You exert 20 newtons of force on the hammer to remove the nail. The hammer exerts 90

Why was Democritus considered a philosopher and not a scientist?

Alika [10]

Answer:

Democritus had no scientific instruments to extend the reach of his senses, so all of his experiments were just 'mind experiments', but because Democritus was a philosopher, he thought more into depth about why we humans are alive which led to the atomic theory.

Explanation:

5 0

3 years ago

What two kinds of crust are involved in a subduction zone

Anarel [89]

Answer:

Oceanic crust and continental crust

Explanation:

A subduction zone is normally between oceanic crust which is made of basalt and continental crust which is made of granite. Oceanic crust is denser than continental crust. So when oceanic crust collides with continental crusts, it subsducts underneath the continental crust since it is denser.

5 0

2 years ago

Please help I'm stuck on this question

Afina-wow [57]

Answer:

increase

decrease

Explanation:

using formula

Vt=mg/6πηr

so if m increases V increases

r is the denominator so if r increases V decreases

8 0

1 year ago

A force of 16 N to the west is applied to each object below. Which object will

Montano1993 [528]

41kg object that is moving east at 5 m s

7 0

2 years ago

Read 2 more answers

A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo

Rasek [7]

Answer:

$t=12.25\ seconds$

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

$x=v_ocos\theta t$

$\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}$

Where vo is the magnitude of the initial velocity, $\theta$ is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

$\displaystyle t=\frac{v_osin\theta}{g}$

There are two times where the value of y is $y_o$ when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making $y=y_o$