How much work is done if I use 5N of force to move an object 2 meters

When you slosh the water back and forth in a tub at just the right frequency, the water alternately rises and falls at each end,

Natali5045456 [20]

Slot in wwatwe to be determined for use

5 0

3 years ago

PLEASE HELP ME !!!

Bogdan [553]

Because the temperature causes it to go the physical change like if water is froze the temperate just caused it to go through another state of matter another example as if the ice got melted you would need high temperature to cause this change physically. Let me know if I help or if their is anything I can change for you to understand better

4 0

3 years ago

16. How much force would it take to accelerate a 75kg object by 5 m/szą

patriot [66]

Answer:

The force experienced is 375 N.

Explanation:

Here,

Mass of the object( M)= 75 kg

Acceleration of the body(a) =5 m/s²

Force required to accelerate the body( F) =?

Then, we know ,

From newton's second law of motion,

$F=M*a$

= $75* 5$

=375 kgm/s²

= 375 Newton

So, The force experienced by the 75 kg body that undergoes an acceleration of 5 m/s² is 375 N.

7 0

3 years ago

While running, a person dissipates about 0.60 J of mechanical energy per step per kilogram of body mass. If a 68 kg person devel

Tresset [83]

Answer:

Speed = 2.25 m/s

Explanation:

(Assume a running step is 1.5 m long)

Given the following data;

Energy = 0.6J

Power = 61 Watts

Mass = 68 kg

To find how fast the person running;

First of all, we would determine the total mechanical energy being dissipated by the person.

Total energy = 0.6 * 68

Total energy = 40.8 Joules

Next, we find the time;

Energy = power * time

40.8 = 61 * time

Time = 61/40.8

Time = 1.5 seconds

Finally, to find the speed;

Speed = distance/time

Speed = number of steps * time

Speed = 1.5 * 1.5

Speed = 2.25 m/s

4 0

3 years ago

Derek and Diane race each other on bicycles. Derek starts 4 seconds before Diane. He goes at a steady speed of 6 m/s. Diane star

ser-zykov [4K]

a. $x(t) = vt' = v(t+4) = 6(t+4) [m]$

Let's call:

$t$ the time calculated from the moment Diane starts her motion and $t'$ the time calculated from the moment Derek starts his motion. Derek starts his motion 4 seconds before Diane: this means that has an "advantage" of 4 seconds, so we can write

$t'=t+4$

Then:

$v=6 m/s$ is the uniform speed of Derek

Derek is moving in a uniform motion, so Derek's position will be given by (assuming that the starting position is zero):

$x(t) = vt' = v(t+4) = 6(t+4) [m]$

We see that this is correct: in fact, when t=0 (instant when Diane starts her motion), Derek has already travelled for

$x(t=0)=v(0+4)=6 (4)=24 m$

b. $x'(t)= \frac{1}{2}at^2 =\frac{1}{2}(2)t^2=t^2 [m]$

t is the time calculated from the moment Diane starts her motion. Diane is moving by accelerated motion, with constant acceleration $a=2 m/s^2$ and initial velocity $v_0=0$ , so its position at time t is given by the law of uniform accelerated motion: