Which type of bond does this molecule have?

In a tug of war, when one team is pulling with a force of 85 N and the other 40 N, what is the net

olya-2409 [2.1K]

85 N - 40 N = 45 N
And depending on direction the greater force is being pulled towards

4 0

3 years ago

Suppose air resistance does not exist, roughly what would be the velocity of a 10 lb object 5-seconds into free fall off of a 50

ira [324]

Answer:

Correct answer : v = 50 m/s

Explanation:

The movement described is a free fall with no initial velocity and we will use the following formula:

v = g · t we take that g = 10 m/s²

v = 10 m/s²· 5 s = 50 m/s

v = 50 m/s

God is with you!!!

4 0

4 years ago

Please help no trolls no links pls

Novosadov [1.4K]

It would move slower/ faster by changing the mass, which would effect the weight and speed.

3 0

3 years ago

Read 2 more answers

The inner planets or call terrestrial because they

AnnyKZ [126]

They are similar to the earth. Not relative to the outer planets

6 0

4 years ago

block of mass 5kgriding on a horizontal frictionlessxy-plane surface is subjected tothree applied forces:→F1= 12√2N[ 45◦]→F2= (8

dsp73

Answer:

(i) See attached image for the drawing

(ii) net force given in component form: (20, 20)N with magnitude: $\sqrt{800} \,\,\,N$

Explanation:

First try to write all forces in vector component form:

The force F1 acting at 45 degrees would have multiplication factors of $\frac{\sqrt{2} }{2}$ on both axes, to take care of the sine and cosine projections. Therefore, the:

x-component of F1 is $F1_x=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N$

y-component of F1 is $F1_y=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N$

As far as force F2, it is given already in x and y components, then:

x-component of F2 = 8 N

y-component of F2 = -6 N (negative meaning pointing down the y-axis)

Force F3 has only component (upwards) in the y-direction

x-component of F3 = 0 N

y-component of F3 =14 N

The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:

x-component of R = 12 + 8 = 20 N

y-component of R = 12 + 14 - 6 = 20 N

Therefore, the acceleration that the mass receives due to this force is given in component form as:

x-component of acceleration: 20 N / 5 kg = $4\,\,\,m/s^2$

y-component of acceleration: 20 N / 5 kg = $4\,\,\,m/s^2$

Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:

x-component of the velocity is: $v_x=4\,*\,2=8\,m/s$

y-component of the velocity is: $v_y=4\,*\,2=8\,m/s$

7 0

4 years ago