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zhannawk [14.2K]

3 years ago

12

-Starts to suck on ur neck giving you hickeys n runs my hand down your body biting my lip against your neck

1 answer:

Hoochie [10]3 years ago

6 0

Answer:

I would shout fore help if I was being raped or try to make him or her stop

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A researcher measured the absorbance and % transmittance of a blue dye at light wavelengths between 425 and 700 nm. the measurem

MA_775_DIABLO [31]

She should plot her <span>data where the y axis contains the absorbance values while the x-axis are the wavelengths. It should be that the dependent variable is always at the y-axis and the independent variable or the measurement that is manipulated is at the y-axis. Hope this helps.</span>

4 0

3 years ago

Read 2 more answers

Is the force of gravity that attracts my body to the Earth related to the force of gravity between the planets and the Sun

pochemuha

Certainly gravity is a force that exists between the Earth and the objects that are near it. As you stand upon the Earth, you experience this force.

8 0

3 years ago

A toy car is moving across a table with a velocity of 7.0 m/s and drives off the end. The table is 1.8 m tall. How long will the

KatRina [158]

Answer:

Option D - 0.2 s

Explanation:

We are given;

Initial velocity; u = 7 m/s

Height of table; h = 1.8m

Now,since we want to find the time the car spent in the air, we will simply use one of Newton's equation of motion.

Thus;

h = ut + ½gt²

Plugging in the relevant values, we have;

1.8 = 7t + ½(9.8)t²

4.9t² + 7t - 1.8 = 0

Using quadratic formula to find the roots of the equation gives us;

t = -1.65 or 0.22

We can't have negative t value, thus we will pick the positive one.

So, t = 0.22 s

This is approximately 0.2 s

7 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

4 years ago

You place an 8 kg ball on the top of your 2 cm^2 finger tip. Calculate the

Rashid [163]

Answer:

the pressure exerted by the object is 392,000 N/m²

Explanation:

Given;

mass of the object, m = 8 kg

area of your finger, A = 2 cm² = 2.0 x 10⁻⁴ m²

acceleration due to gravity, g = 9.8 m/s²

The pressure exerted by the object is calculated as;

$Pressure = \frac{F}{A} = \frac{mg}{A} = \frac{8 \times 9.8}{2\times 10^{-4}} = 392,000 \ N/m^2$

Therefore, the pressure exerted by the object is 392,000 N/m²

7 0

3 years ago