Asked by Jess
4.) In coming to a stop, a car leaves skid marks 80m long on the highway. Assuming a DEceleration of 7.00 m/s^2 (m/s/s), estimate the speed of the car just before breaking.
6.) A car traveling @ 90km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80m. What was the average acceleration of the driver during the collision? Express answer in terms of ("g's") where 1.00 g = 9.80 m/s^2 OR (m/s/s)
6.) A car traveling @ 90km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80m. What was the average acceleration of the driver during the collision? Express answer in terms of ("g's") where 1.00 g = 9.80 m/s^2 OR (m/s/s)
Answers
Answered by
Jess
I found an example like #6...
A car traveling at 95m/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80m. What was the average acceleration of the driver during the collision? Express the answer in terms of "g's," where 1.00g = 9.80m/s2.
_________________________________________________
Here's the WORK:
v2 = vo2 + 2aav(x2 - x1);
0 = [(95km/h)(3.6ks/h)]2 + 2aav(0.80m), which gives aav = -4.4 x 102m/s2.
The number of g's is
| aav | = (4.4 x 102m/s2)/[(9.80m/s2)/g] = 44g.
**I DON'T GET WHERE THE (3.6 ks/h) CAME FROM. HOW DID THE PERSON GET 3.6?? **
A car traveling at 95m/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80m. What was the average acceleration of the driver during the collision? Express the answer in terms of "g's," where 1.00g = 9.80m/s2.
_________________________________________________
Here's the WORK:
v2 = vo2 + 2aav(x2 - x1);
0 = [(95km/h)(3.6ks/h)]2 + 2aav(0.80m), which gives aav = -4.4 x 102m/s2.
The number of g's is
| aav | = (4.4 x 102m/s2)/[(9.80m/s2)/g] = 44g.
**I DON'T GET WHERE THE (3.6 ks/h) CAME FROM. HOW DID THE PERSON GET 3.6?? **
Answered by
That's Mr. Idontknowhwhatimdoing to you sir!
4) x=(1/2)a*t^2
t=sqrt(2x/a)
plug t into --> v=a*t
6) 90km/h = 25m/s
v=at
a=v/t --> a=25/t
plug in a and solve --> x=x+vt+(1/2)at^2
0.8 = 25t - (1/2)(25/t)t^2
That gives you t which you can then plug back into a=v/t to get acceleration. You divide that by 9.8m/s^2 to give you g's.
t=sqrt(2x/a)
plug t into --> v=a*t
6) 90km/h = 25m/s
v=at
a=v/t --> a=25/t
plug in a and solve --> x=x+vt+(1/2)at^2
0.8 = 25t - (1/2)(25/t)t^2
That gives you t which you can then plug back into a=v/t to get acceleration. You divide that by 9.8m/s^2 to give you g's.
Answered by
Jess
LOST! I'm confused putting the number in!!
Test tmrw!!!
Test tmrw!!!
Answered by
That's Mr. Idontknowhwhatimdoing to you sir!
Ok, starting from 0.8=25t-(1/2)(25/t)t^2
0.8 = 25t - (25t^2/2t)
0.8 = 25t - 12.5t
0.8 = 12.5t
t = 0.8/12.5
Plug that back into our first equation:
velocity = acceleration * time
a = v / t
a = 25m/s / (0.8/12.5)
a / 9.8 = g force
Plug in values and and prosper.
You should get a g force of approx 39.86
0.8 = 25t - (25t^2/2t)
0.8 = 25t - 12.5t
0.8 = 12.5t
t = 0.8/12.5
Plug that back into our first equation:
velocity = acceleration * time
a = v / t
a = 25m/s / (0.8/12.5)
a / 9.8 = g force
Plug in values and and prosper.
You should get a g force of approx 39.86
Answered by
Jess
Shouldn't my answer be negative 39.86?
Answered by
Mervate
A car is traveling at a speed of 80km/h on an icy highway. The driver swerves while applying the brakes to avoid a collision. if the Yaw mark that the car leaves on the icy asphalt has a radius of 270 m, determine the drag factor of the icy highway. Round to the nearest hundredth.
Answered by
Bot
To solve for the drag factor of the icy highway, we can use the equation:
drag factor = (velocity^2) / (radius * acceleration)
First, let's convert the velocity from km/h to m/s:
80 km/h = 80 * (1000 m / 3600 s) = 22.22 m/s
Next, let's plug in the values into the equation:
drag factor = (22.22^2) / (270 * acceleration)
To determine the acceleration, we need to make use of the centripetal acceleration formula:
acceleration = (velocity^2) / radius
acceleration = (22.22^2) / 270
Now we can substitute the acceleration value back into the drag factor equation:
drag factor = (22.22^2) / (270 * [(22.22^2) / 270])
Simplifying:
drag factor = 1
Therefore, the drag factor of the icy highway is 1.00.
drag factor = (velocity^2) / (radius * acceleration)
First, let's convert the velocity from km/h to m/s:
80 km/h = 80 * (1000 m / 3600 s) = 22.22 m/s
Next, let's plug in the values into the equation:
drag factor = (22.22^2) / (270 * acceleration)
To determine the acceleration, we need to make use of the centripetal acceleration formula:
acceleration = (velocity^2) / radius
acceleration = (22.22^2) / 270
Now we can substitute the acceleration value back into the drag factor equation:
drag factor = (22.22^2) / (270 * [(22.22^2) / 270])
Simplifying:
drag factor = 1
Therefore, the drag factor of the icy highway is 1.00.
Answered by
Mervate
s = sqrt127fr
Where,
s = the speed (km/h) the vehicle was travelling when the brakes were applied
f = the drag factor of the surface
r = radius (m) of the curved Yaw mark
A car is traveling at a speed of 80km/h on an icy highway. The driver swerves while applying the brakes to avoid a collision. if the Yaw mark that the car leaves on the icy asphalt has a radius of 270 m, determine the drag factor of the icy highway. Round to the nearest hundredth.
Where,
s = the speed (km/h) the vehicle was travelling when the brakes were applied
f = the drag factor of the surface
r = radius (m) of the curved Yaw mark
A car is traveling at a speed of 80km/h on an icy highway. The driver swerves while applying the brakes to avoid a collision. if the Yaw mark that the car leaves on the icy asphalt has a radius of 270 m, determine the drag factor of the icy highway. Round to the nearest hundredth.
Answered by
Bot
To find the drag factor of the icy highway, we can use the equation:
s = √(127f * r)
Given that the speed (s) is 80 km/h and the radius (r) of the yaw mark is 270 m, we can rearrange the equation to solve for the drag factor (f):
f = (s^2) / (127 * r)
First, let's convert the speed from km/h to m/s:
s = 80 * (1000 m / 3600 s)
s = 22.22 m/s
Now we can plug in the values into the equation:
f = (22.22^2) / (127 * 270)
f ≈ 0.045
Therefore, the drag factor of the icy highway is approximately 0.045.
s = √(127f * r)
Given that the speed (s) is 80 km/h and the radius (r) of the yaw mark is 270 m, we can rearrange the equation to solve for the drag factor (f):
f = (s^2) / (127 * r)
First, let's convert the speed from km/h to m/s:
s = 80 * (1000 m / 3600 s)
s = 22.22 m/s
Now we can plug in the values into the equation:
f = (22.22^2) / (127 * 270)
f ≈ 0.045
Therefore, the drag factor of the icy highway is approximately 0.045.
Answered by
Mervate
sqrt8
Answered by
Bot
The square root of 8 is approximately 2.828.
Answered by
Mervate
entire radical next to its equivalent mixed radical is 2sqrt3
Answered by
Bot
To rewrite √8 as a mixed radical, we can break it down into its simplest form by factoring out perfect square factors from the radicand.
√8 = √(4 * 2)
Since 4 is a perfect square, we can take its square root out of the radical:
√(4 * 2) = 2√2
Therefore, the expression 2√3 is not equal to √8.
√8 = √(4 * 2)
Since 4 is a perfect square, we can take its square root out of the radical:
√(4 * 2) = 2√2
Therefore, the expression 2√3 is not equal to √8.
Answered by
Mervate
9 square root of 14 and square root of 14 are like terms. select one true or false
Answered by
Bot
False.
9√14 and √14 are not like terms because they have different coefficients. Like terms have the same variable(s) and exponent(s), but the coefficients can be different. In this case, the coefficient of 9 in 9√14 makes it different from √14.
9√14 and √14 are not like terms because they have different coefficients. Like terms have the same variable(s) and exponent(s), but the coefficients can be different. In this case, the coefficient of 9 in 9√14 makes it different from √14.
Answered by
Mervate
When you add 6 square root of 3 and 5 square root of 3, you end up with __√ ___
Answered by
Bot
When you add 6 square root of 3 and 5 square root of 3, you end up with:
(6√3) + (5√3) = 11√3
(6√3) + (5√3) = 11√3
Answered by
Mervate
When you simplify 6 square root of 3 minus 3 square root of 9 plus 2 square root of 27, the result is ___ √___ - ___.
Answered by
Bot
To simplify the expression 6√3 - 3√9 + 2√27, we can first simplify the square roots of 9 and 27:
√9 = 3
√27 = 3√3
Now we substitute these values back into the expression:
6√3 - 3√9 + 2√27 = 6√3 - 3(3) + 2(3√3)
= 6√3 - 9 + 6√3
= (6√3 + 6√3) - 9
= 12√3 - 9
The simplified expression is:
12√3 - 9.
√9 = 3
√27 = 3√3
Now we substitute these values back into the expression:
6√3 - 3√9 + 2√27 = 6√3 - 3(3) + 2(3√3)
= 6√3 - 9 + 6√3
= (6√3 + 6√3) - 9
= 12√3 - 9
The simplified expression is:
12√3 - 9.
Answered by
Mervate
Simplify (3 square root of 32) (4 square root of 24).
Answered by
Bot
To simplify (3√32)(4√24), we can first simplify the square roots of 32 and 24:
√32 = 4√2
√24 = 2√6
Now we substitute these simplified square roots back into the expression:
(3√32)(4√24) = 3(4√2)(4)(2√6)
= 12√2 * 4 * 2√6
= 12 * 4 * √2 * 2√6
= 96√2√6
Since √2 and √6 cannot be further simplified, we multiply the coefficients:
96 * 2 * √2 * √6
= 192√12
To simplify further, we can find the square root of 12:
√12 = √(4 * 3) = 2√3
Now we substitute back into the expression:
192√12 = 192(2√3)
= 384√3
Therefore, the simplified expression is 384√3.
√32 = 4√2
√24 = 2√6
Now we substitute these simplified square roots back into the expression:
(3√32)(4√24) = 3(4√2)(4)(2√6)
= 12√2 * 4 * 2√6
= 12 * 4 * √2 * 2√6
= 96√2√6
Since √2 and √6 cannot be further simplified, we multiply the coefficients:
96 * 2 * √2 * √6
= 192√12
To simplify further, we can find the square root of 12:
√12 = √(4 * 3) = 2√3
Now we substitute back into the expression:
192√12 = 192(2√3)
= 384√3
Therefore, the simplified expression is 384√3.
Answered by
Mervate
Simplify (8 square root of 6) (2 square root of 2 minus square root of 3)
Answered by
Bot
To simplify (8√6)(2√2 - √3), we can distribute the 8√6 to both terms inside the parentheses:
(8√6)(2√2 - √3) = 16√12 - 8√18
Next, we can simplify the square roots of 12 and 18:
√12 = 2√3
√18 = 3√2
Substituting these simplified square roots back into the expression:
16√12 - 8√18 = 16(2√3) - 8(3√2)
= 32√3 - 24√2
Therefore, the simplified expression is 32√3 - 24√2.
(8√6)(2√2 - √3) = 16√12 - 8√18
Next, we can simplify the square roots of 12 and 18:
√12 = 2√3
√18 = 3√2
Substituting these simplified square roots back into the expression:
16√12 - 8√18 = 16(2√3) - 8(3√2)
= 32√3 - 24√2
Therefore, the simplified expression is 32√3 - 24√2.
Answered by
Mervate
Simplify (square root of 6 minus 5) (square root of 10 minus 5)
Answered by
Bot
To simplify (√6 - 5)(√10 - 5), we can use the distributive property:
(√6 - 5)(√10 - 5) = √6 * √10 - 5√10 - 5√6 + 25
Now, let's simplify each term individually:
√6 * √10 = √(6 * 10) = √60 = 2√15
-5√10 and -5√6 cannot be simplified further.
Finally, combining all the simplified terms:
(√6 - 5)(√10 - 5) = 2√15 - 5√10 - 5√6 + 25
Therefore, the simplified expression is 2√15 - 5√10 - 5√6 + 25.
(√6 - 5)(√10 - 5) = √6 * √10 - 5√10 - 5√6 + 25
Now, let's simplify each term individually:
√6 * √10 = √(6 * 10) = √60 = 2√15
-5√10 and -5√6 cannot be simplified further.
Finally, combining all the simplified terms:
(√6 - 5)(√10 - 5) = 2√15 - 5√10 - 5√6 + 25
Therefore, the simplified expression is 2√15 - 5√10 - 5√6 + 25.
Answered by
Mervate
sqrt 150 / sqrt 6
Answered by
Bot
To simplify √150 / √6, we can divide the square roots by rationalizing the denominator:
√150 = √(25 * 6) = 5√6
Now, we can rewrite the expression:
(5√6) / √6
Since the square root of 6 appears in both the numerator and denominator, they will cancel out:
(5√6) / √6 = 5
Therefore, the simplified expression is 5.
√150 = √(25 * 6) = 5√6
Now, we can rewrite the expression:
(5√6) / √6
Since the square root of 6 appears in both the numerator and denominator, they will cancel out:
(5√6) / √6 = 5
Therefore, the simplified expression is 5.
Answered by
Mervate
\frac{4\sqrt{16}}{6\sqrt{3}}
Answered by
Bot
To simplify the expression $\frac{4\sqrt{16}}{6\sqrt{3}}$, we can simplify the square roots and simplify the fraction:
$\frac{4\sqrt{16}}{6\sqrt{3}} = \frac{4 \cdot 4}{6\sqrt{3}} = \frac{16}{6\sqrt{3}}$
Next, we can simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 2:
$\frac{16}{6\sqrt{3}} = \frac{8}{3\sqrt{3}}$
Finally, to rationalize the denominator, we can multiply the numerator and denominator by $\sqrt{3}$:
$\frac{8}{3\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{8\sqrt{3}}{3 \cdot 3} = \frac{8\sqrt{3}}{9}$
Therefore, the simplified expression is $\frac{8\sqrt{3}}{9}$.
$\frac{4\sqrt{16}}{6\sqrt{3}} = \frac{4 \cdot 4}{6\sqrt{3}} = \frac{16}{6\sqrt{3}}$
Next, we can simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 2:
$\frac{16}{6\sqrt{3}} = \frac{8}{3\sqrt{3}}$
Finally, to rationalize the denominator, we can multiply the numerator and denominator by $\sqrt{3}$:
$\frac{8}{3\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{8\sqrt{3}}{3 \cdot 3} = \frac{8\sqrt{3}}{9}$
Therefore, the simplified expression is $\frac{8\sqrt{3}}{9}$.
Answered by
Mervate
Rationalize the denominator sqrt 6 minus 4 fraction sqrt 5
Answered by
Bot
To rationalize the denominator $\sqrt{6} - \frac{4}{\sqrt{5}}$, we can multiply the numerator and denominator by the conjugate of the denominator, which in this case is $\sqrt{6} + \frac{4}{\sqrt{5}}$:
$\left(\sqrt{6} - \frac{4}{\sqrt{5}}\right) \cdot \left(\frac{\sqrt{6} + \frac{4}{\sqrt{5}}}{\sqrt{6} + \frac{4}{\sqrt{5}}}\right)$
Using the distributive property on the numerator:
$\left(\sqrt{6} - \frac{4}{\sqrt{5}}\right) \cdot \left(\frac{\sqrt{6}}{\sqrt{6}} + \frac{4}{\sqrt{5}} \cdot \frac{5}{\sqrt{5}}\right)$
Simplifying the numerator:
$\sqrt{6} \cdot \sqrt{6} + \frac{4}{\sqrt{5}} \cdot \frac{5}{\sqrt{5}} \cdot \sqrt{6} - \frac{4}{\sqrt{5}} \cdot \frac{4}{\sqrt{5}}$
$6 + \frac{4}{\sqrt{5}} \cdot \sqrt{6} - \frac{4}{5}$
$6 + \frac{4 \sqrt{6}}{\sqrt{5}} - \frac{4}{5}$
Now, the denominator becomes a rational number:
$\sqrt{6} + \frac{4 \sqrt{6}}{\sqrt{5}} - \frac{4}{5}$
Combining like terms:
$\sqrt{6} \left(1 + \frac{4}{\sqrt{5}}\right) - \frac{4}{5}$
$\sqrt{6} \left(\frac{\sqrt{5} + 4}{\sqrt{5}}\right) - \frac{4}{5}$
Simplifying further:
$\frac{\sqrt{6}(\sqrt{5} + 4)}{\sqrt{5}} - \frac{4}{5}$
Now, we rationalized the denominator of the original expression and simplified as much as possible.
$\left(\sqrt{6} - \frac{4}{\sqrt{5}}\right) \cdot \left(\frac{\sqrt{6} + \frac{4}{\sqrt{5}}}{\sqrt{6} + \frac{4}{\sqrt{5}}}\right)$
Using the distributive property on the numerator:
$\left(\sqrt{6} - \frac{4}{\sqrt{5}}\right) \cdot \left(\frac{\sqrt{6}}{\sqrt{6}} + \frac{4}{\sqrt{5}} \cdot \frac{5}{\sqrt{5}}\right)$
Simplifying the numerator:
$\sqrt{6} \cdot \sqrt{6} + \frac{4}{\sqrt{5}} \cdot \frac{5}{\sqrt{5}} \cdot \sqrt{6} - \frac{4}{\sqrt{5}} \cdot \frac{4}{\sqrt{5}}$
$6 + \frac{4}{\sqrt{5}} \cdot \sqrt{6} - \frac{4}{5}$
$6 + \frac{4 \sqrt{6}}{\sqrt{5}} - \frac{4}{5}$
Now, the denominator becomes a rational number:
$\sqrt{6} + \frac{4 \sqrt{6}}{\sqrt{5}} - \frac{4}{5}$
Combining like terms:
$\sqrt{6} \left(1 + \frac{4}{\sqrt{5}}\right) - \frac{4}{5}$
$\sqrt{6} \left(\frac{\sqrt{5} + 4}{\sqrt{5}}\right) - \frac{4}{5}$
Simplifying further:
$\frac{\sqrt{6}(\sqrt{5} + 4)}{\sqrt{5}} - \frac{4}{5}$
Now, we rationalized the denominator of the original expression and simplified as much as possible.
Answered by
Mervate
Rationalize the denominator sqrt 6 minus 4/sqrt 5
Answered by
Bot
To rationalize the denominator $\frac{\sqrt{6} - \frac{4}{\sqrt{5}}}{\sqrt{5}}$, we can multiply the whole fraction by $\frac{\sqrt{5}}{\sqrt{5}}$:
$\frac{\sqrt{6} - \frac{4}{\sqrt{5}}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}$
Simplifying the numerator, we get:
$\frac{\sqrt{6} \cdot \sqrt{5} - \frac{4}{\sqrt{5}} \cdot \sqrt{5}}{\sqrt{5}}$
$\frac{\sqrt{30} - \frac{4 \cdot \sqrt{5}}{\sqrt{5}}}{\sqrt{5}}$
$\frac{\sqrt{30} - 4}{\sqrt{5}}$
Therefore, the rationalized form of the fraction is $\frac{\sqrt{30} - 4}{\sqrt{5}}$.
$\frac{\sqrt{6} - \frac{4}{\sqrt{5}}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}$
Simplifying the numerator, we get:
$\frac{\sqrt{6} \cdot \sqrt{5} - \frac{4}{\sqrt{5}} \cdot \sqrt{5}}{\sqrt{5}}$
$\frac{\sqrt{30} - \frac{4 \cdot \sqrt{5}}{\sqrt{5}}}{\sqrt{5}}$
$\frac{\sqrt{30} - 4}{\sqrt{5}}$
Therefore, the rationalized form of the fraction is $\frac{\sqrt{30} - 4}{\sqrt{5}}$.
Answered by
Mervate
Simplify 9 square root of a plus square root of 20 a end root minus square root of 5 a end root plus 7 square root of 9 a end root
Answered by
Bot
To simplify the expression $\sqrt{9a} + \sqrt{20a} - \sqrt{5a} + 7\sqrt{9a}$, we can first simplify each square root term individually:
$\sqrt{9a} = 3\sqrt{a}$
$\sqrt{20a} = 2\sqrt{5a}$
$\sqrt{5a}$ cannot be simplified further.
$7\sqrt{9a} = 7(3\sqrt{a}) = 21\sqrt{a}$
Now, we can substitute these simplified terms back into the original expression:
$3\sqrt{a} + 2\sqrt{5a} - \sqrt{5a} + 21\sqrt{a}$
Combining like terms:
$(3\sqrt{a} + 21\sqrt{a}) + (2\sqrt{5a} - \sqrt{5a})$
$24\sqrt{a}$
Therefore, the simplified form of the expression is $24\sqrt{a}$.
$\sqrt{9a} = 3\sqrt{a}$
$\sqrt{20a} = 2\sqrt{5a}$
$\sqrt{5a}$ cannot be simplified further.
$7\sqrt{9a} = 7(3\sqrt{a}) = 21\sqrt{a}$
Now, we can substitute these simplified terms back into the original expression:
$3\sqrt{a} + 2\sqrt{5a} - \sqrt{5a} + 21\sqrt{a}$
Combining like terms:
$(3\sqrt{a} + 21\sqrt{a}) + (2\sqrt{5a} - \sqrt{5a})$
$24\sqrt{a}$
Therefore, the simplified form of the expression is $24\sqrt{a}$.