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3 years ago

If jack was traveling at 100 miles in 2 hours what was his velocity in miles per second

Physics

2 answers:

arsen [322]3 years ago

6 0

1 hour is 3600 seconds. In 2 hours, it would be 7200 seconds. Divide the amount of miles by seconds. 100/7200=.01388..

miskamm [114]3 years ago

4 0

We don't know anything about the direction Jack was traveling, so we can't determine his velocity. The best we can do is work with his speed.

Speed = (distance covered) / (time to cover the distance)

Speed = (100 miles) / (2 hours)

Speed = 50 miles/hour

The question has specified a different unit, so we'll need to convert the unit of our answer.

Let's multiply the answer by (1 hour / 3600 seconds). The numerator and denominator of this fraction are both the same quantity, so the value of the fraction is ' 1 ', and multiplying the answer by it won't change the answer in any way except the unit.

Speed = (50 mi / hr) x (1 hr / 3600 sec)

Speed = (50mi · 1hr) / (hr · 3600sec)

Speed = 50 mi / 3600 sec

<em>Speed = 0.01389 mi/sec</em>

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3 years ago

Playing near a road construction site, a child falls over a barrier and down onto a dirt slope that is angled downward at 33° to

Debora [2.8K]

Answer:

$\mu_{k} \approx 0.719$

Explanation:

The equations of equilibrium for the child are: (x' in the direction parallel to slope, y' in the direction perpendicular to slope)

$\Sigma F_{x'} = m\cdot g \cdot \sin \theta - \mu_{k}\cdot N = m\cdot a\\\Sigma F_{y'} = N - m\cdot g\cdot \cos \theta = 0$

After some algebraic manipulation, an expression for the coefficient of kinetic friction is obtained:

$m\cdot g\cdot \sin \theta - \mu_{k}\cdot m \cdot g \cos \theta = m \cdot a$

$g \cdot (\sin \theta - \mu_{k}\cdot \cos \theta) = a$

$\mu_{k}\cdot \cos \theta = \sin \theta - \frac{a}{g}$

$\mu_{k} = \frac{1}{\cos \theta}\cdot (\sin \theta - \frac{a}{g} )$

$\mu_{k} = \frac{1}{\cos 33^{\textdegree}}\cdot \left(\sin 33^{\textdegree}-\frac{(-0.57\,\frac{m}{s^{2}}) }{9.807\,\frac{m}{s^{2}} } \right)$

$\mu_{k} \approx 0.719$

5 0

4 years ago

(3) What is the weight of a 50-kg astronaut (a) on Earth (b) On the Moon ,(g=1.7m/s2), (c) on Mars (g=3.7m/s2) (d)in outer space

artcher [175]

Answer:

a) On Earth

490N

b) On the Moon

85N

c) On Mars

185N

d)in outer space traveling with constant velocity.

Explanation:

The weight is defined as:

$W = mg$ (1)

Where m is the mass and g is the gravity

a) On Earth $g = 9.8m/s^{2}$

Then, equation 1 can be used:

$W = (50Kg)(9.8m/s^{2})$

$W = 490Kg.m/s^{2}$

but 1N = Kg.m/s^{2}

$W = 490N$

Hence, the weight of the astronaut on Earth is $490N$

b) On the Moon $g = 1.7m/s^{2}$

$W = (50Kg)(1.7m/s^{2})$

$W = 85N$

Hence, the weight of the astronaut on the Moon is $85N$

c) On Mars $g = 3.7m/s^{2}$

$W = (50Kg)(3.7m/s^{2})$

$W = 185N$

Hence, the weight of the astronaut on Mars is $185N$

(d) in outer space traveling with constant velocity.

Tanking into consideration that the astronaut is traveling in outer space at a constant velocity, it can be concluded that the acceleration will be zero.

Remember that the acceleration is defined as:

$a = \frac{v_{f} - v_{i}}{t}$

Since the acceleration is the variation of the velocity in a unit of time.

Therefore, from equation 1 is gotten.

$W = (50kg)(0)$

Remember that g is the acceleration that a body experience as a consequence of the gravitational field.

$W = 0$

5 0

3 years ago

In order to decrease the friction an object experiences, you can either change

Kipish [7]

Answer:

Reduce its weight

Explanation:

For an object on a flat surface, the maximum force of friction experienced by the object is given by:

$F = \mu W$

where

$\mu$ is the coefficient of friction

W is the weight of the object

As we can see from the formula, there are two ways of reducing the friction:

- Decreasing $\mu$ --> this coefficient depends on the material and on the roughness of the surface, so this can be changed by changing the properties of the surface

- Decreasing W, the weight --> this can be done by reducing the mass of the object

8 0

3 years ago

If jack was traveling at 100 miles in 2 hours what was his velocity in miles per second​

If jack was traveling at 100 miles in 2 hours what was his velocity in miles per second