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Eva8 [605]
3 years ago
6

A woman parked on a slight incline observes that the car beside her starts moving forward at approximately 0.30 m/s. When she lo

oks at the parking meter on her right, she realizes that it is she herself who is moving. How fast is the woman moving with respect to the ground? Assume that forward is in the positive direction
0.30 m/s
0.60 m/s
-0.30 m/s
-0.60 m/s
Physics
2 answers:
Burka [1]3 years ago
4 0
The lady would be moving at -0.30 m/s as she is moving against the positive direction that is established as forward.
Goshia [24]3 years ago
3 0

<u>Answer:</u> The correct answer is Option -0.30 m/s.

<u>Explanation:</u>

Let the speed in forward direction be positive and the speed in backward direction be negative.

We are given that a woman has parked her car on a slight incline surface and the car beside her was moving with a speed of 0.30m/s in forward direction.

But then she realizes that her car was actually moving in the backward direction and not the car beside her.

Since, the car beside her and the ground are stationary, the magnitude of the relative velocity would remain the same and only the direction changes.

So, her car is moving with the speed of 0.30 m/s in backward direction.

Hence, the correct answer is Option -0.30 m/s.

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A ball moving 2 ft/s rolls off a table (on earth) that is 32 inches high. How long will it take the ball to hit the floor answer
mart [117]

Answer:When the ball rolls off the edge of the table, it will continue moving forward at 2.0 m/s until it hits the floor.

Explanation:This is what I would say is the answer bc I had to do reasearch on a lot of this for my work this year so if its not im veery sorry

3 0
3 years ago
Which types of atoms usually<br> become negative ions?
nalin [4]
Non metal ions gain electrons to become negative ions
5 0
4 years ago
A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between
Oduvanchick [21]

Answer:

Net electric field, E_{net}=91406.24\ N/C

Explanation:

Given that,

Charge 1, q_1=7.5\ nC=7.5\times 10^{-9}\ C

Charge 2, q_2=-2.9\ nC=-2.9\times 10^{-9}\ C

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

E_1=\dfrac{kq_1}{r^2}

E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}

E_1=65917.96\ N/C

Electric field due to charge 2 is given by :

E_2=\dfrac{kq_2}{r^2}

E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}

E_2=25488.28\ N/C

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

E_{net}=E_1+E_2

E_{net}=65917.96+25488.28

E_{net}=91406.24\ N/C

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

3 0
3 years ago
1. A 12 kg barrel is pulled up by a rope. The barrel accelerates at 1.2 m/s2. Find the force exerted by the rope. Show all your
allochka39001 [22]
F-mg=ma
F-(12kg)(9.8m/s^2)=(12kg)(1.2m/s^2)
F-117.6N=14.4N
F=132 Newtons 
5 0
4 years ago
When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees
choli [55]

Answer:

a)  F = 1.26 10⁵ N, b)  F = 2.44 10³ N, c)   F_net = 1.82 10³ N  directed vertically upwards

Explanation:

For this exercise we must use the relationship between momentum and momentum

         I = Δp

         F t = p_f -p₀

a) It asks to find the force

as the man stops the final velocity is zero

         F = 0 - p₀ / t

the speed is directed downwards which is why it is negative, therefore the result is positive

         F = m v₀ / t

         F = 63.5 7.89 / 3.99 10⁻³

         F = 1.26 10⁵ N

b) in this case flex the knees giving a time of t = 0.205 s

          F = 63.5 7.89 / 0.205

          F = 2.44 10³ N

c) The net force is

         F_net = Sum F

         F_net = F - W

         F_net = F - mg

let's calculate

         F_net = 2.44 10³ - 63.5 9.8

         F_net = 1.82 10³ N

since it is positive it is directed vertically upwards

6 0
3 years ago
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