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solong [7]
3 years ago
14

What is the acceleration of a car that goes from zero to 60m/s in 15s?

Physics
1 answer:
bonufazy [111]3 years ago
3 0

Answer:What is the acceleration of a car that moves from rest to 15.0 m/s in 10.0 s? Vi=0, vf= 15.0 m/s,t=10.0s, a=? a= vf =vi/tA=15.0m/s-0m/s/10.0s = 15.0s/10.0s m/s*1/s =1.50 m/s^2 11.

Explanation:

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Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time (when they are not slee
user100 [1]

Answer:

I₁ = (7.78 i ^ - 6.71 j ^) 10⁻³ J s ,  I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s ,  I₃ = (19.1i ^ + 18.6 j ^) 10⁻³ J s  and I₄ = (-9.14i ^ + 7.24 j ^) 10⁻³ J s

Explanation:

The impulse is equal to the variation of the moment, to apply this relationship to our case, we will assume that initially the mouse was at rest

    I = Δp = m v_{f} -m v₀

    I = m (v_{f}  -v₀)

Bold indicates vector quantities, let's calculate the momentum of each mouse in for the x and y axes

We recommend bringing all units to the SI system

Mouse 1.

It has a mass of 22.3 g = 22.3 10⁻³ kg, a final velocity of (v = 0.349 i ^ - 0.301 j ^) m / s with an initial velocity of zero

    Iₓ = m (v_{fx}  - v₀ₓ)

    Iₓ = 22.3 10⁻³ (0.349 -0)

    Iₓ = 7.78 10⁻³ J s

   I_{y} = m (v_{fy}  -v_{oy} )

   I_{y} = 22.3 10⁻³ (-0.301)

   I_{y} = -6.71 10⁻³ J s

   I₁ = (7.78 i ^ - 6.71 j ^) 10⁻³ J s

Mouse 2

Mass 17.9 g = 17.9 10⁻³ kg

Speed ​​(-0.699 i ^ - 0.815 j ^) m / s

    Iₓ = m (v_{fx}  - v₀ₓ)

    Iₓ = 17.9 10⁻³ (-0.699 -0)

    Iₓ = -12.5 10⁻³ J s

    I_{y} = 17.9 10⁻³ (-0.815 - 0)

    I_{y} = -14.6 10⁻³ J s

   I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s

Mouse 3

Mass 19.1 g = 19.1 10⁻³ kg

Speed ​​(0.745i ^ + 0.975 j ^) m / s

    Iₓ = 19.1 10⁻³ (0.745 -0)

    Iₓ = 14.2 10⁻³ J s

    I_{y} = 19.1 10⁻³(0.975 -0)

    I_{y} = 18.6 10⁻³ J s

    I₃ = (19.1i ^ + 18.6 j ^) 10⁻³ J s

Mouse 4

Mass 10.1 g = 10.1 10⁻³ kg

Speed ​​(-0.905i ^ + 0.717j ^) m / s

    Iₓ = 10.1 10⁻³ (-0.905 -0)

    Iₓ = -9.14 10⁻³ J s

    I_{y} = 10.1 10⁻³ (0.717 -0)

    I_{y} = 7.24 10⁻³ J s

   I₄ = (-9.14i ^ + 7.24 j ^) 10⁻³ J s

8 0
4 years ago
The driver accelerates a 330.0 kg snowmobile, which results in a force being exerted that speeds up the snowmobile from 6.00 m/s
just olya [345]

Answer:

(a) 5610 kgm/s

(b) 5610 Ns.

(c)  78. 64 N

Explanation:

a. Change in momentum: This can be defined as the product of the mass of a body to its change in velocity. The S.I unit of change in momentum is kgm/s.

Mathematically, change in momentum is expressed as

ΔM = mΔv......................... Equation 1

Where ΔM = change in momentum, m = mass of snowmobile, Δv = change in velocity.

Given: m = 330 kg, Δv = v₂-v₁ = 23-6 = 17 m/s.

Note: v₁ and v₂ are the initial and the final velocity of the snowmobile.

ΔM = 330(17)

ΔM = 5610 kgm/s.

(b) Impulse: This can be defined as the product and force and time. The S.I unit of impulse is Ns.

Note: From Newton's second law of motion, impulse is equal to change in momentum.

Therefore,

I = ΔM................ Equation 2

Where I = impulse of the force.

Since ΔM = 5610 kgm/s.

Therefore

I = 5610 Ns.

Thus the impulse = 5610 Ns.

(c) Force: This can be defined as the product of the mass of a body and its acceleration. The S.I unit of force is Newton (N).

F = ma ................................. Equation 3

F = force, m = mass of the body, a = acceleration

But,

a = ( v₂-v₁)/t

Where v₂ = 23.0 m/s, v₁ = 6.0 m/s t = 60.0 s.

a = (23-6)/60

a = 0.283 m/s².

Substituting the value a and m into equation 3

F = 330(0.2383)

F = 78.639 N.

F ≈ 78. 64 N

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3 years ago
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Carry capacity determines maximum population size
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