100 POINTS QUESTION. PLEASE PROVIDE EXPLANATION

The final speed of the second puck is 5 m/s.

Speed can be defined as the rate of change of distance with respect to time.

To calculate the final speed of the second puck, we use the formula below.

Formula:

Where:

Make v' the subject of the equation

Assuming,

From the question,

Given:

Substitute these values into equation 2

Hence, The final speed of the second puck is 5 m/s.

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Answer:

30 j

Explanation:

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The kinetic energy of an object is 50 J, and the potential energy is 20 J. The total energy of the object is 30J.

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Answer:

be in front of the club

Explanation:

The ball should be highest off the ground for a driver. The general recommendation is that the bottom of the golf ball on a tee should be level with the top of the driver; for long and mid-irons, push the tee into the ground so that only about a quarter-inch is above ground.

Answer:

Elastic potential energy

Answer:

Elastic potential energy:  increases when you compress a spring.

Explanation:

Answer:

100m

Explanation:

Equation of motion for he truck: s=ut

Equation of motion for the car: s=1/2at^2

the second solution gives , s=2u^2/a = 2*10^2/2 = 100m

Answer:The current focus of climate science involves carbon dioxide (CO2)emissions. Carbon dioxide in the atmosphere acts as a blanket over the planet by trapping long wave radiation, which would otherwise radiate heat away from the planet. As the amount of carbon dioxide increases, so will its warming effect.

Explanation:

Answer:

Explanation:

The work done by an object can be found by using the formula

From the question

force = 600 N

distance = 4 m

We have

work done = 600 × 4

We have the final answer as

Hope this helps you

Answer:

Explanation:

How many atoms of each element are present in the formula for bromoethane

Answer:

a feat

Explanation:

cause the mf already 7ft on the dot there is no such thing as 6 ft 12

The knot is in equilbrium, so there is no net force acting on it. Starting with the unknown force and going clockwise, denote each force by F₁, F₂, F₃, and F₄, respectively. We have

F₁ + F₂ + F₃ + F₄ = 0

Decomposing each force into horizontal and vertical components, we have

F cos(180º - α) + (5.7 N) cos(50º) + (6.2 N) cos(-44º) + (6.7 N) cos(-137º) = 0

F sin(180º - α) + (5.7 N) sin(50º) + (6.2 N) sin(-44º) + (6.7 N) sin(-137º) = 0

Recall that cos(180º - x) = - cos(x) and sin(180º - x) = sin(x), so these equations reduce to

F cos(α) ≈ - 3.22 N

F sin(α) ≈ 4.51 N

(1) Recall that for all x, sin²(x) + cos²(x) = 1. Use this identity to solve for F :

(F cos(α))² + (F sin(α))² = F ² ≈ 30.73 N²   →   F ≈ 5.5 N

(2) Use the definition of tangent to solve for α :

tan(α) = sin(α) / cos(α) ≈ 1.399   →   α ≈ 126º

or about 54º from the horizontal from above on the left of the knot.

(a) The magnitude of the force F acting on the knot is 5.54 N.

(b) The angle α of the force F is 54.4⁰.

The given parameters:

The net vertical force on the knot is calculated as follows;

[tex]F_y = Fsin(\alpha) + 5.7 sin(50) - 6.2 sin(44) - 6.7 sin(43)\\\\F_y = F sin(\alpha) -4.51\\\\Fsin(\alpha) = 4.51[/tex]

The net horizontal force on the knot is calculated as follows;

[tex]F_x = -F cos(\alpha) + 5.7 cos(50) + 6.2cos(44) - 6.7cos(43)\\\\F_x = -Fcos(\alpha) + 3.22\\\\Fcos(\alpha) = 3.22[/tex]

From the trig identity;

[tex]sin^2 \theta + cos^ 2 \theta = 1\\\\[/tex]

[tex](Fsin(\alpha))^2 + (Fcos(\alpha))^2 = (4.51)^2 + (3.22)^2\\\\F^2(sin^ 2\alpha + cos^2 \alpha) = 30.71\\\\F^2(1) = 30.71\\\\F = \sqrt{30.71} \\\\F = 5.54 \ N[/tex]

The angle α of the force F is calculated as follows;

[tex]Fsin(\alpha) = 4.51\\\\sin(\alpha) = \frac{4.51}{F} \\\\sin(\alpha ) = \frac{4.51}{5.54} \\\\sin(\alpha ) = 0.814\\\\\alpha = sin^{-1}(0.814)\\\\\alpha = 54.5 \ ^0[/tex]

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Answer:

The molten rock rises from cracks in the oceanic crust, so option B is correct.

The thinnest and most significant layer on our planet is its crust, which is made up of rocks and forms the planet's outermost layer. Despite making up less than 0.5 percent of the globe's total volume, it is essential to the majority of the natural cycles that take place all over the planet. The crust of the Earth is divided into layers, each of which contains a unique element.

Metamorphic rocks are defined as rocks that change into another rock. Sedimentary rocks are created by the deposition of material at the Earth's surface within water bodies, followed by cementation.

The crust that lies beneath land masses is known as the continental crust, whereas the crust that lies beneath the ocean's surface is known as the oceanic crust. The oceanic crust is younger and much thinner than the continental crust.

Therefore, the molten rock rises from cracks in the oceanic crust.

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Answer:

It depends on the location of the ball during the motion. The string tension are approximately 3.82 N (at the lowest point), 3.06 N (at the highest point), and 3.44 N (at the horizontal point).

Explanation:

Tension in the String can be determined by the Newton's 1st Law of Motion (The ball shouldn't be escaped from the trajectory). The value of [tex]\theta[/tex] indicates the angle that is measured from the vertical lines and the rope of length R)

[tex]\sum F=0\rightarrow \frac{mv^{2}}{R}+mg\cos\theta-T=0[/tex]

[tex]T=mg\cos\theta+\frac{mv^{2}}{R}[/tex]

[tex]T=(38\times10^{-3})(10)(\cos\theta)+(38\times10^{-3})\frac{2^{2}}{0.21^{2}}=0.38\cos\theta+3.44[/tex]

When the ball is at the lowest point, the value of angle [tex]\theta=0[/tex], so the string tension is approximately 3.82 N. If the ball is at the highest point the value of [tex]\theta=180^{0}[/tex], so the string tension is approximately 3.06 N, and at the horizontal point [tex]\theta=90^{0}[/tex], so the string tension is approximately 3.44 N.

Answer:

There is a difference between the words 'drops' and 'throws'

If an object is dropped, it's initial velocity is 0 but if the same object is thrown, it has some initial velocity

Since 'the sister' DROPS the keys, the initial velocity will be 0 m/s

We are given:

u=0 m/s

s = 4.3 m

a = 9.8 m/s/s (gravity)

From the third equation of motion:

v² - u² = 2as

Replacing the known values

v² - (0)² = 2(9.8)(4.3)

v² = 84.28

v = 9.2 m/s (approx)

Hence, the final velocity of the keys is 9.2 m/s

Answer:

They act on different objects

Explanation:

Let's say that I push a cart, the cart moves because the force is also going into the ground. Hope this makes sense.

Answer:

its speed is insignificant before the diver's speed change, so the result does not change

Explanation:

In this exercise of conservation of the momentum, the system is formed by the diver and the Earth

initial instant (before jumping)

       p₀ = 0

final instant (after jumping)

      [tex]p_{f}[/tex] = m v + M v²

how momentum is conserved

        p₀ = p_{f}

        0 = m v + M v²

         v² = m / M v

since the mass of the Earth is M = 10²⁴ kg

   its speed is insignificant before the diver's speed change, so the result does not change

The range of frequencies  of visible light in a vacuum is mathematically given as

Fmin=4.19*10^14Hz  to Fmax=1*10^15Hz

Question Parameters:

The wavelengths of visible light vary from about 300 nm to 700 nm.

Generally, the equation for the frequency   is mathematically given as

F=C/\lambda

Therefore

For Fmax

[tex]Fmax=\frac{300*10^8}{3*10^9}[/tex]

Fmax=1*10^15Hz

Where

[tex]Fmin=\frac{3*10^8}{700*10^9}[/tex]

Fmin=4.19*10^14Hz

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By third equation of motion -

[tex]\green{ \underline { \boxed{ \sf{v^2-u^2=2aS}}}}[/tex]

where

Putting Values to find final velocity of mass before hitting the ground-

[tex]\begin{gathered}\\\implies\quad \sf v^2-(0)^2=2\times g \times 4 \quad (g = acceleration \: due \:to \: gravity) \\\end{gathered} [/tex]

[tex]\begin{gathered}\\\implies\quad \sf v^2=2\times 9.8 \times 4\quad (g= 9.8 \:m/s) \\\end{gathered} [/tex]

[tex]\begin{gathered}\\\implies\quad \sf v=\sqrt{78.6} \\\end{gathered} [/tex]

[tex]\begin{gathered}\\\implies\quad \sf v= 8.86 \:m/s \\\end{gathered} [/tex]

Now finding the momentum of the mass at that moment -

[tex]\green{ \underline { \boxed{ \sf{Momentum= mass \times velocity}}}}[/tex]

[tex]\begin{gathered}\\\implies\quad \sf Momentum= 21.2 \times 8.86 \\\end{gathered} [/tex]

[tex]\begin{gathered}\\\implies\quad \sf Momentum= 187.8 \:kgms^{-1} \\\end{gathered} [/tex]

[tex]\longrightarrow[/tex]The momentum of the mass just before it hits the ground is 187.8 kgm/s

[tex]\\[/tex]

[tex]\therefore \sf Option \: B) \: is \:correct [/tex]✔️

Answer:

First option, third option, fourth option, and the fifth option.

Explanation:

Kinetic energy is energy an object has when it's motion, the greater the speed the greater the kinetic energy. For example, a car moving and increasing in speed is kinetic energy since the object is in motion. If the car stops and parks in a parking lot that is potential energy. Potential energy is the amount of energy an object has when it's at rest or not in motion.

So, the answer for this question is as followed first option or "energy can be stored in the position of an object." Third option or "Energy can be stored in the position of the particles that make up a substance." Fourth option or "Energy exists as movement of the particles of a substance." The last answer will be the fifth option or "Energy is greater in faster-moving particles than in slower-moving particles."

Hope this helps.

The light ray ; ( C ) Bends away from the axis perpendicular to the surface between the materials

Snell's law states that the ratio of the sines of the angles of incidence is equal to the ratio of the refractive indexes of the materials surface through which the light rays pass through. therefore

As the light ray travels from a material with a higher index of refraction into a material with a lower index of refraction at the axis that is perpendicular to the surface which is in between the materials the light ray will bend away due to Snell's law .

Hence we can conclude that The light ray Bends away from the axis perpendicular to the surface between the materials.

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Answer:

The answer that we got will be 22.22....

Answer:

52.6 kg

Explanation:

KE = 1/2mv^2

400 = 1/2 x m x 3.9^2

m = 2(400 J)/ ( 3.9 m/s)^2

m = 52.6 kg

Answer:

Weight Density 1 = 38100 N/m³

Weight Density 2 = 38100 N/m³

b. Yes

Explanation:

The formula for volume of cylinder is:

V = πr²l

where,

V = Volume

r = radius

l = length of cylinder

So, if length has the 3 significant figures which is least in all values, Then the volume must also be in 3 significant figures. The formula for weight density is:

Weight Density = Weight/Volume

Here, the volume has the least significant figures of 3, therefore, the weight densities must also have 3 significant figures:

Weight Density 1 = 38119 N/m³

Weight Density 1 = 38120 N/m³

Weight Density 1 = 38100 N/m³

Weight Density 2 = 38119 N/m³

Weight Density 2 = 38120 N/m³

Weight Density 2 = 38100 N/m³

Hence, the answer is:

b. Yes

Answer:

[tex]OA. \: O \: seconds \: to \: 3 \: seconds[/tex]

The acceleration due to gravity is 9.8 m/s^2.

Answer: A) It's a normal force exerted by the floor. It balances out the gravitational force.

The normal force pushes up which means the box doesn't go through the floor. The up and down forces are balance. If the normal force was larger than the gravitational force, then the box would be pushed up and float/jump in the air. But the box is stationary and not moving, so that's why we don't have any acceleration and the forces effectively cancel each other out. In a way you can think of it like a see-saw.

Friction would only come into play if you applied a force and were trying to move the box. Friction counteracts the applied force. If you push the box to the left, then the friction would push to the right. Friction slows down the box and it allows it to not slide forever.

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Is normal force equal to gravity?

The normal force on an object at rest on a flat surface is equal to the gravitational force on that object.

▀▄▀▄Is friction a gravitational force?

Friction is the resistance to motion of one object moving relative to another. It is not a fundamental force, like gravity or electromagnetism. Instead, scientists believe it is the result of the electromagnetic attraction between charged particles in two touching surfaces.

▀▄▀▄Can normal force be greater than gravity?

Can you imagine the situation when Normal force is greater than mg ? Yes. When an additional downward force F is applied to a mass m resting on a horizontal surface, the normal force is FN=F+mg. ... An example is the normal force on an incline plane with an angle of θ due to a mass m.▄▀▄▀▄▀▄

What is the direction of the force you applied?

An applied force is an interaction of one object on another that causes the second object to accelerate or change velocity or direction. The force can be a push, pull, or drag. The resulting direction of an object depends on the relative direction of the force on the object. A force equation is F = ma.

Explanation:

PE=mgh

5=m(9.8)(2)

m=5/19.6

m=0.2251 kg

m=225.1 grams

Answer:

incident 50°

reflection 90°

refraction 20°

Answer:

a

 [tex]v_r =8.65 \ ft/s [/tex]

b

  [tex] W_{1-2} =  3.24 \  ft \cdot lb[/tex]

Explanation:

From the question we are told that

  The mass of the ball is [tex]m  =  4 \  lb[/tex]

  The radius is  [tex]r= 3 \  ft[/tex]

   The speed is [tex]v_B_1  =  4.8 \ ft /s[/tex]

    The speed of the attached cord is  [tex]v_c =2.2 \  ft[/tex]

   The position that is been considered is  [tex]r_1 =  2 \  ft[/tex]h

Generally according to the law of angular momentum conservation

   [tex]L_a =  L_b[/tex]

Here [tex]L_a[/tex] is the initial  momentum of the ball which is mathematically represented as

      [tex]L_a  =  m*  v_B_1 *  r[/tex]

while  

[tex]L_b[/tex] is the  momentum of the ball  at  r =  2 ft which is mathematically represented as

       [tex]L_a  =  m*  v_B_2 *  r_1[/tex]

So

      [tex]m*  v_B_1 *  r = m*  v_B_2 *  r_1[/tex]

=>      [tex] 4.8 *  3 =  v_B_2 *  2[/tex]

=>     [tex]   v_B_2 =  7.2 \  ft/s [/tex]

Generally the resultant velocity of the ball is  

      [tex]v_r = \sqrt{v_B_2^2 + v_B_1^2   }[/tex]

=>   [tex]v_r = \sqrt{7.2^2 + 4.8^2   }[/tex]

=>   [tex]v_r =8.65 \ ft/s [/tex]

Generally according to equation for principle of work and energy we have that

    [tex]K_1 + \sum W_{1-2} = K_2[/tex]

Here [tex]K_1[/tex] is the initial kinetic energy of the ball which is mathematically represented as

[tex]K_1  =  \frac{1}{2}  *  m* v_B_1^2[/tex]

While  [tex]\sum W_{1-2}[/tex] is the sum of the total  workdone by the ball

and  [tex]K_2[/tex] is the final kinetic energy of the ball  which is mathematically represented as  [tex]K_2  =  \frac{1}{2}  *  m* v_r^2[/tex]

So

     [tex]\sum W_{1-2} =  \frac{1}{2}  *  m  (v_r^2  -  v_B_1^2)[/tex]

Here  m is the mass which is mathematically represented as

     [tex]m = \frac{W}{g}[/tex] here W is the weight in  lb and  g is the acceleration due to gravity which is [tex]g =  32 \ ft/s^2[/tex]

So

    [tex]\sum W_{1-2} =  \frac{1}{2}  *  \frac{4}{32} *   (8.65^2  -  4.8^2)[/tex]

=>   [tex] W_{1-2} =  3.24 \  ft \cdot lb[/tex]