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3 years ago

7

A biker with an initial velocity of 16.0 m/s slows uniformly to 6.0 m/s in 8.0 seconds. What is the acceleration of the car duri

ng this 8.0-second interval?

1 answer:

leonid [27]3 years ago

5 0

Answer:

-1.25 m/s²

Explanation:

Given:

v₀ = 16.0 m/s

v = 6.0 m/s

t = 8.0 s

Find: a

v = at + v₀

6.0 m/s = a (8.0 s) + 16.0 m/s

a = -1.25 m/s²

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A rotating fan completes 1200 revolutions every minute. Consider the tip of a blade, at a radius of 0.15 m. (a) Through what dis

olga55 [171]

Answer:

(a) 0.942 m

(b) 18.84 m/s

(c) 2366.3 m/s²

(d) 0.05 s

Explanation:

(a) In one revolution, it travels through one circumference, 2πr = 2 × 3.14 × 0.15 m = 0.942 m.

(b) Its frequency, f, is 1200 rev/min = $\dfrac{1200}{60}$ rev/s = 20 rev/s.

Its angular frequency, ω = 2πf = 2π × 20 = 40π

The speed is given by

v = ωr = 40π × 0.15 = 6π = 18.84 m/s

(c) Its acceleration is given by, a = ω²r = (40π)² × 0.15 = 2366.3 m/s²

(d) The period is the inverse of the frequency because it is the time taken to complete one revolution.

$T = \dfrac{1}{f}$

T = 1/20 = 0.05 s

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3 years ago

Select all of the following that are dominant alleles.

Dennis_Churaev [7]

Answer:

D E K AND L

Explanation:

HAD TO HAVE POINTS

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3 years ago

how will resistance change when the voltage increases in an electric circuit if the current remains constant

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V=IR
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As V increases, R also increases.

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3 years ago

A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.

kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

$W=Fd cos \theta$

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

$\theta=30^{\circ}$ is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

$W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J$

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

$\Delta K = W$

where

$\Delta K$ is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

$\Delta K=69.3 J$

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

$\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, $\Delta K$ , after it has travelled for d=3 m. Using the work-energy theorem again, we find

$\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J$

And substituting into (1) and re-arrangin the equation, we find

$v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s$

6 0

4 years ago

You are a lifeguard and spot a drowning child 30 meters along the shore and 60 meters from the shore to the child. You run along

stiv31 [10]

Answer:

The lifeguard should run approximately 17.752 meters along the shore, before, jumping in the water

Explanation:

The given parameters are;

The rate at which the lifeguard runs = 5 m/s

The rate at which the lifeguard swims = 1 m/s

The horizontal distance of the child from the lifeguard = 30 meters along the shore

The vertical distance of the child from the lifeguard = 60 meters along the shore

Let x represent the distance the lifeguard runs

We have;

The distance the lifeguard swims = √((30 - x)² + 60²)

Time = Distance/Speed

The time the lifeguard runs = x/5

The time the lifeguard swims = √((30 - x)² + 60²)/1

The total time = √((30 - x)² + 60²) + x/5

The minimum time is given by finding the derivative and equating the result to zero, as follows;

Using an online application, we have;

d(√((30 - x)² + 60²) + x/5)/dx = 1/5 - (30 - x)/(√((30 - x)² + 60²)) = 0

Which gives;

1/5 - (30 - x)/(√(x² - 60·x + 4500) = 0

(30 - x)/(√(x² - 60·x + 4500)) = 1/5

5×(30 - x) = √(x² - 60·x + 4500)

We square both sides to get;

(5×(30 - x))² = (x² - 60·x + 4500)

(5×(30 - x))² - (x² - 60·x + 4500) = 0

25·x² - 1500·x + 22500 - x² + 60·x - 4500 = 0

24·x² - 1440·x + 18000 = 0

Dividing n=by 24 gives;

24/24·x² - 1440/24·x + 18000/24 = 0

x² - 60·x + 750 = 0

By the quadratic formula, we have;

x = (60 ± √((-60)² - 4×1×750))/(2 × 1) =

Using an online application, we have;

x = (60 ± 10·√6)/(2)

x = 30 + 5·√6 or x = 30 - 5·√6

x ≈ 42.25 m and x ≈ 17.752 m

At x = 42.25

Time = √((30 - 42.247)² + 60²) + 42.247/5 ≈ 69.69 seconds

At x = 17.75

Time = √((30 - 17.752)² + 60²) + 17.752/5 ≈ 64.79 seconds

Therefore, the route with the shortest time is when the lifeguard runs approximately 17.752 meters (rounded to three decimal places) along the shore, before, diving in the water

5 0

3 years ago