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3 years ago

Playing soccer on a beach will require more effort because the sand causes a great deal of _______ to the ball.

Physics

2 answers:

34kurt3 years ago

5 0

Answer:

I believe the answer to be B

Explanation:

If you were playing on grass, the ball would be able to roll around much easier rather than it to be on sand. If it's wrong I am so sorry

Gre4nikov [31]3 years ago

5 0

The answer of this question is b

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Name one benefit and one limitation of Comparative investigations

GalinKa [24]

Comparative investigation is a research method that makes comparison across different countries of the world. Limitation : data set in different countries may be defined differently. Benefit : the method is flexible.

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3 years ago

Read 2 more answers

Thomas Newcomen was the first to produce a working steam engine. Why is the work of James Watt more widely known than the work o

LenaWriter [7]

Answer:

The steam engine of James watt is more efficient than Newcomen ans more suitable for the industrial revolution.

Explanation:

James Watt is more widely know for working steam engine because Watt has created better engine which is suitable for the industrial revolution. The steam engine of James watt is more efficient than Newcomen. Watt developed the condensing arrangement by using piston which lessen the initial pressure leading to effectively worked than Newcomen's

4 0

3 years ago

A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c

saw5 [17]

Answer:

The final velocity of the thrower is $\bf{3.88~m/s}$ and the final velocity of the catcher is $\bf{0.029~m/s}$ .

Explanation:

Given:

The mass of the thrower, $m_{t} = 70~Kg$ .

The mass of the catcher, $m_{c} = 55~Kg$ .

The mass of the ball, $m_{b} = 0.0480~Kg$ .

Initial velocity of the thrower, $v_{it} = 3.90~m/s$

Final velocity of the ball, $v_{fb} = 33.5~m/s$

Initial velocity of the catcher, $v_{ic} = 0~m/s$

Consider that the final velocity of the thrower is $v_{ft}$ . From the conservation of momentum,

$&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s$

Consider that the final velocity of the catcher is $v_{fc}$ . From the conservation of momentum,

$&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s$

Thus, the final velocity of thrower is $3.88~m/s$ and that for the catcher is $0.029~m/s$ .

8 0

3 years ago

A person is listening to a tone of 500 Hz sitting from a distance of 450 m from the source of the sound. What is the time interv

VladimirAG [237]

Answer:

ΔX = λ = 0.68 m

Explanation:

Wave speed is related to wavelength and frequency by the equation

v = λ f

where the speed of sound is 340 m / s

λ = v / f

λ = 340/500

λ = 0.68 m

this is the wavelength, it is the minimum distance for which the wave epitates its movement, which is equal to the distance between two consecutive compressions of the sound

ΔX = λ = 0.68 m

5 0

3 years ago

One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has

Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = $\sqrt{6gL}$

Hence, v0 = 6.86 m/s

4 0

3 years ago