All categories

3 years ago

What is the force acting on the object?(g=10m/s^2)

Physics

1 answer:

Vera_Pavlovna [14]3 years ago

3 0

Answer:w=mxg

2x10 =20 N

Explanation:force acting downwards is mg mass into gravitional feild

You might be interested in

A steel tank of weight 600 lb is to be accelerated straight upward at a rate of 1.5 ft/sec2. Knowing the magnitude of the force

VikaD [51]

Answer:

a) the values of the angle α is 45.5°

b) the required magnitude of the vertical force, F is 41 lb

Explanation:

Applying the free equilibrium equation along x-direction

from the diagram

we say

∑Fₓ = 0

Pcosα - 425cos30° = 0

525cosα - 368.06 = 0

cosα = 368.06/525

cosα = 0.701

α = cos⁻¹ (0.701)

α = 45.5°

Also Applying the force equation of motion along y-direction

∑Fₓ = ma

Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)

525sin45.5° + F + 212.5 - 600 = 27.95

374.46 + F + 212.5 - 600 = 27.95

F - 13.04 = 27.95

F = 27.95 + 13.04

F = 40.99 ≈ 41 lb

8 0

3 years ago

What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL

lozanna [386]

Answer:

$\displaystyle \Delta x=1.74\ m$

Explanation:

Elastic Potential Energy

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

$\displaystyle PE = \frac{1}{2}k(\Delta x)^2$

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

$\displaystyle \Delta x=\swrt{\frac{2PE}{k}}$

Substituting:

$\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}$

Calculating:

$\displaystyle \Delta x=\sqrt{3.0175}$

$\boxed{\displaystyle \Delta x=1.74\ m}$

6 0

3 years ago

A student wants to start a small business in school. Write down six items that

Fiesta28 [93]

Answer:

packets of pen

packets of pencil

copies

books

bottles

mask

3 0

3 years ago

A ball is projected with an initial velocity of 40 meter per second and reached maximum height of 160 meters calculate tge angle

Andru [333]

There's a problem with the question as given. Even with a maximum projection angle of θ = 90°, the initial velocity is not large enough to get the ball up in the air 160 m. With angle 90°, the ball's height y at time t would be

y = (40 m/s) t - 1/2 g t ²

Set y = 160 m, and you'll find that there is no (real) solution for t, so the ball never attains the given maximum height.

From another perspective: recall that

v ² - v₀² = 2a ∆y

where

• v₀ = initial velocity

• v = final velocity

• a = acceleration

• ∆y = displacement

At its maximum height, the ball has zero vertical velocity, and ∆y = maximum height = 160 m. The ball is in free fall once it's launched, so a = -g.

So we have

0² - (40 m/s)² = -2g (160 m)

but this reduces to

(40 m/s)² = 2 (9.8 m/s²) (160 m)

1600 m²/s² ≠ 3136 m²/s²

7 0

3 years ago

A boy on a skateboard is pushed down the sidewalk by a friend. The skateboard rolls for a while but eventually comes to a stop.

Ket [755]

Friction is causing the skateboard to stop rolling.

4 0

3 years ago

Read 2 more answers

What is the force acting on the object?(g=10m/s^2)​

What is the force acting on the object?(g=10m/s^2)