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2 years ago

What is the weight of a 3.5 kg dog on the moon? acceleration of gravity is 1.63 m/s?)

Physics

2 answers:

Sav [38]2 years ago

5 0

1.98 kilograms

weight on the moon = (weight on Earth / acceleration of Earth's gravity) x acceleration of the moon's gravity

weight on the moon = (12 kg. / 9.81 m/s^2) x 1.63 m/s^2

weight on the moon = 1.98 kg.

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boyakko [2]2 years ago

5 0

1.98 kilograms. I just answered that question like an hour ago and got it right

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A train at a constant 60.0 km/h moves east for 40.0 min, then in a direction 50.0 degrees east of due north for 20.0 min, and th

son4ous [18]

Answer:

Part a)

$v = 7.57 km/h$

Part b)

$\theta = 67.5 degree$ North of East

Explanation:

Speed of train towards East = 60 km/h

displacement towards East is given as

$d_1 = 40 km$

now it turns towards 50 degree East of North

so its distance is given as

$d_2 = 20 km(sin50 \hat i + cos50\hat j)$

$d_2 = 15.3 \hat i + 12.8 \hat j$

then finally it moves towards west for 50 min

$d_3 = -50 \hat i$

Now the total displacement of the train is given as

$d = d_1 + d_2 + d_3$

$d = (40 + 15.3 - 50)\hat i + 12.8 \hat j$

$d = 5.3\hat i + 12.8 \hat j$

now total time duration of the motion is given as

$T = 40 min + 20 min + 50 min$

$T = 1.83 h$

now average velocity is given as

$v_{avg} = \frac{5.3\hat i + 12.8\hat j}{1.83}$

$v_{avg} = 2.89\hat i + 6.99\hat j$

Part a)

magnitude of the average velocity is given as

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{2.89^2 + 6.99^2}$

$v = 7.57 km/h$

Part b)

Direction of the velocity is given as

$tan\theta = \frac{v_y}{v_x}$

$tan\theta = \frac{6.99}{2.89}$

$\theta = 67.5 degree$ North of East

6 0

4 years ago

Please help! i'm horrible at this

frez [133]

Answer:

a = Δv/t = (vf - vi)/t = (0 - 5)/4 = -1.25 m/s²

Explanation:

You may or may not need the negative sign, depending on how the question designer was thinking about the problem.

4 0

3 years ago

An underground gasoline tank can hold 1.07 103 gallons of gasoline at 52.0°F. If the tank is being filled on a day when the outd

Damm [24]

Answer:

1069.38 gallons

Explanation:

Let V₀ = 1.07 × 10³ be the initial volume of the gasoline at temperature θ₁ = 52 °F. Let V₁ be the volume at θ₂ = 97 °F.

V₁ = V₀(1 + βΔθ) β = coefficient of volume expansion for gasoline = 9.6 × 10⁻⁴ °C⁻¹

Δθ = (5/9)(97°F -52°F) °C = 25 °C.

Let V₂ be its final volume when it cools to 52°F in the tank is

V₂ = V₁(1 - βΔθ) = V₀(1 + βΔθ)(1 - βΔθ) = V₀(1 - [βΔθ]²)

= 1.07 × 10³(1 - [9.6 × 10⁻⁴ °C⁻¹ × 25 °C]²)

= 1.07 × 10³(1 - [0.024]²)

= 1.07 × 10³(1 - 0.000576)

= 1.07 × 10³(0.999424)

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4 years ago

Density of water 1000kgper m^3 What will be the volume of 3500 kg water

skad [1K]

Answer:

The volume of water is 3.5 cubic meter

Explanation:

Density

The density of a substance or material is the mass per unit volume. The density varies with temperature and pressure.

The formula to calculate the density of a substance of mass (m) and volume (V) is:

$\displaystyle \rho=\frac{m}{V}$

We are given the density of water as $\rho=1,000\ kg/m^3$ .It's required to find the volume of m=3,500 kg of water. Solving for V:

$\displaystyle V=\frac{m}{\rho}$

$\displaystyle V=\frac{3,500}{1,000}$

$V=3.5\ m^3$

The volume of water is 3.5 cubic meter

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3 years ago

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Alexus [3.1K]

Answer: Runofff

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3 years ago