How much work is accplished when a force of 250 n pushes a box accross the floor for a distance of 50 meters

A 55.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300. What h

MariettaO [177]

Answer:

161.86 N

Explanation:

mass of box m= 55.0 kg

weight of the box, mg= 55×9.81

g here is acceleration due to gravity =9.81 m/sec^2

coefficient of friction between the box and the surface μ= 0.3

the friction force F_s= μmg= 0.3×55×9.81

=161.86 N

to move the ball horizontal force required is 161.86 N

8 0

3 years ago

List two of the number of strategies you have learned

jolli1 [7]

Answer:

what?

Explanation:

does that means also i learned Ratios and multiplication i keared them like 5 years ago?

7 0

2 years ago

1. A student practicing for a track meet ran 263 m in 30 sec. What was her average speed?

faltersainse [42]

Answer: By 47

Explanation: subtract 310 and 263

6 0

2 years ago

A boat takes 3.0 hours to travel 50 km down a river, then 5.4 hours to return. Determine the speed of the water in the river.

Nutka1998 [239]

Answer:

3.7 km/h

Explanation:

Let's call v the proper speed of the boat and v' the speed of the water in the river.

When the boat travels in the direction of the current, the speed of the boat is:

v + v'

And it covers 50 km in 3 h, so we can write

$v+v' = \frac{50 km}{3 h}=16.7 km/h$ (1)

When the boat travels in the opposite direction, the speed of the boat is

v - v'

And it covers 50 km in 5.4 h, so

$v-v'=\frac{50 km}{5.4 h}=9.3 km/h$ (2)

So we have a system of two equations: by solving them simultaneously, we find the value of v and v':

$v+v'=16.7 \\v-v'=9.3$

Subtracting the second equation from the first one we get:

$(v+v')-(v-v')=16.7-9.3\\2v'=7.4\\v'=3.7$

So, the speed of the water is 3.7 km/h.

5 0

3 years ago

During the spin cycle of a washing machine, the clothes stick to the outer wall of the barrel as it spins at a rate as high as 1

Darya [45]

To answer the two questions, we need to know two important equations involving centripetal movement:

v = ωr (ω represents angular velocity <u>in radians</u>)

a = $\frac{v^{2}}{r}$

Let's apply the first equation to question a:

v = ωr

v = ((1800*2π) / 60) * 0.26

Wait. 2π? 0.26? 60? Let's break down why these numbers are written differently. In order to use the equation v = ωr, it is important that the units of ω is in radians. Since one revolution is equivalent to 2π radians, we can easily do the conversion from revolutions to radians by multiplying it by 2π. As for 0.26, note that the question asks for the units to be m/s. Since we need meters, we simply convert 26 cm, our radius, into meters. The revolutions is also given in revs/min, and we need to convert it into revs/sec so that we can get our final units correct. As a result, we divide the rate by 60 to convert minutes into seconds.

Back to the equation:

v = ((1800*2π)/60) * 0.26

v = (1800*2(3.14)/60) * 0.26

v = (11304/60) * 0.26

v = 188.4 * 0.26

v = 48.984

v = 49 (m/s)

Now that we know the linear velocity, we can find the centripetal acceleration:

a = $\frac{v^{2}}{r}$

a = $\frac{49^{2}}{0.26}$

a = 9234.6 (m/ $s^{2}$ )

Wow! That's fast!

<u>We now have our answers for a and b:</u>

a. 49 (m/s)

b. 9.2 * $10^{3}$ (m/ $s^{2}$ )

If you have any questions on how I got to these answers, just ask!

- breezyツ

5 0

2 years ago