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kompoz [17]
3 years ago
6

Lindsey started biking to the park traveling 15 mph, after some time the bike got a flat so Lindsey walked the rest of the way,

traveling 4 mph. If the total trip to the park took 5 hours and it was 53 miles away, how long did Lindsey travel at each speed
Physics
1 answer:
-BARSIC- [3]3 years ago
5 0

Answer:

Lindsey biked 45 miles for 3 hours at 15 mph and walked 8 miles for 2 hours at 4 mph.

Explanation:

Speed = distance/time

Let the distance that Lindsey biked through be x miles and the time it took her to bike through that distance be t hours

Then, the rest of the distance that she walked is (53 - x) miles

And the time she spent walking that distance = (5 - t) hours

Her biking speed = 15 mph = 15 miles/hour

Speed = distance/time

15 = x/t

x = 15 t (eqn 1)

Her walking speed = 4 mph = 4 miles/hour

4 = (53 - x)/(5 - t)

53 - x = 4 (5 - t)

53 - x = 20 - 4t (eqn 2)

Substitute for X in (eqn 2)

53 - 15t = 20 - 4t

15t - 4t = 53 - 20

11t = 33

t = 3 hours

x = 15t = 15 × 3 = 45 miles.

(53 - x) = 53 - 45 = 8 miles

(5 - t) = 5 - 3 = 2 hours

So, it becomes evident that Lindsey biked 45 miles for 3 hours at 15 mph and walked 8 miles for 2 hours at 4 mph.

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A rope is wrapped three and a half times around a cylinder. Determine the range of force T exerted on the free end of the rope f
Taya2010 [7]

The range of force exerted at the end of the rope is 285.7 N to 1,000 N.

<h3>Net horizontal force of the cylinder</h3>

The net horizontal force of the cylinder when it is at equilibrium position is determined by applying Newton's second law of motion.

∑F = 0

F - μFn = 0

F - 0.2(5,000) = 0

F - 1,000 = 0

F = 1,000 N

The strength of the applied force increases as the number of turns of the rope increases.

minimum force = total force/number of turns of rope

minimum force = 1,000/3.5

minimum force = 285.7 N

Thus, the range of force exerted at the end of the rope is 285.7 N to 1,000 N.

Learn more about Newton's second law of motion here: brainly.com/question/3999427

8 0
2 years ago
Siruis, the brightest star in the night sky, has a luminosity of 22. This means that Sirius: A, B, C, D QUESTION
suter [353]

Answer:

i think C

Explanation:

4 0
3 years ago
When was the first electric typewriter produced and by what company
LiRa [457]
In 1920, after returning from Army service, he produced a successful model and in 1923 turned it over to the Northeast Electric Company of Rochester for development.
6 0
4 years ago
Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the ot
almond37 [142]

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

\Sigma F_{x} = P+T-\mu_{k}\cdot N = 0 (Ec. 1)

\Sigma F_{y} = N - W = 0 (Ec. 2)

Where:

P - Pushing force, measured in newtons.

T - Tension, measured in newtons.

\mu_{k} - Coefficient of kinetic friction, dimensionless.

N - Normal force, measured in newtons.

W - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

P+T -\mu_{k}\cdot W = 0

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

\mu_{k} =\frac{P+T}{m\cdot g}

If we know that P = 400\,N, T = 290\,N, m = 300\,kg and g = 9.807\,\frac{m}{s^{2}}, then:

\mu_{k} = \frac{400\,N+290\,N}{(300\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\mu_{k} = 0.235

The kinetic coefficient of friction of the crate is 0.235.

5 0
3 years ago
An engineering firm is designing a ski lift. The wire rope needs to travel with a linear velocity of 2.0 meters per second, and
Mazyrski [523]

Answer:

The diameter of the bull-wheel is 3.82

Explanation:

Given that,

Velocity = 2.0 m/s

Angular velocity = 10 rev/m

\omega=10\times\dfrac{2\pi}{60}

\omega=1.0472\ rad/s

We need to calculate the diameter of bull-wheel

Using formula of angular velocity

v= r\omega

r=\dfrac{v}{\omega}

Put the value into the formula

r=\dfrac{2.0}{1.0472}

r=1.91\ m

The diameter of the bull-wheel

D=2r

D=2\times1.91

D=3.82\ m

Hence, The diameter of the bull-wheel is 3.82 m.

6 0
3 years ago
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