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QveST [7]
2 years ago
11

In the same year, the average finishing time for U.S. women was 4 hours and 41 minutes. What was the average female speed in mil

es per hour?
Physics
1 answer:
Nana76 [90]2 years ago
8 0

We have that for the Question "In the same year, the average finishing <em>time</em> for U.S. women was 4 <u>hours </u>and 41 <em>minutes</em>. What was the average female speed in miles per <em>hour</em>?" it can be said that

The average female speed in miles per hour

V_a=\frac{D}{4.683hours}

From the question we are told

In the same year, the <u>average</u> finishing <em>time</em> for U.S. women was 4 <u>hours </u>and 41 <em>minutes</em>. What was the average female speed in <u>miles</u> per <em>hour</em>?

<h3>Average speed </h3>

This is defined as the collective <u>mean</u> of different travel <em>intervals </em>

Generally the equation for Speed   is mathematically given as

V=\frac{Distance}{Time}

Therefore

For  4 <u>hours </u>and 41 <em>minutes=4.683hours</em>

This travels time was <em>obtained </em>over a  series of distance

<em>Hence</em>

<em>Distance D</em>

The average female <em>speed</em> in miles per hour

V_a=\frac{D}{4.683hours}

For more information on speed visit

brainly.com/question/7359669

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a uniform metre rule is pivoted at its centre and weights of 5 N and 12 N are hung at the 3 cm and 5 cm marks respectively. how
schepotkina [342]

In order to balance the stick on the pivot, the total "moments" must be equal on both sides.  A "moment" is (a weight) x (its distance from the center).

for the 5N weight: Moment = (5N) x (3 cm) = 15 N-cm

for the 12N weight: Moment = (12N) x (5 cm) = 60 N-cm

Sum of the moments trying to pull the stick down on that side = 75 N-cm

Whatever we hang on the other side has to provide a moment of 75 N-cm in the other direction.  We have a 25N weight. Where should we hang it ?  

(25N) x (distance from the pivot) = 75 N-cm

Distance from the pivot = (75 N-cm) / (25 N)

<em>Distance from the pivot =  3 cm </em>

8 0
3 years ago
A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turn
kumpel [21]

Answer:

a) The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

Explanation:

Statement is incomplete. The complete description is now described below:

<em>A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.250 m/s2 in the x direction. The acceleration lasts for 45.0 s, at which point the thruster turns off. </em>

<em>(a) What is the magnitude of the satellite's velocity when the thruster turns off</em>

<em>(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis. ° counterclockwise from the +x-axis</em>

Let be x and y-directions orthogonal to each other and the satellite is accelerated uniformly from rest in the +x direction and moves at constant velocity in the +y direction. The velocity vector of the satellite (\vec{v}_{S}), measured in meters per second, is:

\vec{v}_{S} = (v_{o,x}+a_{x}\cdot t)\,\hat{i}+v_{y}\,\hat{j}

Where:

v_{o,x} - Initial velocity in +x direction, measured in meters per second.

a_{x} - Acceleration in +x direction, measured in meter per square second.

t - Time, measured in seconds.

v_{y} - Velocity in +y direction, measured in meters per second.

If we know that v_{o,x} = 0\,\frac{m}{s}, a_{x} = 0.250\,\frac{m}{s^{2}}, t = 45\,s and v_{y} = 21.4\,\frac{m}{s}, the final velocity of the satellite is:

\vec{v}_{S} = \left[0\,\frac{m}{s}+\left(0.250\,\frac{m}{s^{2}} \right)\cdot (45\,s) \right]\,\hat{i}+\left(21.4\,\frac{m}{s} \right)\,\hat{j}

\vec{v_{S}} = 11.25\,\hat{i}+21.4\,\hat{j}\,\,\left[\frac{m}{s} \right]

a) The magnitud of the satellite's velocity can be found by the resource of the Pythagorean Theorem:

\|\vec {v}_{S}\| = \sqrt{\left(11.25\,\frac{m}{s} \right)^{2}+\left(21.4\,\frac{m}{s} \right)^{2}}

\|\vec{v}_{S}\| \approx 24.177\,\frac{m}{s}

The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is determined with the help of trigonometric functions:

\tan \alpha = \frac{v_{y}}{v_{x}} = \frac{21.4\,\frac{m}{s} }{11.25\,\frac{m}{s} }

\tan \alpha = 1.902

\alpha = \tan^{-1}1.902

\alpha \approx 62.266^{\circ}

The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

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3 years ago
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a spring gun initially compressed 2cm fires a 0.01kg dart straight up into the air. if the dart reaches a height it 5.5m determi
Vikki [24]

Answer:

2697.75N/m

Explanation:

Step one

This problem bothers on energy stored in a spring.

Step two

Given data

Compression x= 2cm

To meter = 2/100= 0.02m

Mass m= 0.01kg

Height h= 5.5m

K=?

Let us assume g= 9.81m/s²

Step three

According to the principle of conservation of energy

We know that the the energy stored in a spring is

E= 1/2kx²

1/2kx²= mgh

Making k subject of formula we have

kx²= 2mgh

k= 2mgh/x²

k= (2*0.01*9.81*5.5)/0.02²

k= 1.0791/0.0004

k= 2697.75N/m

Hence the spring constant k is 2697.75N/m

7 0
3 years ago
Please answer this question.
Len [333]

Answer:

Explanation:

need more work

6 0
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