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Airida [17]
3 years ago
14

When you push with a horizontal 15-N force on a book that slides at constant

Physics
1 answer:
Marat540 [252]3 years ago
8 0

Answer:

f = 15 N

Explanation:

It is given that, when you push with a horizontal 15-N force on a book that slides at constant  velocity, a frictional force also acts on it. Frictional force is an opposing force. The magnitude of applied force and frictional forces are same. So, the force of friction on the book is equal to 15 N.

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Thus, area is the product of two lengths and so has dimension L2, or length squared.

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A motorcycle that is slowing down uniformly covers 2.0 successive km in 80 s and 120 s, respectively. Calculate (a) the accelera
8090 [49]

Answer:

acceleration = -0.042 m/s²

velocity at beginning =  14.167 m/s

velocity at  end = 5.7183 m/s

Explanation:

given data

distance d1 = 1 km

distance d2 = 2 km

time  t1 = 80 s

time t2 = 120 s + 80s = 200 s

to find out

acceleration and velocity at beginning and end

solution

we apply here law of motion that is

d = vt + 1/2×at²    

put value

1000 = v(80) + 1/2×a(80)²           ........................1

and

2000 = v(200) + 1/2×a(200)²      ........................2

so from equation 1 and 2 we get a and v

a = -0.042 m/s² and

v = 14.167 m/s

so by kinematic final velocity will be

V² = v² + 2ad

V² = (14.167)² + 2×(-0.042)×(2000)

V²  = 32.70

V = 5.7183 m/s

so

acceleration = -0.042 m/s²

velocity at beginning =  14.167 m/s

velocity at  end = 5.7183 m/s

8 0
3 years ago
Do all substances radiate heat
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I don't believe they do
6 0
3 years ago
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Consider the data collected in science class. Different masses were thrown with varied amounts of
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Answer:

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Explanation:

I know this because its a clear obvious answer not only that it was one of my USA TESTPREP questions and it was right.

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3 years ago
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A heavy rope, 60 ft long, weighs 0.5 lb/ft and hangs over the edge of a building 130 ft high. (Let x be the distance in feet bel
Nataliya [291]

Answer:

Riemann sum

W = lim n→∞ Σ 0.5xᵢΔx (with the summation done from i = 1 to n)

Integral = W = ∫⁶⁰₀ 0.5x dx

Workdone in pulling the entire rope to the top of the building = 900 lb.ft

Riemann sum for pulling half the length of the rope to the top of the building

W = lim n→∞ Σ 0.5xᵢΔx (but the sum is from i = 1 to n/2)

Integral = W = ∫⁶⁰₃₀ 0.5x dx

Work done in pulling half the rope to the top of the building = 675 lb.ft

Step-by-step explanation:

Using Riemann sum which is an estimation of area under a curve

The portion of the rope below the top of the building from x to (x+Δx) ft is Δx.

The weight of rope in that part would be 0.5Δx.

Then workdone in lifting this portion through a length xᵢ ft would be 0.5xᵢΔx

So, the Riemann sum for this total work done would be

W = lim n→∞ Σ 0.5xᵢΔx (with the summation done from i = 1 to n)

The Riemann sum can easily be translated to integral form.

In integral form, with the rope being 60 ft long, we have

W = ∫⁶⁰₀ 0.5x dx

W = [0.25x²]⁶⁰₀ = 0.25 (60²) = 900 lb.ft

b) When half the rope is pulled to the top of the building, 60 ft is pulled up until the length remaining is 30 ft

Just like in (a)

But the Riemann sum will now be from the start of the curve, to it's middle

Still W = lim n→∞ Σ 0.5xᵢΔx (but the sum is from i = 1 to n/2)

W = ∫⁶⁰₃₀ 0.5x dx

W = [0.25x²]⁶⁰₃₀ = 0.25 (60² - 30²) = 675 lb.ft

Hope this Helps!!!

7 0
3 years ago
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