grrr i don't have much time, help please T^T
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The mass of the string is 0.515 kg.

Given that:

To solve for the mass of the string,

Therefore, the mass would be[tex]15.9*0.03242= 0.515448= 0.515[kg][/tex]

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Answer:

F = 750N

Explanation:

We know that F(force) = mass(m) × acceleration(a)

And that Acceleration(a) = velocity(V) ÷ time(t)

So we'll first calculate the acceleration

a = V = 8m/s = 1m/s²

     t        8

Next we'll calculate the force

F = ma = 750kg × 1m/s² = 750N

Answer:

  3/70 kg ≈ 0.04286 kg

  2.8 m/s

Explanation:

In a perfectly elastic collision, the total momentum of the system is constant, and the difference in velocities changes sign.

We assume the other ball is at rest initially, and that the collision is "head-on." That is, final velocities are in the same direction as the initial velocity of the cue ball. We further assume no energy loss, and that the balls do not rotate.

__

Assuming the ball being hit is initially at rest, it contributes no momentum to the system. The total momentum is ...

  (0.1 kg)(2 m/s) = 0.2 kg·m/s

__

If the cue ball has velocity v1 and the hit ball has velocity v2, the initial difference of velocities is ...

  v1 -v2 = 2 m/s -0 m/s = 2 m/s

__

After the collision, we must have ...

  v1 -v2 = -2 m/s . . . . the velocity difference changes sign

  v2 = v1 +2 m/s = 0.8 m/s +2 m/s

  v2 = 2.8 m/s

The velocity of the other ball is 2.8 m/s after the collision.

__

The total momentum after the collision is the same as before, so we must have ...

  m1·v1 +m2·v2 = initial momentum

  (0.1 kg)(0.8 m/s) +(m2)(2.8 m/s) = 0.2 kg·m/s

  m2 = (0.2 kg·m/s -0.08 kg·m/s)/(2.8 m/s) = (0.12/2.8) kg = 3/70 kg

The mass of the other ball is 3/70 ≈ 0.04286 kg.

Answer:

Propagation of Sound Waves:

Sound waves move (propagate) through a medium by means of particle-to-particle interaction. The source disturbs the immediately surrounding particles, causing them to vibrate about their fixed position at the same frequency as the source.

Explanation:

formula is given by

h/mv

where h is 6.626 × 10 ^-34

and c = 3 × 10^8

m = mass

by plugging in the values in the formula, the de Brogile wavelength is equal to 7.6424 × 10^(-15)

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When oxygen comes in contact with NOx and VOCs They combine to create ozone to replenish the ozone layer. Option A is correct.

The layer form on the stratosphere region of the earth, which absorbs most of the ultraviolet rays.

The ultraviolet rays come from the sun are very harmful, increases global warming when enter inside the ozone layer due to its deplection.

When oxygen comes in contact with NOx and VOCs They combine to create ozone to replenish the ozone layer.

Hence, option A is correct.

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Answer:

Because some of the energy escaped or was used.

Explanation:

AnswerAnswerAnswerAnswerAnswer:0.3571m

Explanation:

Answer:

D.) A carpet transfers more energy to heat than the floor.



The frictional force between carpet and ball is more due to the rough surface of the carpet as compared to the friction between floor and ball, so the ball on the carpet would come to rest first before a ball on the floor.

Explanation:

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The far point from the mirror can a person be and still see an upright, magnified image is focus i.e. 35cm from mirror.

The mirror is made out of a sphere which has its radius called the radius of curvature.

The focal length f is half of the radius of curvature R of the lens.

Given the radius of curvature is 70cm, then focal length will be 70/2 = 35cm.

The makeup mirror is a concave mirror which forms real and upright images. The magnified images are obtained when the object is placed between focus and mirror.

Thus, the far point is focus i.e. 35cm from mirror.

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Answer:

Reflection involves a change in direction of waves when they bounce off a barrier. Refraction of waves involves a change in the direction of waves as they pass from one medium to another.

Explanation:

Frictional force is reduced in smooth surfaces but is large in rough surfaces.

Force is defined as an agent which causes a change in the motion or state of an object.

Forces are of two types:

Contact forces include frictional force.

Frictional forces refers to the force which acts between two surfaces in contact by opposing the relative motion of one over the other

Frictional force depends on the nature of surfaces in contact where rough surfaces produce greater frictional force than smooth surfaces.

Therefore, frictional force is reduced in smooth surfaces but is large in rough surfaces.

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Answer:

electromagnetic waves

Answer:

5 meters

Explanation:

4  and  3   are legs of a right triangle with the hypotenuse = displacement

Use Pythagorean theorem

4 ^2 + 3^2 = (displacement)^2

displacement = 5 m

The displacement of the object that moved 4 meters north, then 3 meters east is 5 meters.

Displacement is change in the position of an object. It can be described as the shortest distance between the initial and final position of an object.

The displacement of the object is calculated as follows;

d² = 4² + 3²

d² = 25

d = 5 m

Thus, the displacement of the object that moved 4 meters north, then 3 meters east is 5 meters.

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Answer:

0.2597hertz

Explanation:

first of all we have to calculate time.

time is given by speed/distance

hence=30.67/7.96

=3.85s

frequency is given by 1/time

=1/3.85

=0.2597hertz

Inboard gasoline boats built after July 31st, 1981 must have a mechanical type of ventilation system.

A mechanical ventilation system can be defined as a device in which circulates fresh/cold currents of air.

A mechanical ventilation system is based on the use of systems of ducts capable of ventilating the device.

In conclusion, inboard gasoline boats built after July 31st, 1981 must have a mechanical type of ventilation system.

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Answer:

150w

Explanation:

power = workdone/time

9000/60

150w

Answer:

discovered as obisdan

Explanation:

me ginus brooo

Answer:26.7 and 7.4

Explanation:

So, first you add all the time up then the area travelled.

Firstly; 45 mins + 300 minutes(5 hours) + 165 minutes(2.75 minutes)= 510 minutes

Secondly; 42.2KM + 180.5KM + 3.86KM= 226.56KM

Finally: For hourly you would do: 510 divided by 60 to get 510 mins in terms of hours: so, 8.5 hours then you would do 226.56KM divided by 8.5 hours to get: 26.7Km/hr

For m/s: 7.4m/s

Unguarded live parts operating at 30,000 volts located above a working space must be elevated at least 9.5' above the working space.

The voltage is the potential difference between the end of the conductor.
The place where electricity supply wires and power stations are present, the lives must protected. They must live far above the working space.

Thus, it must be 9.5' above the working space.

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This is a uniform rectilinear motion (MRU) exercise.

We get the data:

d = ?, v = 55 km/h, t = 2 hr

To calculate the distance, multiply the speed by the time.

d = v * t

The distance traveled is 110 km.

There are two systems that contain only the fundamental frequency and they are the systems C and D.

The term fundamental frequency refers to the lowest frequency of a vibrating body. It does not include the overtones or the harmonics.

From the image shown in the question, there are two systems that contain only the fundamental frequency and they are the systems C and D.

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Answer:

Usual speed: [tex]500\; {\rm km \cdot h^{-1}}[/tex].

Explanation:

Let the usual duration of the trip be [tex]t[/tex] hours ([tex]t > 0[/tex].) The actual duration of this trip would be [tex](t - (1/2))[/tex] hours.

The usual speed would be:

[tex]\begin{aligned}\frac{1500}{t}\end{aligned}[/tex].

The actual speed would be:

[tex]\begin{aligned}\frac{1500}{t - (1/2)}\end{aligned}[/tex].

The actual speed is [tex]100\; {\rm km \cdot h^{-1}}[/tex] greater than the usual speed. In other words:

[tex]\begin{aligned}\frac{1500}{t - (1/2)} - \frac{1500}{t} = 100\end{aligned}[/tex].

Simplify this equation and solve for [tex]t[/tex] ([tex]t > 0[/tex]):

[tex]\begin{aligned}\frac{15}{t - (1/2)} - \frac{15}{t} = 1\end{aligned}[/tex].

[tex]\begin{aligned}15\, t - 15\, \left(t - \frac{1}{2}\right) = t^{2} - \frac{1}{2}\, t\end{aligned}[/tex].

[tex]\begin{aligned}t^{2} - \frac{1}{2}\, t - \frac{15}{2} = 0\end{aligned}[/tex].

[tex]2\, t^{2} - t - 15 = 0[/tex].

[tex](2\, t + 5)\, (t - 3) = 0[/tex].

Since [tex]t > 0[/tex], the only possible value for [tex]t[/tex] would be [tex]t = 3[/tex].

Thus, the usual duration of this trip would be [tex]3\; {\rm h}[/tex]. The usual speed of this trip would be:

[tex]\begin{aligned}v &= \frac{s}{t} = \frac{1500\; {\rm km}}{3\; {\rm h}} = 500\; {\rm km \cdot h^{-1}}\end{aligned}[/tex].

A mouse have the same momentum as an elephant when the mouse has a large velocity and the elephant has a small velocity. The correct option is A.

The momentum is the product of the mass of the object and its velocity.

Mouse has mass very less than the elephant. In order to make their momentum same, the velocity of the mouse must be greater than the elephant.

Mm x Vm = Me x Ve

A mouse have the same momentum as an elephant when the mouse has a large velocity and the elephant has a small velocity.

Thus,  the correct option is A.

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Answer:

infrasound

Explanation:

The horizontal force be to keep the sled moving at constant velocity is 392 N.

The horizontal force applied to the sled to prevent it from moving is calculated from Newton's second law of motion.

F - Ff = ma

where;

at constant velocity, a = 0

F - μmg = m(0)

F - μmg  = 0

F = μmg

F = 0.2 x 200 x 9.8

F = 392 N

Thus, the horizontal force be to keep the sled moving at constant velocity is 392 N.

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Answer: yes

Explanation: it's fueled by nuclear reactions that fuse hydrogen atoms together into helium and other heavier elements, releasing huge amounts of energy in the process.

The two factors that may affect the forces exerted by the pieces of tape are  the charge on the tape and distance between them.

The effect of one of the factors could be tested by altering the distance and calculating the forces between them.

A force is the push or pull which make a body to move or stop it.

a. When two objects are kept apart. they are always attracted when uncharged kept at any distance.

When the objects are charged, the force will be attractive or repulsive.

The distance between those charged objects increase or decrease in strength depending on the distance between them.

Thus, the two factors that may affect the forces exerted by the pieces of tape are the magnitude of charge on the tape and distance between them.

b. By increasing the distance between the tapes, we can find the strength of forces.

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