pulls" on the arrow with a force F2. 2. The tip of the arrow T moves slightly to the left.

pls thank me and brainliest me

Answer:

The free end of the blade has a tangential velocity of about 88.19 m/s

Explanation:

The angular velocity of the blades is  [tex]2 \pi /5.7\,\,rad/sec[/tex]

since the blades are 80 m long, then the tangential velocity of the free end of the blade is:

[tex]v_{tan} \approx 88.19\,\,m/s[/tex]

2 conditioner are

Explanation:

force applied and displacement produced

Answer:

0.02 Ns

Explanation:

From the question given above, the following data were obtained:

Force (F) = 5 N

Time (t) = 4 ms

Change in momentum =?

Next, we shall convert 4 ms to s. This can be obtained as follow:

1 ms = 1×10¯³ s

Therefore,

4 ms = 4 ms × 1×10¯³ s / 1 ms

4 ms = 4×10¯³ s

Thus, 4 ms is equivalent to 4×10¯³ s

Change in momentum = Impulse

Impulse (I) = Force (F) × time (t)

Change in momentum = Force (F) × time (t)

Force (F) = 5 N

Time (t) = 4×10¯³ s

Change in momentum =?

Change in momentum = 5 × 4×10¯³

Change in momentum = 0.02 Ns

Therefore, the change in the momentum of the ball is 0.02 Ns.

Answer:

The increase of kinetic energy is 108,000 J

Explanation:

Kinetic Energy

Is the energy an object has due to its state of motion. It's proportional to the square of the speed and the mass.

The equation for the kinetic energy is:

[tex]\displaystyle K=\frac{1}{2}mv^2[/tex]

Where:

m = mass of the object

v = speed at which the object moves

The kinetic energy is expressed in Joules (J)

A car has a total mass of m=1,200 kg and travels at v1=12 m/s. Then it increases its speed at v2=18 m/s.

It's required to compute the increase of kinetic energy. We'll calculate both energies K1 and K2 and then subtract them.

[tex]\displaystyle K_1=\frac{1}{2}1,200*12^2=86,400\ J[/tex]

[tex]\displaystyle K_2=\frac{1}{2}1,200*18^2=194,400\ J[/tex]

The increase of kinetic energy is:

[tex]\Delta K=K_2-K_1 =194,400\ J-86,400\ J[/tex]

[tex]\Delta K=108,000\ J[/tex]

The increase of kinetic energy is 108,000 J

Answer:

The common speed is 1.95 m/s

Explanation:

Law Of Conservation Of Linear Momentum

It states that the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and velocity v is  

P=mv.  

If we have a system of bodies, then the total momentum is the sum of all of them:

[tex]P=m_1v_1+m_2v_2+...+m_nv_n[/tex]

If a collision occurs, the velocities change to v' and the final momentum is:

[tex]P'=m_1v'_1+m_2v'_2+...+m_nv'_n[/tex]

In a system of two masses, the law of conservation of linear momentum

is written as:

[tex]m_1v_1+m_2v_2=m_1v'_1+m_2v'_2[/tex]

If both masses stick together after the collision at a common speed v', then:

[tex]m_1v_1+m_2v_2=(m_1+m_2)v'[/tex]

The common velocity after this situation is:

[tex]\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}[/tex]

The truck of m1=1200 N (weight) travels at v1=2 m/s and hits a stationary mass (v2=0) of m2=30 N (weight). After the bodies collide, they keep moving together. Before we can calculate the common speed, we need to calculate the masses of the bodies, since they are given as weights.

[tex]m_1=\frac{P_1}{g}=\frac{1200}{9.8}=122.45 Kg[/tex]

[tex]m_2=\frac{P_2}{g}=\frac{30}{9.8}=3.06 Kg[/tex]

Now calculate the common speed:

[tex]\displaystyle v'=\frac{122.45 * 2+3.06 * 0}{122.45+3.06}[/tex]

[tex]\displaystyle v'=\frac{244.9}{125.51}=1.95\ m/s[/tex]

The common speed is 1.95 m/s

Answer:

To calculate the absolute pressure we used the depth as 2 m instead of 2 mm because the given dimensions for the swimming pool do not make sense (they are too small).  

The absolute pressure on the bottom of a swimming pool is 1.21x10⁵ Pa.

Explanation:

The absolute pressure is given by:

[tex]P_{A} = P_{g} + P_{atm}[/tex]

Where:

[tex]P_{g}[/tex]: is the gauge pressure

[tex]P_{atm}[/tex]: is the atmospheric pressure = 1 atm = 101325 Pa

The gauge pressure can be found as follows:

[tex]P_{g} = \rho gh[/tex]

Where:

ρ: is the density of water = 1000 kg/m³

g: is the gravity = 9.81 m/s²

h: is the height or the depth

Since the given dimensions in the statement for the swimming pool do not make sense (they should be in meters instead of millimeters), we will use h = 2 m.

[tex]P_{g} = \rho gh = 1000 kg/m^{3}*9.81 m/s^{2}*2.0 m = 19620 Pa[/tex]

Hence, the absolute pressure is:

[tex]P_{A} = P_{g} + P_{atm} = 19620 Pa + 101325 Pa = 120945 Pa = 1.21 \cdot 10^{5} Pa[/tex]        

Therefore, the absolute pressure on the bottom of a swimming pool is 1.21x10⁵ Pa.

I hope it helps you!                                                  

Answer:

Explanation:

The formula method is used to calculate termination payments on a prematurely ended swap, where the terminating party compensates the losses borne by the non-terminating party due to the early termination.