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kotegsom
2 months ago
10

On a caterpillars map all distances are marked in kilometers . The caterpillars map shows the distance between two milkweed plan

ts is 4,012 milimeters what is the distance in kilometers
Physics
1 answer:
ValentinkaMS [3.4K]2 months ago
5 0

Answer:

The equivalent distance in kilometers is 4012 ×10^{-6} km.

Explanation:

It's known that 1 millimeter converts to 10^{-3} meters. Then, 1 meter converts to 10^{-3} kilometers. Therefore, the conversion for 1 millimeter to kilometers can be stated as

1 mm = 10^{-3} m

1 m = 10^{-3} km

Thus, 1 mm = 10^{-3}×10^{-3} km = 10^{-6} km.

Given the distance of 4012 mm, the corresponding distance in kilometers will be

4012 mm = 4012 ×10^{-6} km.

The distance therefore is 4012 ×10^{-6} km.

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The approximate answer is 6.84. Reasoning: (2.78^2 + 6.25^2)^(1/2) = 6.84 approx.
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A 55 kg gymnast wedges himself between two closely spaced vertical walls by pressing his hands and feet against the walls. Part
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answer:

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1 month ago
A 0.050-kg lump of clay moving horizontally at 12 m/s strikes and sticks to a stationary 0.15-kg cart that can move on a frictio
Ostrovityanka [3204]

Answer:

Explanation:

According to the parameters provided,

mass of the clay lump, m₁ = 0.05 kg

initial velocity of the lump, u₁ = 12 m/s

mass of the cart, m₂ = 0.15 kg

initial speed of the cart, u₂ = 0

As the clay adheres to the cart, we have an inelastic collision scenario. Let v represent the combined speed of both the cart and lump post-collision. Given that momentum is conserved, we have:

m_1u_1+m_2u_2=(m_1+m_2)v

v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}

v=\dfrac{0.05\ kg\times 12\ m/s+0}{0.05\ kg+0.15\ kg}

The resultant speed is v = 3 m/s.

Thus, the final speed of both cart and lump following the collision is 3 m/s. This concludes the solution.

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Ability of the muscles to function effectively and efficiently without undue fatigue
Keith_Richards [3271]

Response:

Physical well-being

Clarification:

7 0
2 months ago
A person is rowing across the river with a velocity of 4.5 km/hr northward. The river is flowing eastward at 3.5 km/hr (Figure 4
Yuliya22 [3333]

Answer: Her velocity magnitude (v) relative to the shore is 5.70 km/h.

Explanation:

Let Q be the speed of the boat, and P be the speed of the river flow.

R represents the resultant velocity combining boat velocity and river current.

According to vector addition using the law of triangles:

R=\sqrt{P^2+Q^2+2PQCos\theta}

From the diagram:

P = 3.5 km/h, Q = 4.5 km/h

\theta= 90^o

R=\sqrt{P^2+Q^2+2PQCos\theta}=\sqrt{(3.5)^2+(4.5)^2+3.5\times 4.5\times cos90^o}=5.70

(Cos90^o=0),(sin 90^o=1)

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Therefore, her velocity magnitude relative to the shore is 5.70 km/h.

8 0
2 months ago
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