If a ball rolls at constant velocity, it is rolling at constant speed? explain

1 answer:

7 0

Velocity is speed with direction. So, if velocity varies directly with speed, that statement would be true. A constant velocity would resort in a constant speed. They are connected and are dependant on each other.

I hope this helps!
~kaikers

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A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 21.0 N/m. The

mestny [16]

Answer:

a. Δx = 2.59 cm

Explanation:

mb = 0.454 kg , mp = 5.9 x 10 ⁻² kg , vp = 8.97 m / s , k = 21.0 N / m

Using momentum conserved

mb * (0) + mp * vp = ( mb + mp ) * vf

vf = ( mp / mp + mb) * vp

¹/₂ * ( mp + mb) * (mp / mp +mb) ² * vp ² = ¹/₂ * k * Δx²

Solve to Δx '

Δx = √ ( mp² * vp² ) / ( k * ( mp + mb )

Δx = √ ( ( 5.9 x 10⁻² kg ) ² * (8.97 m /s) ² / [ 21.0 N / m * ( 5.9 x10 ⁻² kg + 0.454 kg ) ]

Δx = 0.02599 m ⇒ 2.59 cm

8 0

3 years ago

20 cubic inches of a gas with an absolute pressure of 5 psi is compressed until its pressure reaches 10 psi. What is the new vol

Alborosie

Initial volume of the gas (V1) = 10 inches^3
Initial pressure (P1) = 5 psi
Final pressure after compression of the gas (P1) = 10 psi
Let us assume the final volume of the gas (V2) = x
According to Boyle's Gas law, the pressure and volume of a gas remains constant under ideal condition. Then
P1V1= P2V2
5 * 10 = 10 * x
50 = 10x
x = 50/10
= 5 cubic inches
So the volume of the gas after it was compressed was 5 cubic inches. I hope the procedure is clear enough for you to understand.

8 0

2 years ago

What two traits must a viable hypothesis have

I am Lyosha [343]

<span>The hypothesis must be testable and falsifiable. The first stipulation is that the hypothesis must be able to be tested and be phrased as such. In addition, the second stipulation holds that the statement must be able to be falsified; that is, the statement can be showed to not be the case.</span>

8 0

2 years ago

A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickn

KengaRu [80]

Answer:

88 %

Explanation:

The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by

P = KAΔT/d

The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m

The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

= 1/(0.003927/0.0288 + 1)

= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.