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3 years ago

An object is traveling 15 m/s East and accelerates at a rate of 2.5 m/s. How long was in motion?

Physics

1 answer:

abruzzese [7]3 years ago

5 0

Answer:

Explanation:

An object begins from rest and accelerates at a rate of 2.5 meters ... at a rate of +1.6 meters per second' while traveling +5.0 ... Calculate the block's rate of acceleration. 4. Vf² V, ²+2 ad. (12 m/s)?-(Omis) + 2(a)(50m) ... represents the motion of a car during a 6.0 second time interval. ... A cat moves 15 meters east in 5 seconds.

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Three small balls of the same size but different masses are hung side-by-side in parallel on the strings of same length. They to

andrey2020 [161]

Answer:

m1/6 ( c )

Explanation:

since all the balls starts having the same momentum after the two collisions we will apply the principal of conservation of energy

After first collision

m1v = m1v1 + m2v2 --- ( 1 )

After second collision

m2v2 = m2v2 + m3v3 ---- ( 2 )

combining equations 1 and 2

m1v = m1v1 + m2v2 + m3v3 ----- ( 3 )

All balls moving at the same momentum ( p ) = m1v1 = m2v2 = m3v3

note ; 3p = m1v ∴ m3 = $\frac{m1v}{3v3}$ ----- ( 4 )

applying conservation of energy

3v = v1 + v2 + v3 ------- ( 5 )

also 3m1v1 = m1v = v1 = v/3 =

v2 + v3 = 8/3 v ----- ( 6 )

next eliminate V3 for equation 6 by applying conservation of energy and momentum

m1 = 2m2 ------ ( 7 )

now using p1 = p2 = m1v1 = 1/2 m1v1 hence v2 = 2v1 where v1 = 1/3 v

hence ; v2 = 2/3 v ------- ( 8 )

solving with equation 6 and 8

v3 = 2v ------ ( 9 ) ∴ v/v3 = 1/2 ---- ( 10 )

solving with equation 9 and 10

m3 = m1/3 * 1/2 = m1/6

8 0

3 years ago

A sound wave of the form s = sm cos(kx - ?t + f) travels at 343 m/s through air in a long horizontal tube. At one instant, air m

Naddik [55]

Answer:

960.24 Hz

Explanation:

Here is the complete question

A sound wave of the form

S=Smcos(kx−ωt+Φ)

travels at 343 m/s through air in a long horizontal tube. At one instant, air molecule A at x = 2.000 m is at its maximum positive displacement of 6.00 nm and air molecule B at x = 2.070 m is at a positive displacement of 2.00 nm. All the molecules between A and B are at intermediate displacements. What is the frequency of the wave?

Solution

Given x₁ = 2.0 m, x₂ = 2.070 m, maximum positive displacement s = 6.00 nm at x₁, positive displacement s = 2.00 nm at x₂, velocity of wave v = 343 m/s, maximum positive displacement s₀ = 6.00 nm

Let t₀ = 0 at x₁ = 2.0 m for maximum displacement.

So, s = s₀cos(kx−ωt+Φ)

6 = 6cos(2k - 0 + Φ) = 6cos(2k + Φ)⇒ cos(2k + Φ) = 6/6 = 1

cos(2k + Φ) = 1 ⇒ (2k + Φ) = cos⁻¹ (1) = 0 ⇒ 2k + Φ = 0

Let t₀ = 0 at x₂ = 2.070 m for displacement s = 2.00 nm.

So, s = s₀cos(kx−ωt+Φ)

2 = 6cos(2.070k - 0 + Φ) = 6cos(2.070k + Φ)⇒ cos(2.070k + Φ) = 2/6 = 1/3

cos(2.070k + Φ) = 1/3 ⇒ (2.070k + Φ) = cos⁻¹ (1/3) = 70.53 ⇒ 2.070k + Φ = 70.53.

We now have two simultaneous equations.

2k + Φ = 0 (1)

2.070k + Φ = 70.53. (2)

Subtracting (2) -(1)

2.070k - 2k = 70.53

0.070k = 70.53

k = 70.53/0.070 = 1007.554π/180 rad/m = 17.59 rad/m

k = 2π/λ ⇒ λ = 2π/k

and frequency, f = v/λ = v/2π/k = kv/2π = 17.59 × 343/2π = 960.24 Hz

4 0

3 years ago

Convective cooling cools rocks much more rapidly than heat conduction. hydrothermal circulation represents convective cooling at

Luda [366]

When geophysicists measure the geothermal gradient in areas on the ridges where there is no activity hydrothermally, the gradient is far below than what is predicted theoretically, but when measured near hydrothermal vents it is more than what is predicted. This is because most of the heat is being carried through convection by hydrothermal systems so that the average gradient when measured far from the circulation would be depressed or lower.

7 0

3 years ago

Approximately how much electrical energy does a 5-W lightbulb convert to radiant and thermal energy in one hour?

k0ka [10]

Answer:

18,000 j

Explanation:

the lightbulb dissipates 5W of power

P = ΔE / Δt

rearrange to solve for energy

ΔE = PΔt

P = 5W

t = 1 hour = 60 minutes = 3600 seconds

ΔE = 5 * 3600

ΔE = 18000 J

8 0

3 years ago

A 0.45 kg ball is moving horizontally with a speed of 5.3 m/s when it strikes a vertical wall. The ball rebounds with a speed of

qaws [65]

Answer:

1.31Kgms-1

Explanation:

∆p = m∆v

Where:

∆p= change in momentum

m= mass of the ball

∆v= change in velocity of the ball= (5.3-2.4)= 2.9ms-1

Therefore, substituting appropriately with the values above:

∆p= 0.45×2.9= 1.31Kgms-1

6 0

4 years ago

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