A 7500 kg truck traveling at 5.0 m/s east collides with a 1500 kg car moving
at 20 m/s in a direction 30 o

south of west. After collision, the two vehicles
remain tangled together. Find the speed and state the direction of car.

The distance traveled by the object during the time interval from t = 2 s to t = 3s is 3m. Option C.

From the formula d = ½ at2 the displacement is proportional to the square of time. If you run twice as far from the rest, the displacement will quadruple (or 4m). Since the object has already moved 1 m in the first second, the remaining 3 m have moved in time intervals from 1 to 2 seconds.

This means that the velocity and net force in the direction normal to the plane must be zero. Assuming the plane is frictionless means that the plane exerts no force on the block parallel to the surface. For angular frictionless tilt, the acceleration is the result of the gravitational acceleration multiplied by the sine of the angle.

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The volume flow rate at which the water flows is 0.0188 m³/s

Since the horizontal pipe 10.0 cm in diameter has a smooth reduction to a pipe 5.00 cm in diameter and If the pressure of the water in the larger pipe is 8.00 × 10⁴ Pa and the pressure in the smaller pipe is 6.00 × 10⁴ Pa, since we require the rate at which water flows through the pipe, we use Bernoulli's equation at constant height to find the speed of flow.

Bernoulli's equation governs flows through pipes. It is given by

P₁ + ρgh₁ + 1/2ρv₁² = P₂ + ρgh₂ + 1/2ρv₂² where

With h₁ = h₂, the Bernoulli's equation becomes

P₁ + 1/2ρv₁² = P₂ + 1/2ρv₂²

Using the volume flow rate equation

Q = A₁v₁ = A₂v₂ where

So, A₁v₁ = A₂v₂

(πd₁²/4)v₁ = (πd₂²/4)v₂

d₁²v₁ = d₂²v₂

Making v₂ subject of the formula, we have

v₂ = d₁²v₁/d₂²

v₂ = (d₁/d₂)²v₁

Substituting v₂ into the Bernoulli equation, we have

P₁ + 1/2ρv₁² = P₂ + 1/2ρv₂²

P₁ + 1/2ρv₁² = P₂ + 1/2ρ(d₁/d₂)⁴v₁²

Making v₁ subject of the formula, we have

1/2ρv₁² - 1/2ρ(d₁/d₂)⁴v₁² = P₂ - P₁

1/2ρ[1 - (d₁/d₂)⁴]v₁² = P₂ - P₁

v₁² = √{2(P₂ - P₁)/ρ[1 - (d₁/d₂)⁴]}

Given that

Substituting the values of the variables into the equation, we have

v₁ = √{2(P₂ - P₁)/ρ[1 - (d₁/d₂)⁴]}

v₁ = √{2(6.00 × 10⁴ Pa - 8.00 × 10⁴ Pa)/1000 kg/m³[1 - (10.0 cm/5.0 cm)⁴]}

v₁ = √{2(- 2.00 × 10⁴ Pa)/1000 kg/m³[1 - 2⁴]}

v₁ = √{(- 4.00 × 10⁴ Pa)/1000 kg/m³[1 - 8]}

v₁ = √{(- 4.00 × 10⁴ Pa)/1000 kg/m³[-7]}

v₁ = √{(40.0 × 10³ Pa)/7 × 10³ kg/m³}

v₁ = √{(40.0/7 Pam³/kg}

v₁ = √{5.7143 Pam³/kg}

v₁ = 2.39 m/s

Since the volume flow rate Q = A₁v₁ = (πd₁²/4)v₁

Given that

Substituting the values of the variables into the equation, we have

Q = (πd₁²/4)v₁

Q = π(0.1 m)²/4 × 2.39 m/s

Q = π × 0.01 m²/4 × 2.39 m/s

Q = π × 0.0239 m³/s/4

Q = 0.0751 m³/s ÷ 4

Q = 0.0188 m³/s

So, the water flows at a rate of 0.0188 m³/s

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The mass of the lighter star is 2.15 x 10³⁰ kg.

The mass of the lighter star is obtained by applying the formula force gravitational potential energy between the two stars.

According to Newton's law of universal gravitation, every two objects in the universe attracts each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects.

U = F x r

U = Gm₁m₂/r²   x   r

U = Gm₁m₂/r

where;

r = 3 x 10⁸ m/s  x 3.154 x 10⁷ s/yr

r = 9.46 x 10¹⁵ m

let the lighter star = m₁

then, the heavier star = 2m₁

U = (G x m₁ x 2m₁)/r

U = 2Gm₁²/r

Ur = 2Gm₁²

m₁² = Ur/2G

m₁ = √(Ur/2G)

m₁ = √(6.5 x 10³⁴ x 9.46 x 10¹⁵ / 2 x 6.626 x 10⁻¹¹)

m₁ =  2.15 x 10³⁰ kg

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Answer:

P = P1 + P2     where P is a vector quantity

P1 = P2 = .10 kg * 30.0 m/s = 3 kg-m/sec

P = (p1^2 + p2^2)^1/2      add vectors and find magnitude of resultant

p = (3.0^2 + 3.0^2)^1/2 = 4.24      length of resultant vector

Since vectors are at right angles p is the length of the hypotenuse

tan θ = 3.0 / 3.0 = 1      where theta is measured with respect to the positive x-axis and will be below that axis

θ = tan^-1 1 = 45 deg      (45 degrees below positive x-axis)

One can find the value of tan θ from the length of the x-y vectors and from tan θ obtain θ

The amount of current flowing in the wire, given that it has a resistance of 25 Ω and connected to a 3.0 V is 0.12 A

From Ohm's law, we understood that the voltage, current and resistance are related according to the following formula:

Voltage (V) = Current (I) × resistance (R)

V = IR

With the above formula, we can obtain the current in wire. Details below.

The following data were obtained from the question:

Thus, the current can be obtained as follow:

Voltage (V) = Current (I) × resistance (R)

3 = Current × 25

Divide both sides by 25

Current = 3 / 25

Current = 0.12 A

Thus, we can conclude that the current in the wire is 0.12 A

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The age of the rock is 9 * 10^9 years and there is no rock that can be as old as this age on earth.

The term radioactive decay has to do with the process by which a radioactive nuclide disintegrates so as to obtain the daughter nucleus. The radioactive decay may occur by alpha or by beta radiation.

In this case, we are told that only 25% of the original uranium 238 isotope of half life 4.5 * 10^9 years still remains.

Using the formula;

0.693/[tex]t_{\frac{1}{2}[/tex] = 2.303/t log (No/N)

[tex]t_{\frac{1}{2}[/tex] = half life of the isotope

N= amount of isotope at time t

No = amount of isotope initially present

Then;

0.693/4.5 * 10^9 = 2.303/t log(No/0.25No)

1.54 * 10^-10 =  1.39/t

t = 1.39/1.54 * 10^-10

t = 9 * 10^9 years

There is no rock that is this much old since the average age of the earth is 4.5 billion years.

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Answer:

b. Is the answer because It depends on what Galileo Galilea thinks :,-)

Galileo most likely have to do this in order to convince other scientists that his observations were correct, is repeats the experiment multiple times in front of witnesses. So the correct option is C.

Galileo probably had to repeat his tests multiple times in front of witnesses to present convincing proof to other scientists rather than keeping the results to himself, making an object fall faster, or just publicizing his results.

Scientific certainty is produced by repeating an experiment and observing the outcomes, which is still regarded as a key component of the scientific method in modern times.

Galileo would have benefited from documenting his procedures and conclusions in a way that allowed other scientists to replicate his experiment and confirm his findings, even though it wasn't mentioned in your multiple-choice answers.

This concept is largely covered by the peer-review procedure used in contemporary science. So the correct option is C.

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Answer:If the momentum is large on the other hand, you will cause turbulences that make the water splash , So Yes

Explanation:

Answer:

M1 V1 + M2 V2 = (M1 + M2) V

Conservation of momentum where V is speed of combined vehicles

V = (1500 * 10 + 1500 * 5) / (1500 + 1500) = 7.5 m/s

We first apply the data to the problem.

Data:

Then, we apply the formula that is.

Formula:

Finally we develop the problem.

Developing:

The car travels a distance of 50 meters.

The kinetic energy of the blocks that is lost due to the collision is 8.64 J.

The initial speed of the smaller block is calculated by applying the following equation.

m₁u₁  +  m₂u₂ = v(m₁ + m₂)

where;

9u₁ + 27(0.5) = 0.9(9 + 27)

9u₁ + 13.5 = 32.4

9u₁ = 18.9

u₁ = 18.9/9

u₁ = 2.1 m/s

The total initial kinetic energy of the blocks is calculated as follows;

K.Ei = ¹/₂m₁u₁²  +  ¹/₂m₂u₂²

K.Ei = ¹/₂(9)(2.1)²    +    ¹/₂(27)(0.5)²

K.Ei = 23.22 J

The final kinetic energy of the blocks is calculated as follows;

K.Ef = ¹/₂(m₁ + m₂)v²

K.Ef = ¹/₂(9 + 27) x (0.9)²

K.Ef = 14.58 J

The change in the kinetic energy of the block is calculated as;

ΔK.E = K.Ef  -  K.Ei

ΔK.E = 14.58 J   -   23.22 J

ΔK.E = -8.64 J

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Answer:

it has water

Explanation:

water is good

The physical weathering of rock by the repeated thawing of water is likely to happen more quickly than the physical weathering of rock by plant roots.

How does repeated thawing of water causes the rock to weather?

First, the physical weathering of rock is usually caused by water (which can come from rain or erosion), ice, plant roots, and even animals. These agents ensure the rock is broken down into smaller particles or sizes.

So, the thawing of water ensures continuous freezing and melting of water which allows water in the form of ice to build up in the cracks of rocks after expanding on freezing. This makes the rock breaks apart due to the pressure that is built in the crack.

However, physical weathering from the root of plants is slower compared to that of the repeated thawing of water. Though both processes needed time, but it will take more time for the roots in a crack. For the plant roots to get cracks to grow into, it needs repeated thawing of water to create that crack or other agents of rock weathering to create the crack.

In summary, repeated thawing of water makes rocks more susceptible to chemical and biological weathering. The weathering of rock by plant roots will need the plant to grow and survive in the first place, which might take more time.

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Answer:

Explanation:

Given:

m₁ = 8 kg

V₁ = 3 m/s

m₂ = 2kg

V₂ = 0

V₂' = 6 m/s

__________

V₁' - ?

m₁V₁ + m₂V₂ = m₁V₁' + m₂V₂'

8·3 + 2·0 = 8·V₁'  + 2·6

24 = 8·V₁' + 12

24 - 12 = 8·V₁'

12 = 8·V₁'

V₁' = 12 / 8 = 1.5 m/s

The fact that Pedro likes to talk to people means that Pedro is an extrovert.

We define an extrovert as a person that is able to interact very much freely with other people. It is one of the kinds of temperaments that a person can have. We know that people that can have very different temperament types and these are the things that determine the behavior of a person.

Thus, a person acts in accordance with the kind of temperaments that a person has. Just as we can see in the person of Pedro. We are told in his case that the person Pedro would always like to talk with is friends and the team mates which suggests that he must like to associate with other people.

From our definition of the person of Pedro in the lines that are above, we can say the Pedro is an extrovert.

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The  argument by analogy (regarding other minds) and the objection against it is that:

Russell thinks there are other minds due to the fact that he can see behaviors in others that are similar to his own without giving them any conscious thought. He contends that thoughts are the root of all behavior, but what happens when a thought results in a reaction that one forces upon oneself.

Therefore, the traditional philosophical defense of the existence of other minds is the argument from analogy, which, as persuasively put forth by John Stuart Mill, a 19th-century empiricist, claims that since one's body and outward behavior are observably similar to those of others, one is justified by analogy in thinking that others have feelings similar to one's own and are not just automatons with similar bodies and outward behaviors.

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The behavior of a clock sitting on the surface of a planet with a surface gravity of one g says that rim clock always runs slower

Gravity is the principal force acting on a body in projectile motion when it is hurled at a specific angle. The power with which the body is originally thrown determines maximum height and how far it will travel. When an object moves in a projectile motion, there are numerous instances in daily life. For instance, a basketball player will obliquely toss the ball at a specific angle towards the direction of the hoop. The ball then rises and descends like a missile in the direction of the basket. Other sports like cricket, football, and soccer may demonstrate this.

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P = I²R = (10 A)² (120 Ω) = 120 000 Watts = 120 kW

The work that is done by a forklift raising a 583 kg box a distance of 1.2m is 699.6N.

Work refers to the measure of energy expended in moving an object. It is most commonly calculated by multiplying the force by the distance.

This means that no work is done if the object does not move. This means that the work done can be calculated as follows:

Work done = Force × distance

According to this question, a forklift raises a 583kg box over a distance of 1.2m. The work done can be calculated as follows:

Work done = 583kg × 1.2m

Work done = 699.6N

Therefore, 699.6N is the force required to lift the box.

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Answer:

123.6 N

Explanation:

I will try to answer this

Find height of the ball

  9. 5   - 9.5 cos 35  = 1.72 m   of potential energy

   PE = mgh    will be converted to KE at the bottom of the swing

     mgh = 1/2  m v^2

    sqrt( 2 gh )  = v      at the bottom = 5.81  m/s

then the tension is the mass times the acceleration of gravity PLUS the acceleration of circular motion  v^2 / r

F = ma

F = 9.25 kg ( 9.81 m/s^2 + 5.81^2 / 9.5 m/s^2 ) = tension = 123.6 N

(Let me know if this is incorrect, and I will research a bit more)

The length of the axes of the eclipse are 1.25cm and 10cm.

With the formula, F = (mv²)/r

where F = force

m = mass

v = velocity and

r = diameter/length

To find how long the axes of the eclipse are at an outward force of 10N and a similar force of 1.25N:

For F₁ = 10N

10 = (0.5 x 5²)/r

10 = 12.5/r, cross multiply

r = 12.5/10 = 1.25cm

For F₂ = 1.25N

1.25 = (0.5 x 5²)/r

1.25 = 12.5/r, cross multiply

r = 12.5/1.25 = 10cm

The lengths of the eclipse are found by dividing the product of the mass and velocity by the first and second forces to get 1.25cm and 10cm

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The angular acceleration of the wheel 0.16 rad/sec².

The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration.

The net acceleration that objects get as a result of the combined action of gravity and centrifugal force is known as the Earth's gravity, or g. It is a vector quantity whose strength or magnitude is determined by the norm and whose direction correlates with a plumb bob.

Given in the question,

angular acceleration = 4/25

angular acceleration = 0.16 rad/sec²

The angular acceleration of the wheel 0.16 rad/sec².

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With the use of formulas, the maximum height reached by the cannonball is 635.4 meters

The motion of an object under gravity is the vertical motion of the object under the influence of acceleration due to gravity.

If I shoot a cannonball (m = 4.5 kg) straight up from a cannon that exerts a force of 7000 N through a barrel length of 4.00 meters, the acceleration of the cannonball in the barrel can be calculated by using the formula

F = ma

Where

Substitute all the parameters into the formula

7000 = 4.5a

a = 7000/4.5

a = 1555.6 m/s²

By using the below formula, we will find the velocity at which the ball leaves the barrel.

Where

v² = u² + 2as

v² = 0 + 2 × 1555.6 × 4

v² = 12444.4

v = √12444.4

v = 111.6 m/s

The same speed will be the initial velocity at the entrance of the barrel. How high will the cannonball go can be known by using the formula

v² = u² - 2gH

Where

Substitute all the parameters into the formula

0 = 111.6² - 2 × 9.8 × H

19.6H = 12454.56

H = 12454.56/19.6

H = 635.4 m

Therefore, the cannonball will go as high as 635.4 m

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The coefficient of static friction needed for a drag racer to cover the given distance is 1.27.

The magnitude of the acceleration of the slipping tires is calculated by applying the kinematic equation for average distance travelled by the slipping tires as shown below.

s = ut + ¹/₂at²

where;

s = 0 + ¹/₂at²

s = ¹/₂at²

at² = 2s

a = 2s/t²

a = (2 x 1,400 m) / (15 s)²

a = 12.44 m/s²

The coefficient of static friction needed for a drag racer to cover the given distance is calculated as;

μ = a/g

where;

μ = 12.44 / 9.81

μ =  1.27

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mvr³  is the angular momentum of the hoop about its center of mass

L=mvr=Iω

L= angular momentum

m= mass

v=velocity

r= radius

L=mvr

L=mr²×r×v

L=mvr³

A characteristic known as angular momentum describes the rotating inertia of an item or set of objects when they are moving along an axis that may or may not pass through them. The Earth possesses spin angular momentum from its daily rotation around its axis and orbital angular momentum from its yearly revolution around the Sun. Since angular momentum is a vector quantity, its full representation calls for the identification of both a magnitude and a direction. An object in orbit has an angular momentum that is proportional to its linear momentum times the perpendicular distance r from the center of rotation to a line drawn through the center of gravity of the object, or just mvr.

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Answer:

mean extremely affect by the two level of values

highest and lowest points

The horizontal component of the force pushing him forward is 37 N.

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Frictional force due to ice, F =   μN

Normal force N = mg = 75*9.8 = 735 N

Frictional force = 0.05*735 = 36.75 N

Coefficient of friction, μ for ice is 0.05

The horizontal component of the force pushing him forward is 37 N.

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Three reasons why it may be difficult to accelerate the bookcase are;

1) Inertia that acts on the book case

2) Frictional force that prevents the book case from moving

3) The weight of the book  case.

We know that the Newton law states that a body would tend to remain at a state of rest or uniform motion unless it is acted upon by unbalanced forces. When the book lies without movement it is acted upon by its weight and the reaction which are exactly balanced. The inertia is the reluctance of the book to move.

As the book begins to move when pull, the friction that is acting on the pull would also militate against the forward motion of the book as it acts in the opposite direction.

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