What is the net force on the purple ring in the picture below. ________

Let w(x)=3x-7.If w(x)=14, find x

dlinn [17]

Answer:

7

Explanation:

We are given:

w(x) = 3x - 7

w(x) = 14

The problem here entails us to solve for x;

To solve for x; equate the two expressions:

3x - 7 = 14

3x = 14 + 7

3x = 21

x = 7

So the value of x = 7

6 0

2 years ago

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

2 years ago

What do you mean by kinetic energy and potential energy ?

Verizon [17]

Kinetic Energy of an object is the measure of the work an object can do by virtue of its motions..

Formula - KE = 1/2 mv^2

Where KE is the kinetic energy, m is the body’s mass, and v is the body’s velocity.

Potential energy is the stored energy in any object or system by virtue of its position or arrangement of parts..

Formula - W = mgh

Where,

. m is the mass in kilograms

. g is the acceleration due to gravity

. h is the height in meters

Hope it helpz~ uh..

4 0

1 year ago

Que es una ondaaaa?? no me deja escribiiiirrrrrr

UkoKoshka [18]

Answer:

yo queiro escribir mi hermano

7 0

2 years ago

What is the formula for Ba+2 and F-?

Alinara [238K]

Answer:

BaF2

Explanation:

since you got the valence numbers just do the scissors move where you:

give the F the 2 and give the Ba the 1 so it be like

BaF2 here is the chemical compound

5 0

2 years ago

What is the net force on the purple ring in the picture below. _________