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frosja888 [35]
3 years ago
12

A roller coaster car starts at a height of 90 M at Point a and is moving at a speed of 20 m per second on a later Hill at Point

C assuming friction does not affect the motion of the car what is its height at Point C
Physics
2 answers:
ankoles [38]3 years ago
7 0
Hi there!

The answer you are looking for.....



20.4 m!





Hope this helps you!

Have a great day!
valentina_108 [34]3 years ago
3 0
The answer is 20.4m in height. You can use the kinematics equation Vf^2=V0^2+2ad  Therefore we have solving for d or height, we get d=Vf^2/2g
(20m/s)^2/(2*9.8m/s^2)=20.4m
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Mrs. Smith can walk 1.4 m/s. If it takes her 8.5 seconds to get to the teacher lounge, how far is the teacher lounge from her ro
AlexFokin [52]

Answer:

6 meters away

Explanation:

6*1.4= 8.4 which is pretty close

5 0
3 years ago
A bird flies 3.6 km due west and then 1.8 km due north. Another bird flies 1.8 km due west and 3.6 km due north. What is the ang
kondor19780726 [428]
Define unit vectors as follows:
\hat{i} is in the eastern direction.
\hat{j} is in the northern direction.

The position of the first bird is
\vec{a} = -3.6 \, \hat{i} + 1.8 \, \hat{j}

The position of the second bird is
\vec{b} = - 1.8 \, \hat{i} + 3.6 \, \hat{j}

Let θ = the angle between the net displacement vector for the two birds.
By definition,
\vec{a} . \vec{b} = |a| |b| cos\theta \\\\ \theta = cos^{-1} ( \frac{\vec{a}.\vec{b}}{|a||b|} )

\vec{a}.\vec{b} = (-3.6)(-1.8)+(3.6)(1.8) = 12.96
|a| =  \sqrt{3.24+12.96} =4.025
Similarly,
|b| = 4.025

Therefore
\theta = cos^{-1}  \frac{12.96}{4.025^{2}} =36.9^{o}

Answer:  36.9°
5 0
3 years ago
Your physics textbook is sliding to the right across the table draw the vectors starting at the black dot. the location and orie
liberstina [14]
You would take a black dot at the top right of the y axis and dray it to the the far right of your x axis. hope this helps have a nice day and God bless
6 0
3 years ago
PLS ANSWER FAST WILL GIVE BRAINLY TIMED TEST
solong [7]
A=F/m
a=(3000000)/(20000)
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4 0
2 years ago
plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 85.0 nC . The plates are in vacuu
tensa zangetsu [6.8K]

Answer:

12500 V

Explanation:

The electric field in the gap of a parallel-plate capacitor is uniform, so the following relationship between electric field strength, potential difference and distance can be used:

\Delta V = E d

where

\Delta V is the potential difference between the plates

E is the electric field strength

d is the distance between the plates

For the capacitor in this problem, we have

E=5.00\cdot 10^6 V/m

d = 2.50 mm = 2.50\cdot 10^{-3} m

Substituting, we find

\Delta V = (5.00\cdot 10^6)(2.50\cdot 10^{-3})=12500 V

3 0
3 years ago
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