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12345 [234]
3 years ago
8

Bicycle in its writers have combined a mass of 80 kg in a speed of 6.0 meters per Second what is the magnitude of the average fo

rce needed to bring bicycle in its writers to stop in 4.0 seconds please help !!!
Physics
1 answer:
erik [133]3 years ago
7 0

Answer:

F = 120 N

Explanation:

Force x distance = energy

The bike has energy 1/2 . 80 . 6^2   = 1440 J

You are looking at an example of not reading the question properly.

Impulse = Force . time  = change in momentum

F . 4  = 80 .6

F = 120 N

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Mercury is the 80th position in the periodic table how many protons does it have
snow_lady [41]
Errrrr, it has 80.




80 is the correct answer
4 0
3 years ago
Read 2 more answers
Answer plzz<br>How do we apply Interval Training to our lower body for a workout?
weqwewe [10]
Interval training is simply alternating short bursts (about 30 seconds) of intense activity with longer intervals (about 1 to 2 minutes) of less intense activity. For instance, if your exercise is walking and you're in good shape, you might add short bursts of jogging into your regular brisk walks.
5 0
3 years ago
Water is contained in a closed, rigid 0.2 m 3 tank at an initial pressure of 5 bar and a quality of 50%. Heat transfer occurs un
elena55 [62]

Answer:

Final mass=0.89kg

Final pressure=5.6bar

Explanation:

To find mass,m=v/v1

But v1=vf + x(vg-vf)

Vf= 0.001093m^3/kg

Vg= 0.3748m^3/kg

V1= 0.001093+0.5(0.3748-0.001093)

V1= 0.225m^3/kg

M= 0.20/0.225 =0.89kg

Final pressure will be:

V/V1= P/P1

Cross multiply

VP1=V1P

P1= 0.225×5/0.2

P1=:5.6 bar

7 0
3 years ago
How fast would 40 Newtons of force accelerate a 2 kg object?
Digiron [165]

Answer:

20 m/s^2

Explanation:

We can solve this problem by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between its mass and its acceleration:

F=ma

where

F is the net force on the object

m is its mass

a is its acceleration

In this problem:

F = 40 N is the force on the object

m = 2 kg is its mass

Therefore, the acceleration of the object is

a=\frac{F}{m}=\frac{40}{2}=20 m/s^2

8 0
3 years ago
When Jim and Rob ride bicycles, Jim can only accelerate at three-quarters the acceleration of Rob. Both startfrom rest at the bo
Natali5045456 [20]

Answer:

46.4 s

Explanation:

5 minutes = 60 * 5 = 300 seconds

Let g = 9.8 m/s2. And \theta be the slope of the road, s be the distance of the road, a be the acceleration generated by Rob, 3a/4 is the acceleration generated by Jim .  Both of their motions are subjected to parallel component of the gravitational acceleration gsin\theta

Rob equation of motion can be modeled as s = a_Rt_R^2/2 = a300^2/2 = 45000a[/tex]

Jim equation of motion is s = a_Jt_J^2/2 = (3a/4)t_J^2/2 = 3at_J^2/8

As both of them cover the same distance

45000a = 3at_J^2/8

t_J^2 = 45000*8/3 = 120000

t_J = \sqrt{120000} = 346.4 s

So Jim should start 346.4 – 300 = 46.4 seconds earlier than Rob in other to reach the end at the same time

7 0
3 years ago
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