A horizontal line on a displacement-time graph means the object is

What is the most effective way to clean the science desks? <br><br> Hypothesis:

lukranit [14]

The most effective way to clean a science desks is to use a soapy water solution to clean tables using a clean disposable paper towel. Second, after cleaning the table surface with soap or detergent and rinsing with water, disinfect tables by using a diluted bleach water solution

6 0

3 years ago

Consider three force vectors F~ 1 with magnitude 43 N and direction 38◦ , F~ 2 with magnitude 26 N and direction −140◦ , and F~

Rama09 [41]

Answer:

34.70 N

Explanation:

Given :

F~ 1 = 43 N in direction 38◦

F~ 2 = 26 N in direction −140◦

F~ 3 = 27 N in direction 110◦

Therefore,

F~x = 43 cos (38) + 26 cos (-140) + 27 cos (110)

= 43 (0.7) + 26 (-0.7) + 27 (-0.3)

= 3.8

F~y = 43 sin (38) + 26 sin (-140) + 27 sin (110)

= 43 (0.6) + 26 (-0.6) + 27 (0.9)

= 34.5

so, F~ = $\sqrt{3.8^2 + 34.5^2}$

= 34.70 N

6 0

3 years ago

A skydiver has reached terminal velocity—she now falls at a constant speed, so her acceleration is zero. Is there a net force on

Taya2010 [7]

Answer:

No, the net force on the skydiver is zero

Explanation:

According to Newton's Second Law, the net force on an object is equal to the product between the mass of the object and its acceleration:

$F=ma$

where

F is the net force

m is the mass of the object

a is the acceleration

In this problem, the acceleration of the skydiver is zero:

a = 0

This implies that also the net force on the skydiver is zero, according to the previous equation:

F = 0

So, the net force on the skydiver is zero. This occurs because the air resistance, which points upward, exactly balances the force of gravity on the skydiver, acting downwards.

6 0

4 years ago

A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p

kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

$y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2$

$V(t) = V_0 + a * t$ ,

Where $y_0$ its our initial height, $V_0$ our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

$y(t) = 38 \frac{m}{s} * t - \frac{1}{2} g t^2$