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Citrus2011 [14]
3 years ago
14

You are pushing a wooden crate across the floor at constant speed. You decide to turn the crate on end, reducing by half the sur

face area in contact with the floor. In the new orientation, to push the same crate across the same floor with the same speed, the force that you apply must be aboutA. Four times as greatB. Twice as greatC. Equally greatD. Half as greatE. One-fourth as greatas the force required before you changed the crate's orientation.
Physics
1 answer:
borishaifa [10]3 years ago
3 0

Answer:

D. Half as great

Explanation:

Since we know that the friction force between the surface of crate and ground is given as

F_f = \mu F_n

so here we know that

\mu = friction coefficient between two surfaces which depends on the effective contact area between two surfaces

F_n = normal force due to the object

So when we turn the object on another side such that the surface area is half then the friction coefficient will become also half

So here the friction force will also reduce to half

so correct answer will be

D. Half as great

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A student standing on a cliff that is a vertical height d = 8.0 m above the level ground throws a stone with velocity v0 = 22 m/
Alina [70]

Answer:

V = 25.3 , θf = 36.7° below the horizontal

Explanation:

Given Vo = 22m/s , θ = 23°

Vx = VoCosθ

Vy = VoSinθ – gt

t = total time of flight

5 0
3 years ago
Convert 1 mm to meters using scientific notation. Include units in your answer.
miskamm [114]

Answer:

1.0 x 10^-3 m

Explanation:

1mm = (1/1000) m = 0.001m = 1.0 x 10^-3m (because to get 1.0, you have to  movi the decimal point in the "0.001" three spots over to the right)

3 0
3 years ago
Which image shows the correct way of lining up vectors to add them
Marina CMI [18]

Answer:

it's A

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wen aligning the vectors the head and the tail should meet

4 0
1 year ago
Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t
marshall27 [118]

Answer:

        x = 240 m

Explanation:

This is a kinematics exercise

Let's fix our frame of reference on car A

           x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

         

the initial position of car a is zero

           x = 0 + v₀ₐ t + ½ 0.8 t²

for car B

          x = x_{ob} + v_{ob} t - ½ a_b t²

     

car B's starting position is 30 m

         x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

         0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²

let's reduce the speeds to the SI system

        v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

        v_{ob} = 23.4 km / h = 6.5 m / s

        4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

        0.2 t² - 2.5 t - 30 = 0

        t² - 12.5 t - 150 = 0

we solve the quadratic equation

       t = \frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150}  }{2}

       t = \frac{12.5 \  \pm 27.5}{2}

       t₁ = 20 s

       t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

        x = 4 t + ½ 0.8 t²

        x = 4 20 + ½ 0.8 20²

        x = 240 m

8 0
3 years ago
You sit "at rest" in front of your computer to answer this question. But you sit on the surface of a planet that spins, so even
igomit [66]

Answer: Linear speed is 1,670 Kph.

Explanation:

If we assume that the earth is a perfect sphere, and that is spinning itself once every roughly 24 hr, we can get the angular velocity of the Earth, in magnitude, as follows:

ω = 2π / 24 Hr

Now, by definition, an angle is the relationship between the arc s, and the radius r, so we can replace these values in the angular velocity expression, as follows:

ω = (Δs / r) . 1/Δt ⇒ ω = (Δs/Δt). 1/r

But, by definition, Δs/At, is just the linear velocity, v, so we can conclude the following;

ω = v/r ⇒ v = ω. r

So, we can get v, as follows:

v = 2π /24 hr . 6378 Km = 1,670 Km/hr.

4 0
3 years ago
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