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MrRissso [65]
3 years ago
5

A 1430 kg car speeds up from 7.50 m/s to 11.0 m/s in 9.30 s. Ignoring friction, how much power did that require?(unit=W)PLEASE H

ELP
Physics
1 answer:
Aleks04 [339]3 years ago
3 0

Answer:

Kinetic energy = (1/2) (mass) (speed²)

Original KE = (1/2) (1430 kg) (7.5 m/s)²  =  40,218.75 joules

Final KE  =  (1/2) (1430 kg) (11.0 m/s)²  =   86,515 joules

Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules

Power = work/time = 46,296.25 joules / 9.3 sec  =  4,978.1 watts .

Explanation:

Dont report my answer please

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Richard Julius once made a model plane that could travel a max speed of 110 m/s. Suppose the plane was held in a circular path b
hjlf

Answer:

85.8 m/s

Explanation:

We know that the length of the circular path, L the plane travels is

L = rθ where r = radius of path and θ = angle covered

Now,its speed , v = dL/dt = drθ/dt = rdθ/dt + θdr/dt

where dθ/dt = ω = angular speed = v'/r where v' = maximum speed of plane and r = radius of circular path

Now, from θ = θ₀ + ωt where θ₀ = 0 rad, ω = angular speed  and t = time,

θ = θ₀ + ωt = 0 + ωt = ωt

So, v = rdθ/dt + θdr/dt

v = rω + ωtdr/dt

v = (r + tdr/dt)ω

v = (r + tdr/dt)v'/r

v = v' + tv'/r(dr/dt)

v = v'[1 + t(dr/dt)/r]

Given that v' = 110 m/s, t = 33.0s, r = 120 m and dr/dt = rate at which line is shortened = -0.80 m/s (negative since it is decreasing)

So, v = 110 m/s[1 + 33.0 s(-0.80 m/s)/120 m]

v = 110 m/s[1 + 11.0 s(-0.80 m/s)/40 m]

v = 110 m/s[1 + 11.0 s(-0.02/s)]

v = 110 m/s[1 - 0.22]

v = 110 m/s(0.78)

v = 85.8 m/s

8 0
3 years ago
Why do different substances have different melting and boiling points?
azamat
Because the atoms and molecules all have different properties
3 0
3 years ago
Forces always act alone. true or false
rjkz [21]
The statement "<span>Forces always act alone" is false. Force do not always act alone. It is always accompanied by another force that is always opposite to it. </span>
7 0
3 years ago
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Find the ratio of the lengths of the two mathematical pendulums, if the ratio of periods is 1.5​
juin [17]

Answer:

The ratio of lengths of the two mathematical pendulums is 9:4.

Explanation:

It is given that,

The ratio of periods of two pendulums is 1.5

Let the lengths be L₁ and L₂.

The time period of a simple pendulum is given by :

T=2\pi \sqrt{\dfrac{l}{g}}

or

T^2=4\pi^2\dfrac{l}{g}\\\\l=\dfrac{T^2g}{4\pi^2}

Where

l is length of the pendulum

l\propto T^2

or

\dfrac{l_1}{l_2}=(\dfrac{T_1}{T_2})^2 ....(1)

ATQ,

\dfrac{T_1}{T_2}=1.5

Put in equation (1)

\dfrac{l_1}{l_2}=(1.5)^2\\\\=\dfrac{9}{4}

So, the ratio of lengths of the two mathematical pendulums is 9:4.

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3 years ago
A boy is pushing a box on the ground to the right. What four forces is he using?
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