Would a pair of scissors with really long blades and a short handle be practical? Why or why not?

An 800 N box is pushed across a level floor for a distance of 5.0 m with a force of 400 N. How much work was done on this box.

AveGali [126]

Answer: 2000 J.

Explanation: Since work is force*displacement, we just have to multiply the force by the distance: w = f*d = 400 N*5.0 m = 2000 J.

4 0

3 years ago

Two rams run toward each other. One ram has a mass of 49 kg and runs west

olga nikolaevna [1]

Answer: (d)

Explanation:

Given

Mass of the first ram $m_1=49\ kg$

The velocity of this ram is $v_1=-7\ m/s$

Mass of the second ram $m_2=52\ kg$

The velocity of this ram $v_2=9\ m/s$

They combined after the collision

Conserving the momentum

$\Rightarrow m_1v_1+m_2v_2=(m_1+m_2)v\\\Rightarrow 49\times (-7)+52\times (9)=(52+49)v\\\Rightarrow v=\dfrac{125}{101}\ m/s \quad[\text{east}]$

Momentum after the collision will be

$\Rightarrow 101\times \dfrac{125}{101}=125\ kg-m/s\ \text{East}$

Therefore, option (d) is correct

4 0

3 years ago

What is the main difference between the Schrödinger model and the Bohr atomic model?

netineya [11]

Answer:

Schrödinger believed that electrons could only exist in orbits, but Bohr stated that electrons could be found anywhere in the atom

Explanation: I got it right on test

4 0

3 years ago

A tennis ball with a velocity of +10.0 m/s to the right is thrown perpendicularly at a wall. After striking the wall, the ball r

netineya [11]

Answer:

-1500 m/s2

Explanation:

So the ball velocity changes from 10m/s into the wall to -8m/s in a totally opposite direction within a time span of 0.012s. Then we can calculate the average acceleration of the ball as the change in velocity over a unit of time.

$a = \frac{\Delta v}{\Delta t} = \frac{-8 - 10}{0.012} = \frac{-18}{0.012} = -1500 m/s^2$

6 0

3 years ago

a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the

ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: $F=m*a$ with this we can get both accelerations; solving for acceleration $a=\frac{F}{m}$ . Now $a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2})$ and $a_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2})$ . Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, $x=\frac{1}{2}*a_{girl}*t^{2}$ and $15.0-x=\frac{1}{2}*a_{sled}*t^{2}$ , solving for the time we get: $t=\sqrt{\frac{2x}{a_{girl} } }$ and $t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } }$ with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get: $\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }$ . Finally we get: $\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} }$ and replacing the values we have got: $\frac{x}{0.14} =\frac{(15.0-x)}{0.73}$ so $5.33*x=15-x$ so x=2.37 (meters).

5 0

3 years ago