A light spring obeys Hooke's law. The spring's unstretched length is 34.0 cm. One end of the spring is attached to the top of a doorframe and a weight with mass 7.00 kg is hung from the other end. The final length of the spring is 44.5 cm. (a) Find its spring constant (in N/m).

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

Explanation:

Glomar Challenger studies about the "age of rocks in various places in the ocean" in 1968. EXPLANATION: Glomar Challenger was a "deep sea research vessel" for marine geology and oceanography studies.

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Answer:

c no cappp :)

Explanation:

Answer: 6 J

Explanation:

Total force applied = 47 N Assuming that direction of movement of pencil and applied force is same. Work done by force in moving the pencil W = Force × Distance through which force moves ⇒ W = 47 × 0.25 = 11.75 J .-.-.-.-.-.-.-.-. In case we are asked useful work done then we calculate net force used for pushing the pencil: Net force used for pushing the pencil = 47 − 23 = 24 N Assuming that direction of movement of pencil and net force is same. Useful Work done by force in moving the pencil W u = Force × Distance ⇒ W u = 24 × 0.25 = 6 J

The work done the pushing force is required.

The work done by the pushing force is [tex]6\ \text{J}[/tex]

[tex]F_1[/tex] = Pushing force = 47 N

[tex]F_2[/tex] = Opposing force = 23 N

m = Mass of object = 0.025 kg

s = Displacement = 0.25 m

Work done is given by

[tex]W=F_ns[/tex]

[tex]\Rightarrow W=(47-23)\times 0.25[/tex]

[tex]\Rightarrow W=6\ \text{J}[/tex]

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Answer:

Motion with constant velocity of magnitude 1 m/s (uniform motion) for 4 seconds in a positive direction and then for 2 seconds uniform motion with constant velocity of magnitude 3 m/s in reverse direction .

Explanation:

The graph shows a constant velocity of 1 m/s for 4 seconds in the positive direction. After that, between 4 seconds and 6 seconds, the object reverses its motion with constant velocity of magnitude 3m/s.

Answer:

  orbiting the Sun in an elliptical orbit

Explanation:

When orbits are circular, the center and the focus are the same point. For an elliptical orbit, the orbited object is at one focus of the ellipse. Kepler found planetary orbits to be elliptical with the sun at one focus.

__

As with a lot of scientific theories, it only explained some of the motion. As with a lot of scientific theories, explaining the discrepancies resulted in new discoveries.

Answer:

at one focus of the elliptical orbit of the planets.

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:

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.

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:

8.8 kN

Explanation:

V = 2 m³, W = 40 kN, SG = 1.59

Bouyant force N = 1.59 * 1000 kg/m³ * 9.81 N/kg * 2 m³ = 31.2 kN

So the weight becomes 40 - 31.2 = 8.8 kN

three charged particals are located at the corners of an equil triangle shown in the figure showing let (q 2.20 Uc) and L 0.650

Answer:

Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.

Explanation:

PE=mgh

5=m(9.8)(2)

m=5/19.6

m=0.2251 kg

m=225.1 grams

Answer:

C, slightly less than 243 N

Explanation:

Road friction and air resistance aren't that much on a force. Try pushing something and see how much friction there is. Not that much.

a car moving at a speed shows that the force applied to the car is greater than the frictional force and air resistance

c. Slightly less than 243 N.

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:

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

The periodic time of the campus shuttle is = 1 sec

The periodic time is also known as  revolution per second is the time it takes object in motion ( revolving ) to pass through a point twice ( usually the starting position ) of the object and it calculated as :

For a simple pendulum = 2π√L/g

But a campus shuttle bus in motion

since it takes

1 revolution = 1 sec therefore the time period is = 1 sec

Hence we can conclude that the periodic time of the campus shuttle = 1 second.

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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:

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.

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: A : 250 is the answer

B; The frequency of a wave is the number of complete oscillations (cycles) made by the wave in one second.

Instead, the wavelength is the distance between two consecutive crests (highest position) or 2 troughs (lowest position) of the wave.

In this problem, we are told that the leaf does two full up and down bobs: this means that it completes 2 full cycles in one second. Therefore, its frequency is

where  is called Hertz (Hz). So, the correct answer is

Explanation:

#Wavespeed

#1

[tex]\\ \rm\Rrightarrow v=\nu\lambda=500(3)=1500m/s[/tex]

#2

[tex]\\ \rm\Rrightarrow v=2(0.4)=0.8m/s[/tex]

#Wavelength

#1

[tex]\\ \rm\Rrightarrow \lambda=\dfrac{v}{\nu}=\dfrac{3}{2}=1.5m[/tex]

#2

[tex]\\ \rm\Rrightarrow \lambda= \dfrac{1450}{40000}=0.03625m[/tex]

#Frequency

[tex]\\ \rm\Rrightarrow \nu=\dfrac{v}{\lambda}=\dfrac{8}{20}=0.4Hz[/tex]

#2

[tex]\\ \rm\Rrightarrow \nu=\dfrac{3\times 10^8}{15\times 10^{-2}}=0.2\timee 10^{10}=2\times 10^9Hz[/tex]

Answer:

i think its the top one

Explanation:

pls tell me if im wrong  

Answer:

V2 = 15.53 [m/s]]

Explanation:

In order to solve this problem we must use the principle of energy conservation, where potential energy is transformed into kinetic energy. At the bottom is taken as a reference level of potential energy, where the value of this energy is equal to zero.

Above the inclined plane we have two energies, kinetics and potential. While when the sled is at the reference level all this energy will have been transformed into kinetic energy.

[tex]E_{1}=E_{2}\\ m*g*h+(\frac{1}{2} )*m*v_{1} ^{2}=\frac{1}{2}*m*v_{2} ^{2} \\(9.81*8.21)+(0.5*8.96^{2} )=(0.5*v_{2}^{2} )\\(0.5*v_{2}^{2} )=120.68\\v_{2} ^{2}=241.36\\v_{2} =\sqrt{241.36}\\ v_{2} =15.53[m/s][/tex]

Answer:

Higher denisty

Explanation:

High pressure=high denisty

Answer:

A. right before it hits the ground.

Explanation:

because all the way down, it is building kinetic energy.

Answer:

h = 2.49 [m]

Explanation:

In order to solve this problem we must use the definition of potential energy, which tells us that energy is equal to the product of mass by gravity by height.

The potential energy can be calculated by means of this equation:

Ep = m*g*h

where:

Ep = potential energy = 980 [J]

m = mass = 40 [kg]

g = gravity acceleration = 9.81 [m/s^2]

h = elevation [m]

Now replacing:

980 = 40*9.81*h

h = 2.49 [m]

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