All categories

3 years ago

The threatened or actual use of physical force to get someone to have sex without consent is called

Physics

2 answers:

Nimfa-mama [501]3 years ago

5 0

Wouldn't it be r a p e?

kaheart [24]3 years ago

4 0

Its r a p e hope that helps btw u cant spell it without spaces

You might be interested in

How are solar energy, wind energy and hydroelectric energy related?

spin [16.1K]

Solar energy, wind energy, and hydroelectric energy are all renewable energy

5 0

4 years ago

Read 2 more answers

Every atom of the ____ carbon has 6 protons.

miss Akunina [59]

Hi, the answer for this question is element!

Hope this helped!

Plz mark brainliest!

~Nate

7 0

3 years ago

Smialarities between kinetic friction and static friction?

kiruha [24]

Kinetic and static friction are both resistive forces

6 0

3 years ago

Read 2 more answers

A cubical box, 5.00 cm on each side, is immersed in a fluid. The gauge pressure at the top surface of the box is 594 Pa and the

Zolol [24]

Answer:

The density of the fluid is 1100 kg/m³.

Explanation:

Given that,

Height = 5.00 cm

Pressure at top =594 Pa

Pressure at bottom = 1133 Pa

We need to calculate the change in pressure

Using formula of change in pressure

$\Delta P=P_{b}-P_{t}$

Where, $P_{b}$ = Pressure at bottom

$P_{t}$ = Pressure at top

put the value into the formula

$\Delta P=1133-594$

$\Delta P=539\ Pa$

Using formula of pressure for density

$\Delta P = \rho g h$

$\rho =\dfrac{\Delta P}{gh}$

Where, $\rho$ = density

P = pressure

h = height

Put the value in to the formula

$\rho =\dfrac{539}{5.00\times10^{-2}\times9.8}$

$\rho =1100\ kg/m^3$

Hence, The density of the fluid is 1100 kg/m³.

4 0

4 years ago

A 2.0 kg wooden block is slid along a concrete surface (μk = 0.21) with an initial speed of 15 m/s. How far will the block slide

exis [7]

Answer:

The distance is 54.6 m

Explanation:

Given that,

Mass = 2.0 kg

Frictional coefficient = 0.21

Initial velocity = 15 m/s

We need to calculate the acceleration

Using formula of frictional force

$F = \mu mg$

$F=0.21\times2.0\times9.8$

$F = 4.12\ N$

We need to calculate the acceleration

$F = ma$

$a = \dfrac{F}{m}$

$a =\dfrac{4.12}{2.0}$

$a=2.06\ m/s^2$

We need to calculate the initial velocity

Using equation of motion

$v^2=u^2-2as$

Put the value

$0=15^2-2\times2.06\times s$

$s = \dfrac{15^2}{2\times2.06}$

$s=54.6\ m$

Hence, The distance will be 54.6 m.

3 0

3 years ago