A constant 20 N force is applied to a 7 kg box to push it along the ground. How

lys-0071 [83]

It's a virtual force ... one that seems to be there but isn't really there.

When something tries to fly or flow straight, through a rotating neighborhood, it ends up flying or flowing in a curved path, AS IF there were a force acting on it to make it curve. THAT apparent force is the Coriolis force.

It's what makes air, flowing away from high pressure or into low pressure, form the big rotating pressure systems on the rotating Earth.

7 0

2 years ago

50 POINTS AND BRAINLIST!!!PLEASE HELP FOR BRAINLIST! Give answer and explanation/ work

UkoKoshka [18]

20 joule is your answer

Answer:

here

mass m =100kg

distance d=50m

acceleration due to gravity a =10m/s²

work =force×displacement

= ma/d=100×10/50=20joule

4 0

2 years ago

Read 2 more answers

A Ping-Pong ball has a diameter of 2.55 cm and average density of 0.116 g/cm3 . What force would be required to hold it complete

Mrac [35]

Answer:

F = 0.075 N

Explanation:

given,

diameter of ping pong ball = 2.55 cm

average density of the ball = 0.116 g/cm³

Force require to hold the ball completely submerged= ?

Volume of the ball

$V = \dfrac{4}{3} \pi r^3$

$V = \dfrac{4}{3} \pi (\dfrac{2.55}{2})^3$

V = 8.63 cm³

density of water = 1 g/cm³

force required to keep the ball submerged is equal to

$F = \rho_w V_w g - \rho_b V g$

$F = 1\times 8.63\times g - 0.116 \times 8.63 \times g$

F = 7.63 x 9.8 x 10⁻³

F = 0.075 N

Force required to keep the ball inside the water is equal to 0.075 N

7 0

3 years ago

A paint brush is dropped from the top of a tall ladder and free falls to the ground. What changes, if any, would be observed of

muminat

Answer:

Explanation:

As the paint brush is dropped from the top of a tall ladder it started free falling

i.e. velocity of brush increases as it falls down due to the acceleration provided by gravity.

We know acceleration due to gravity is the attraction experienced by the object due to earth attraction

Value of acceleration due to gravity is constant i.e. $9.8\ m/s^2$

Therefore the correct choice is

velocity is increasing and acceleration is constant

4 0

3 years ago

a 0.145 kg pitched baesball moving horizontally at 35 m/s strikes a bat and is popped straight up to a height of 55.6 m before t

nadezda [96]

Answer:

The average force on the ball during the contact is 13,951.9 N.

Explanation:

Given;

mass of the ball, m = 0.145 kg

initial horizontal velocity, uₓ = 35 m/s

the contact time, t = 0.5 ms = 0.5 x 10⁻³ s.

vertical height reached by the ball, h = 55.6 m

The initial vertical velocity of the ball is calculated as;

v² = u² - 2gh

where;

v is the final vertical velocity at maximum height = 0

u is the initial vertical velocity

$0 = u_y^2 - 2(9.8 \times 55.6)\\\\0 = u_y^2 - 1089.76\\\\u_y^2 = 1089.76\\\\u_y = \sqrt{1089.76} \\\\u_y = 33.01 \ m/s$

The resultant velocity is calculated as;

$u = \sqrt{u_x^2 + u_y^2} \\\\u = \sqrt{35^2 + 33.01^2} \\\\u = 48.11 \ m/s$

The acceleration of the ball is calculated as;

$a = \frac{u}{t} \\\\a = \frac{48.11}{0.5\times 10^{-3}} \\\\a = 96220 \ m/s^2\\\\$

The average force on the ball during the contact is calculated as;

F = ma

F = (0.145)(96220)

F = 13,951.9 N

Therefore, the average force on the ball during the contact is 13,951.9 N.

5 0

2 years ago