In many places on Earth, humans are responsible for the removal of grasses, shrubs, trees, and other plants with roots that hold soil in place. This activity is best described by which of the following? *
A) deforestation
B) urbanization
C) air pollution
D) rise in sea level

A 4.0 kg block is moving at 5.0 m/s along a horizontal frictionless surface toward an ideal spring that is attached to a wall, the maximum speed of the block when the spring is compressed to one-half of the maximum distance is 4.33 m/s

From the conservation of energy; the kinetic energy of the mass is equal to the work done on the spring.

i.e.

[tex]\mathbf{\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2_{max}}[/tex]

Given that:

∴

From the equation above, multiplying both sides with 2, we have:

[tex]\mathbf{mv^2 =kx^2_{max}}[/tex]

Making (k) the subject of the formula;

[tex]\mathbf{k = \dfrac{mv^2}{x^2_{max}}}[/tex]

[tex]\mathbf{k = \dfrac{4 \times 5^2}{0.68^2}}[/tex]

k = 216.26 N/m

However, when compressed to one-half of the maximum distance; the speed is computed as follows:

x = 0.68/2 = 0.34 m

∴

[tex]\mathbf{\dfrac{1}{2}mv_o^2 - \dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2}[/tex]

[tex]\mathbf{m(v_o^2 -v^2) =kx^2}[/tex]

[tex]\mathbf{(v_o^2 -v^2) =\dfrac{kx^2}{m}}[/tex]

[tex]\mathbf{(5^2 -v^2) =\dfrac{216.26 \times 0.34^2}{4.0}}[/tex]

25 - v² = 6.25

25 -6.25 = v²

v² = 18.75

[tex]\mathbf{ v= \sqrt{18.75 }}[/tex]

v = 4.33 m/s

Therefore, we can conclude that the speed of the block when the spring is compressed to only one-half of the maximum distance is 4.33 m/s

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Answer:

true

Explanation:

Answer:

I'd say D

Explanation:

because not all weather happens within the atmosphere, and most weather depends on region (lile if your near the equator or not)

Answer:

8 kV

Explanation:

Here is the complete question

Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500 μF capacitors and an 800−V charging source?

Solution

Since the capacitors are initially connected in parallel, the same voltage of 800 V is applied to each capacitor. The charge on each capacitor Q = CV where C = capacitance = 500 μF and V = voltage = 800 V

So, Q = CV

= 500 × 10⁻⁶ F × 800 V

= 400000 × 10⁻⁶ C

= 0.4 C

Now, when the capacitors are connected in series and the voltage disconnected, the voltage across is capacitor is gotten from Q = CV

V = Q/C

= 0.4 C/500 × 10⁻⁶ F

= 0.0008 × 10⁶ V

= 800 V

The total voltage obtained across the ten capacitors is thus V' = 10V (the voltages are summed up since the capacitors are in series)

= 10 × 800 V

= 8000 V

= 8 kV

Answer:

Option C

Explanation:

Options for the question are as follows -

A. At a faucet close to entrance valve

B. At a faucet away from the entrance valve

C. It will be the same at all faucets

D. There will be no increase in the pressure at the faucets

Solution -

The static force will be the same at all faucets and also the area of the faucets be same.

Thus, the pressure created at all faucets will be the same.

Thus, option C is correct

Answer:

Distance travel by go-cart = 500 meter

Explanation:

Given:

Speed of go cart = 25 m/s

Time travel = 20 seconds

Find:

Distance travel by go-cart

Computation:

Distance = Speed x time

Distance travel by go-cart = Speed of go cart x Time travel

Distance travel by go-cart = 25 x 20

Distance travel by go-cart = 500 meter

Answer:

[tex]0.886[/tex] N buoyant force is exerted by air

Explanation:

My weight is [tex]75[/tex] Kg

Weight = mass * gravity

As we know

Buoyant Force is equal to the product of density * acceleration due to gravity and volume of the body

Assuming weight density is a bit less than that of water or equal to water i.e [tex]997.77[/tex] kg/m3

Volume is equal to mass / density

[tex]= 75[/tex] Kg * g/[tex]997.777[/tex]

[tex]= 0.0751[/tex] * g

Buoyant Force

= Volume * g * density

[tex]= 0.0751 * 9.8 * 1.2041[/tex]kg/m3

[tex]= 0.886[/tex] N

The final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.

The final angular velocity of the carnival ride is determined by applying third kinematic equation as shown below;

ωf = ωi + 2αθ

where;

ωf = 0 + 2(2.0) x 6.3

ωf = 25.2 rad/s

Thus, the final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.

​

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Answer: 5.0 rad/s

Explanation: Because that’s what khan said so try it out.

Answer:

U=30cm

Explanation:

All you have to do is to put

Mirror formula , 1/f=1/u + 1/v

You should be careful in sign convention .

Virtual image is negative

we take focal length of convex lens negative even if its not given and so on...

Explanation:

mass, m = 5kg

initial velocity, u = 16m/s

final velocuty, v = -22m/s

change in momentum, ∆p = ?

∆p = m (v-u)

5(-22-16)

5(38)

∆p = 190kgm/s

check the calculations!

Answer:

1)   P₁ = -2 D,   2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

          [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

where f is the focal length, p and q are the distance to the object and the image, respectively

1) to see a distant object it must be at infinity (p = ∞)

          [tex]\frac{1}{f_1} = \frac{1}{q}[/tex]

           q = f₁

2) for an object located at p = 25 cm

            [tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}[/tex]

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

2)  

        [tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}[/tex]

        [tex]\frac{1}{f_2}[/tex] = 0.06

         f₂ = 16.67 cm

the expression for the power of the lenses is

          P = [tex]\frac{1}{f}[/tex]

where the focal length is in meters

1)       P₁ = 1/0.50

        P₁ = -2 D

2)     P₂ = 1 /0.16667

        P₂ = 6 D

The two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.

A double replacement reaction have two ionic compounds that are exchanging anions or cations.

From the given options, we can choose the following based on their exchange of anions or cations.

Thus, the two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.

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#SPJ2

Answer:

2

Explanation:

Answer:

work done by friction = 5889 J

Explanation:

We are given;

Mass of car; m = 130 kg

Speed at point A; v1 = 25 m/s

Speed at point B: v2 = 8 m/s

Since radius is 12 m

At point A, distance is; y1 = 12 m

At point B, distance is; y2 = -12 m

Now, formula for work done by all the forces is given by the equation;

Total work;

W_gravity + W_others = K2 - K1

Where W_others is work done by other forces which is equal to work done by friction

Where K2 - K1 is change in kinetic energy.

W_grav is also change in potential energy and is expressed as;

W_grav = mgy1 - mgy2

K2 - K1 = ½m(v1)² - ½m(v2)²

Thus;

mgy1 - mgy2 + W_others = ½m(v1)² - ½m(v2)²

Making W_others the subject;

W_others = ½m(v1)² - ½m(v2)² + mgy2 - mgy1

Plugging in the relevant values;

W_others = (½ × 130 × 25²) - (½ × 130 × 8²) + (130 × 9.8 × -12) - (130 × 9.8 × 12)

W_others = 5889 J

Recall that I earlier said W_others = work done by friction.

Thus, work done by friction = 5889 J

Answer:

W = 1.875 J

Explanation:

For this exercise let's use the relationship between work and kinetic energy

          W = ΔK

The kinetic energy of rotational motion is

         K₀ = ½ I w²

we can assume that the box is small, so it can be treated as a point object, with moment of inertia

          I = m rₐ²

angular and linear velocity are related

          v = w r

          w = v / r

we substitute in the equation, for point A

         K₀ = ½ (m rₐ²) (v / rₐ)²

         K₀ = ½ m v²

For the final point B, as the system is isolated the angular momentum is conserved

initial        L₀ = Io wo

final          L_f = I_f w_f

                L₀ = L_f

                 I₀ w₀ = I_f w_f

                (m rₐ²) w₀ = (m  [tex]r_{b} ^2[/tex]) w_f

                 w_f = (rₐ/r_b)² w₀

with this value we find the final kinetic energy

         K_f = ½ I_f w_f²

         K_f = ½ (m [tex]r_{b}^2[/tex]) ( (rₐ / r_b)²  w₀) ²

         K_f = ½ m [tex]\frac{r_a^4}{r_b^2} \ w_o^2[/tex]

we substitute in the realcion of work

          W = K_f - K₀

          W = ½ m  [tex]( \( \frac {r_a^2 }{r_b} )^2[/tex] w₀² - ½ m v²

          W = ½ m  [tex]\frac{r_a^4}{r_b^2} ( \frac{v}{r_a} ) ^2[/tex] - ½ m v²

           W = ½ m [tex]\frac{r_a^2}{r_b^2} \ v^2[/tex] - ½ m v2

          W = ½ m v² (([tex]( \ (\frac{r_a}{r_b})^2 -1)[/tex]

let's calculate

           W = ½ ( [tex]\frac{8}{32}[/tex] ) 5 ((2/1)² -1)

           W = 0.625 (3)

           W = 1.875 J

Answer:

We apply an electric field in the negative y direction

Explanation:

Since A uniform magnetic field is in the positive z direction and A positively charged particle is moving in the positive x direction through the field, the magnetic force acting on the positively charged particle is in the positive y direction according to Fleming's right-hand rule.

For the net force on the particle to be zero, we apply an electric field in the negative y direction to create an electric force on the positively charged particle, so as to cancel out the magnetic force.

Answer:

[tex]3.49\times 10^{-8}\ \Omega\text{m}[/tex]

Explanation:

r = Radius = [tex]\dfrac{2}{2}=1\ \text{mm}[/tex]

B = Magnetic field = 3 mT

1 mm = Distance from the surface of the wire

V = Voltage

x = Distance from the probe = [tex]r+1=1+1=2\ \text{mm}[/tex]

R = Resistance

L = Length of wire = 4.5 m

Magnetic field is given by

[tex]B=\dfrac{\mu_0I}{2\pi x}\\\Rightarrow I=\dfrac{B2\pi x}{\mu_0}\\\Rightarrow I=\dfrac{3\times 10^{-3}\times 2\times \pi 2\times 10^{-3}}{4\pi 10^{-7}}\\\Rightarrow I=30\ \text{A}[/tex]

Voltage is given by

[tex]V=IR\\\Rightarrow R=\dfrac{V}{I}\\\Rightarrow R=\dfrac{1.5}{30}\\\Rightarrow R=0.05\ \Omega[/tex]

Resistivity is given by

[tex]\rho=\dfrac{RA}{L}\\\Rightarrow \rho=\dfrac{0.05\times \pi (1\times 10^{-3})^2}{4.5}\\\Rightarrow \rho=3.49\times 10^{-8}\ \Omega\text{m}[/tex]

The resistivity of the material is [tex]3.49\times 10^{-8}\ \Omega\text{m}[/tex].

Answer:

475.7 m/s

Explanation:

Given,

Frequency ( f ) = 67 Hz

Wavelength ( λ ) = 7.1 m

To find : Speed ( v ) = ?

Formula : -

v = f λ

v

= 67 x 7.1

= 475.7 m/s

Therefore,

the speed of the wave is 475.7 m/s.

Answer:

B i think

Explanation:

...

Answer:

B. it increases

Explanation:

As shown in the table provided, the speed of sound in water (1493 m/s) is greater than the speed of sound in air (346 m/s).

Answer:

B is the correct answer.

Explanation:

Answer:

the length of the wire is 134.62 m.

Explanation:

Given;

resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm

cross-sectional area of the wire, A  = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²

resistance of the wire, R = 10Ω

The length of the wire is calculated as follows;

[tex]R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m[/tex]

Therefore, the length of the wire is 134.62 m.

Answer:

question #1 is A

Question #2 is C

Explanation:

Answer:

The equation says that due to variation in temperature is

delt T = .59 m/s / C = 16 C * .59 m/s = 9.44 m/s

So v = 332 m/s + 9.44 m/s = 341 m/s (to three significant figures)

Answer:

W = 1/2 K x^2

x^2 = 2 * W / K = 2 * 49 J / (N/m)  = .218 / m^2

x =  .467 m

Answer:

I think its d

Explanation:

I'm not sure I'm sorry if I'm wrong