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2 years ago

Muscle surround your body

Physics

1 answer:

Elis [28]2 years ago

8 0

<span>Muscle surrounds your body organs and bones.</span>

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A car turns a certain curve of radius 24.98 m with constant linear speed of

Anastaziya [24]

Answer:

3525.19 kg

Explanation:

The computation of the mass of the car is shown below:

As we know that

Fc = m × V^2 ÷ R

m = Fc × R ÷ V^2

Provided that:

Fc = 34.652 kN = 34652 N

R = Radius = 24.98 m

V = speed = 15.67 m/s

So,

m = 34652 × 24.98 ÷ 15.67^2

= 3525.19 kg

7 0

2 years ago

A child of mass 40.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m. at point a the speed of the car is 1

pav-90 [236]

I attached the missing picture.
The force of seat acting on the child is a reaction the force of child pressing down on the seat. This is the third Newton's law. The force of a child pressing down the seat and the force of the seat pushing up on the child are the same.
There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part A
For point A we have:
$F_a=F_cf-F_g$
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
$F_a=m\frac{v^2}{r}-mg=179 $N$
Part B
At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
$F=F_{cf}\cos(30)-mg=m\frac{v^2}{r}\cos(30)-mg=153.2$N$
Part C
The child will stay in place at point A when centrifugal force and force of gravity are in balance:
$F_g=F_{cf}\\
mg=m\frac{v^2}{r}\\
gr=v^2\\
v=\sqrt{gr}=8.29\frac{m}{s}$

6 0

2 years ago

A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the

mylen [45]

Answer:

Angular acceleration, is $708.07\ rad/s^2$

Explanation:

Given that,

Initial speed of the drill, $\omega_i=0$

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, $\omega_f=28940\ rev/min=3030.58\ rad/s$

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

$\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2$

So, the drill's angular acceleration is $708.07\ rad/s^2$ .

4 0

2 years ago

Compared with a force, a torque involves

ycow [4]

A rotational movement.

6 0

3 years ago

How long would it take 2.0x10^20 electrons to pass through a point in a conductor if the current was 10.0A?

inysia [295]

1 coulomb of electric charge is carried by 6.25 x 10^18 electrons

1 Ampere = 1 coulomb per second
10 A = 10 coulombs per second

(2.0 x 10^20 electrons) x (coul / 6.25 x 10^18 electrons) / (10 coul/sec) =

(2.0 x 10^20) / (6.25 x 10^18 x 10) sec = <em>3.2 seconds</em>

6 0

2 years ago