What actions can be explained by physics?

A pendulum clock is taken to the moon (g = 1.6 N/kg). How long must the pendulum be in order for the clock to

mamaluj [8]

Answer:

0.041 m

Explanation:

T = 2π√(L/g)

1 s = 2π√(L / (1.6 N/kg))

L = 0.041 m

4 0

3 years ago

8) A racecar accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the

Vlada [557]

8)Acceleration of the car is 11.17 m/s².

9) Acceleration of skater is 0.57 m/s².

10) The deceleration is 1.8 m/s².

11) The average acceleration of the sprinter is 2 m/s².

12) The velocity of the stroller after 4.75 minutes is 171 m/s.

13) The skier will be moving as 11 m/s speed.

14) 0.56 seconds is required by the rattle snake to reach the speed of 28 m/s from rest.

Answer:

Explanation:

8) Acceleration of the car is determined by finding the ratio of change in velocity to the time taken for the change in velocity. As here the initial velocity is 18.5 m/s and the final velocity is 46.1 m/s, then in 2.47 seconds, the acceleration is

Acceleration = (46.1-18.5)/2.47 = 11.17 m/s².

Acceleration of the car is 11.17 m/s².

9)Similarly, the acceleration of the skater with initial velocity zero to final velocity as 5.7 m/s in 10 seconds is

Acceleration = (5.7-0)/10=0.57 m/s².

So acceleration of skater is 0.57 m/s².

10) In this case, the shuttle bus is stopping so its speed is decreasing from high value to low value. This kind of acceleration related to the decrease in velocity is termed as deceleration as the value of acceleration will be coming as negative.

Acceleration = (0-9)/5=-1.8 m/s².

So the deceleration is 1.8 m/s².

11) Average acceleration of sprinter = (Final velocity-Initial velocity)/Time

Average acceleration = (7.5-5)/1.25=2 m/s².

So the average acceleration of the sprinter is 2 m/s².

12)Since the acceleration of the stroller is given as 0.6 m/s². And the initial velocity is given as zero. So for the time of 4.75 minutes, the velocity will be equal to the final velocity.

As per equations of motion,

v = u +at

As u =0, a = 0.6 m/s² and t = 4.75 ×60 s = 285 s

So v = 0.6×285 = 171 m/s

Thus, the velocity of the stroller after 4.75 minutes is 171 m/s.

13) Similarly, in this case, u = 0, a = 2.2 m/s² and t = 5 s

Then v = u+at=0+(2.2×5)=11 m/s

So the skier will be moving as 11 m/s speed.

14) Here a = 50 m/s² and v = 28 m/s with u = 0 so time taken to reach the speed of 28 m/s is

v = u +at = 0+ (50 t)

28 = 50 t

t = 28/50 = 0.56 s

So 0.56 seconds is required by the rattle snake to reach the speed of 28 m/s from rest.

4 0

3 years ago

A horizontal force of 350N is exerted on a 2.5 kg ball as it rotates uniformly in a horizontal circle of radius of 0.90m. Calcul

harkovskaia [24]

F=mv^2/R
----> V^2=FR/m=(350x0.9)/2.5=126
----- V=11.22 m/s

5 0

3 years ago

HELLLP PLEASE || the graph below shows a conversion of energy for a skydive jumping out of a plane and landing safely on the gro

fenix001 [56]

Answer: I maybe wrong but i'm pretty sure its C) Kinetic energy

5 0

3 years ago

A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. comput

Nikitich [7]

There is a missing data in the text of the problem (found on internet):
"with force constant k=450N/m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
 $E=U+K= \frac{1}{2}kx^2 + \frac{1}{2} mv^2$
where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
 $E=K_{max} = \frac{1}{2}mv_{max}^2$
Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
 $E=U_{max}= \frac{1}{2}k A^2$

Since the mechanical energy must be conserved, we can write
 $\frac{1}{2}mv_{max}^2 = \frac{1}{2}kA^2$
from which we find the maximum speed
 $v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }= 1.2 m/s$

b) the speed of the glider when it is at x= -0.015m

We can still use the conservation of energy to solve this part.
The total mechanical energy is:
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
 $U= \frac{1}{2}kx^2 = \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J$
And since
 $E=U+K$
we find the kinetic energy when the glider is at this position:
 $K=E-U=0.36 J - 0.05 J = 0.31 J$
And then we can find the corresponding velocity:
 $K= \frac{1}{2}mv^2$
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s$

c) the magnitude of the maximum acceleration of the glider;

For a simple harmonic motion, the magnitude of the maximum acceleration is given by
$a_{max} = \omega^2 A$
where $\omega= \sqrt{ \frac{k}{m} }$ is the angular frequency, and A is the amplitude.
The angular frequency is:
$\omega = \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s$
and so the maximum acceleration is
$a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2$

d) the acceleration of the glider at x= -0.015m

For a simple harmonic motion, the acceleration is given by
 $a(t)=\omega^2 x(t)$
where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find
 $a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2$

e) the total mechanical energy of the glider at any point in its motion. 

we have already calculated it at point b), and it is given by
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

8 0

3 years ago