HELP URGENT!!!!!!!!!!!!!!!!!!!!!!!!

A light spring has a force constant of 70 N/m and is used to pull a 10 kg box on a horizontal frictionless surface. If the box h

Sergio039 [100]

Answer:

82.4 cm

Explanation:

∑F = ma

kx cos θ = ma

x = ma / (k cos θ)

x = (10 kg) (5 m/s²) / (70 N/m cos 30.0°)

x = 0.824 m

x = 82.4 cm

8 0

3 years ago

If a magnet is attracted to a metal object what can you say about the pole of the magnet and the pole of a object ?

Sauron [17]

Space surrounding magnet is called the magnetic field. In this field we can encounter magnetic force. Magnet attraction occurs when poles are unlike and is caused by magnetic force. It means that poles should have opposite values.

7 0

3 years ago

What is the effect in the picture

stepan [7]

The effect of this problem is that negative particles and positive particles contract to each other caused by electrical force.

8 0

3 years ago

As the shuttle bus comes to a sudden stop to avoid hitting a dog, it accelerates uniformly at -4.1 m/s^2 as it slows from 9.0 m/

zhuklara [117]

Acceleration = (change in speed) / (time for the change)

- 4.1 m/s² = (-9 m/s) / (time for the change)

Time for the change = (-9 m/s) / (-4.1 m/s²) = 2.2 seconds

3 0

3 years ago

A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/ s. A I.O-kg stone is thrown

nadya68 [22]

(a) 296.6 m

The motion of the stone is the motion of a projectile, thrown with a horizontal speed of

$v_x = 15.0 m/s$

and with an initial vertical velocity of

$v_{y0} = -20.0 m/s$

where we have put a negative sign to indicate that the direction is downward.

The vertical position of the stone at time t is given by

$y(t) = h + v_{0y} t + \frac{1}{2}gt^2$ (1)

where

h is the initial height

g = -9.81 m/s^2 is the acceleration due to gravity

The stone hits the ground after a time t = 6.00 s, so at this time the vertical position is zero:

y(6.00 s) = 0

Substituting into eq.(1), we can solve to find the initial height of the stone, h:

$0 = h + v_{0y} y + \frac{1}{2}gt^2\\h = -v_{0y} y - \frac{1}{2}gt^2=-(-20.0 m/s)(6.00 s) - \frac{1}{2}(9.81 m/s^2)(6.00 s)^2=296.6 m$

(b) 176.6 m

The balloon is moving downward with a constant vertical speed of

$v_y = -20 m/s$

So the vertical position of the balloon after a time t is

$y(t) = h + v_y t$

and substituting t = 6.0 s and h = 296.6 m, we find the height of the balloon when the rock hits the ground:

$y(t) = 296.6 m + (-20.0 m)(6.00 s)=176.6 m$

(c) 198.2 m

In order to find how far is the rock from the balloon when it hits the ground, we need to find the horizontal distance covered by the rock during the time of the fall.

The horizontal speed of the rock is

$v_x = 15.0 m/s$

So the horizontal distance travelled in t = 6.00 s is

$d_x = v_x t = (15.0 m/s)(6.00 s)=90 m$