An ant travels 2.78 cm [W] and then turns and travels 6.25 cm [S 40 degrees E]. What is the ant's total displacement?

(a) A 15.0 kg block is released from rest at point A in the figure below. The track is frictionless except for the portion betwe

serg [3582]

Answer:

(a) the coefficient of friction is 0.451

This was derived using the energy conservation principle (the total energy in a closed system remains constant).

(b) No, the object stops 5.35 m away from point B. This is due to the spring's expansion only performing 43 J of work on the block, which isn't sufficient compared to the 398 J required to overcome friction.

Explanation:

For more details on how this issue was resolved, refer to the attached material. The solution for part (a) separates the body’s movement into two segments: from point A to B, and from B to C. The total system energy originates from the initial gravitational potential energy, which transforms into work against friction and into work compressing the spring. A work of 398 J is needed to counteract friction over the distance of 6.00 m. The energy used for this is lost since friction is not a conservative force, leaving only 43 J for spring compression. When the spring expands, it exerts a work of 43 J back on the block, which is only sufficient to move it through a distance of 0.65 m, stopping 5.35 m short of point B.

Thank you for your attention; I trust this is beneficial to you.

4 0

2 months ago

At a location where the acceleration due to gravity is 9.807 m/s2, the atmospheric pressure is 9.891 × 104 Pa. A barometer at th

serg [3582]

Answer:

$8616.7468 \ kg/m^3$

Explanation:

The measurement of pressure is indicated as $p=\rho gh$ where p denotes the pressure, $\rho$ signifies density, and h represents height

Given values include pressure $p=9.891\times 10^4\ Pa$ , gravity's acceleration $g=9.9870\ m/sec^2$ , and height =1.163 m

$\rho =\frac{p}{gh}=\frac{9.891\times 10^4}{9.870\times 1.163}=8616.7468 \ kg/m^3$

5 0

1 month ago

Read 2 more answers

A ball rolls up a slope. At the end of three seconds its velocity is 20 cm/s; at the end of eight seconds its velocity is 0. Wha

serg [3582]

Answer:

$a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}$

Explanation:

Data provided

initial velocity v₀=20 cm/s at time t=3s

final velocity vf=0 at time t=8 s

Required

Average Acceleration for the interval from 3s to 8s

Solution

Acceleration can be defined as the first derivative of velocity concerning time

$a_{acceleriation} =\frac{dv_{velocity}}{dt_{time}}\\a_{acceleriation} =\frac{v_{f}-v_{o} }{dt}\\ a_{acceleriation} =\frac{0-20cm/s }{8s-3s}\\ a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}$

8 0

1 month ago

You are camping in the wilderness. After a few days, you are horrified to discover that you did not pack as many batteries as yo

Keith_Richards [3271]

Angular speed is calculated as 2.5 rev/sec.

8 0

1 month ago

Three balls of equal mass are fired simultaneously with equal speeds from the same height h above the ground. Ball 1 is fired st

Ostrovityanka [3204]

Answer:

d) v1 = v2 = v3

Explanation:

This can be determined through the principle of energy conservation. We assess the total mechanical energy E=K+U (the sum of kinetic energy and gravitational potential energy) at both the initial and final positions, ensuring they remain constant.

<pInitially, for the three spheres, we have:

$E_i=K_i+U_i=\frac{mv_i^2}{2}+mgh_i=\frac{mv^2}{2}+mgh$

Finally, for the three spheres, we see:

$E_f=K_f+U_f=\frac{mv_f^2}{2}+mgh_f=\frac{mv_f^2}{2}$

<pGiven that $E_i=E_f$ , and since $E_i$ remains identical for all spheres, it follows that $E_f$ is identical for all spheres, indicating that $v_f$ , the final velocity, is equal for each ball.

4 0

2 months ago