A vector has an x component of 25.0 units and a y

Mr.Z holds 200 kg above his head for 5 seconds. What is the work done on the weights?

Alenkasestr [34]

m = mass held by mr. Z above his head = 200 kg

g = acceleration due to gravity = 9.8 m/s²

F = force applied by mr. Z to hold the mass

Using equilibrium of force , force equation is given as

F = mg

F = (200) (9.8)

F = 1960 N

Since the mass is not moved,

d = displacement of the mass = 0 m

we know that , work done is given as

W = F d

inserting the values

W = (1960) (0)

W = 0 J

8 0

3 years ago

you spray your sister with water from a garden hose. the water is supplied to the hose at a rate of 0.649x10-3 m^3/s and the dia

Ivanshal [37]

Answer:

27.82 m/s

Explanation:

The radius of the hose is half of its diameter

$r = d/2 = 5.45\times10^{-3}/2 = 0.002725 m$

So its area must be

$A = \pi r^2 = \pi 0.002725^2 = 2.33\times10^{-5} m^2$

The speed of water coming out of the hose is its flow rate divided by the cross-section area of the hose

$v = \dot{V}/A =0.000649 / 2.33\times10^{-5} = 27.82 m/s$

7 0

2 years ago

Answer this question correctly for a pat on the back.

ss7ja [257]

At Korona100 I don’t think it’s 10, becz I remember having this question on my test/ quiz but I got it incorrect so yeah.

7 0

2 years ago

What is the magnitude (size) and direction of the cumulative force acting on the car shown in the picture above?

zaharov [31]

Answer:

B

Explanation:

.....50N-20N =30N ...

3 0

2 years ago

Read 2 more answers

A small rock is thrown straight up from the edge of the roof of an 8.00 m tall building with an initial speed vo . The speed of

Serjik [45]

Answer:

20.47 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{24^2-0^2}{2\times 9.81}\\\Rightarrow s=29.3577\ m$

Total height of the fall is 29.3577 m

Height the ball reached above the building is $29.3577-8=21.3577\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 21.3577=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{21.3577\times 2}{9.81}}\\\Rightarrow t=2.0866\ s$

Time taken to reach the point from where the ball was thrown is 2.0866 s

This will also be the time it takes the ball to reach the maximum height

$s=ut+\frac{1}{2}at^2\\\Rightarrow u=\dfrac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\dfrac{21.3577-\frac{1}{2}\times -9.81\times 2.0866^2}{2.0866}\\\Rightarrow u=20.47\ m/s$

The initial velocity with which the rock was thrown was 20.47 m/s

7 0

2 years ago