In which circuit does charge reverse direction many times per second?

marshall27 [118]

It is 800 N FN = 600N + 200 N = 800 N Answer to your question: The net force is all Newton's second law. It is the force that acts on a body or a particle. for example: It is the force we make when we push a car or something heavy that is in a straight line. .

3 0

3 years ago

You are doing an experiment to determine how many passengers would fit into a full-sized plane. If a scale model plane has space

romanna [79]

The real place should theoretically have space for 87 passengers if it is an exact model and doesn't have modifications in the seat numbers.

5 0

3 years ago

Read 2 more answers

A rocket ship has several engines and thrusters. While the Solid Rocket Booster (SRB) and main engines only work together during

Katarina [22]

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

A rocket ship has several engines and thrusters. We can divide its initial movement into 2 parts:

From t = 0 min to t = 2.0 min, the SRB and the main engines act together and the speed goes from 0 m/s (rest) to 1341 m/s.
From t = 2.0 min to t = 8.5 min, the main engines alone accelerate the ship form 1341 m/s to 7600 m/s.

We want to know the acceleration in the first part (first 2.0 minutes). We need to consider that:

The speed increases from 0 m/s to 1341 m/s.
The time elpased is 2.0 min.
1 min = 60 s.

The acceleration of the ship during the first 2.0 minutes is:

$a = \frac{\Delta v }{t} ) \frac{(1341m/s-0m/s)}{2.0min} \times \frac{1min}{60s} = 11 m/s^{2}$

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

Learn more: brainly.com/question/16274121

3 0

2 years ago

Can anyone help me with this question please

JulsSmile [24]

Explanation:

V=u+at

where,

v=final speed

u=initial speed,(starting speed)

a=acceleration

t=time

v=u+at = 6=2+a*2

6=2+2a

2a=6-2

2a=4

a=4/2 = 2

a =2

2. to find time taken

v=u+at

25=5*2t

2t=25-5

2t=20

t=20/2

t=10sec

3. finding final speed

v=u+at

v=4+10*2

=4+20

v=24m/sec

5.v=u+at

=5+8*10

=5+80

V=85m/sev

6. v=u+at

8=u+4*2

8=u+8

U=8/8

u=1

these are your missing values

5 0

3 years ago

a catcher "gives" with the ball when he catches a 0.196 kg baseball moving at 31 m/s. if he moves his glove a distance of 5.32 c

Aloiza [94]

Answer:

3540.5N

Explanation:

Step one:

given data

mass m= 0.196kg

speed v= 31m/s

distance r= 5.32cm = 0.0532m

Step two

The expression relating force, mass, velocity and distance is

F= mv^2/r

substitute we have

F=0.196*31^2/0.0532

F=0.196*961/0.0532

F=188.356/0.0532

F=3540.5N

6 0

3 years ago