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katrin2010 [14]
3 years ago
13

A bungee jumper who is about to jump has her energy stored entirely as

Physics
1 answer:
Morgarella [4.7K]3 years ago
3 0

Answer:

it will get slower and eventually she will stop jumping because there isnt enough force on the gravity causing her to go up and down

Explanation:

!!

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A pilot flies in a straight path for 1 h 30 min. She then makes a course correction, heading 10 degrees to the right of her orig
r-ruslan [8.4K]

Answer:

2406 miles

Explanation:

Let A be the starting position, B the junction position and C the final position after flying the 3.5 hrs. Also, let b be the distance from the starting point:

\angle ABC =(180-10) \textdegree=170\textdegree

#Distance traveled in 1.5hrs is;

c=690x1.5\\=1035mi

#Distance traveled in next two hrs:

a=690\times 2\\=1380mi

#Now using the Cosine Rule:

b^2=a^2+c^2-2ab\cos B\\\\=1380^2+1035^2-2(1380)(1035)cos170\textdegree\\\\b^2=5788.83\\\\b\approx 2406.00 \ mi

Hence, the pilot is 2406 miles from her starting position.

4 0
3 years ago
An elevator car has a mass of 750 kg, and its three passengers have a combined mass of 135 kg. If the elevator and its passenger
joja [24]

Answer:

The change in gravitational potential energy is -1.80x10⁵ J.

Explanation:

The change in gravitational potential energy is given by:

\Delta E_{p} = E_{p_{f}} - E_{p_{i}}

\Delta E_{p} = mgh_{f} - mgh_{i}

Where:

"i" is for final and "f" for final

m: is the mass

g: is the gravity = 9.81 m/s²

h: is the height

For the car and the passengers we have:

\Delta E_{p} = m_{T}g(h_{f} - h_{i}) = (750 kg + 135 kg)9.81 m/s^{2}(0 - 20.7 m) = -1.80 \cdot 10^{5} J      

The minus sign is because when the elevator car and the passengers are up they have a bigger gravitational potential energy than when they are in the ground.

Therefore, the change in gravitational potential energy is -1.80x10⁵ J.

I hope it helps you!                                                

3 0
3 years ago
There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co
mars1129 [50]

Answer:

0.208 N

Explanation:

We are given that

q_1=q_2=2.06\mu C=2.06\times 10^{-6} C

q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C

Distance,d=0.41 m

The magnitude of the net electrostatic force experienced by any charge at point 4

Net force,F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2

F_1=F_3=F

F_{net}=\sqrt{F^2+F^2+0}-F_2

F_{net}=\sqrt 2F-F_2

F=\frac{kq^2}{d^2}

F_2=\frac{Kq^2}{2d^2}

F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})

Where k=9\times 10^9

F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})

F_{net}=0.208 N

3 0
3 years ago
At what altitude above the earth's surface would the acceleration due to gravity be 4.9 m/s^2? Assume the mean radius of the ear
Mandarinka [93]
We apply the gravity calculation expressed in the formula: g=GM/r2 
where G is the gravitational constant, m is the mass and r is the radius
r=√GM/g
(1)       Radius = √6.674e-11*5.972e24/8             = 7058 kms    Earth radius or surface of earth from center of earth= 6400 kmsSo r= 658 kms from surface of earth.
Gravity 8m/s2 will be at 658 kms from surface of earth.
(2) half gravity= 9.8/2= 4.9 m/s2     Radius=√6.674e-11*5.972e24/4.9                 = 9019 kms        Half Gravity will exist at 9019-6400= 2619 kms from surface of earth.
4 0
3 years ago
A car with a mass of 1100kg is moving in a circular curve at a uniform velocity of 15m/s
Dmitrij [34]
Centripetal force = (mv^2)/r
so r = (mv^2)/ force = 246500 / 1100 = 224 m
7 0
3 years ago
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