All categories

3 years ago

When an object in simple harmonic motion is at its maximum displacement, its____________ is also at a maximum.

Physics

1 answer:

Ivan3 years ago

7 0

When an object in simple harmonic motion is at its maximum displacement, its acceleration is also at a maximum.

Reason: The speed is zero when the simple harmonic motion is at its maximum displacement, however, the acceleration is the rate of change of velocity. The velocity reverses the direction at that point therefore its rate of change is maximum at that moment. thus the acceleration is at its maximum at this point

Hope that helps!

You might be interested in

What chemicals does the body produce to keep neutral pH?

AleksandrR [38]

The body produces Buffers that can neutralize acids

6 0

3 years ago

A GPS receiver can tell you your exact latitude and longitude?

Dmitry_Shevchenko [17]

That is false. A GPS cant tell latitude and longitude

3 0

3 years ago

A rectangular coil lies flat on a horizontal surface. A bar magnet is held above the center of the coil with its north pole poin

Llana [10]

Answer:

There is no induced current on the coil.

Explanation:

Current is induced in a coil or a circuit, when there is a break of flux linkage. A break in flux linkage is caused by a changing magnetic field, and must be achieved by a relative motion between the coil and the magnet. Holding the magnet above the center of the coil will cause no changing magnetic filed since there is no relative motion between the coil and the magnet.

7 0

3 years ago

A rocket in orbit just above the atmosphere is moving in uniform circular motion. The radius of the circle in which it moves is

Zinaida [17]

Centripetal acceleration is given by the formula

$a_{cen}= \frac{v^2}{r}$

Since we know the value of the acceleration is 9.8 m/s² and we know r, we can solve for the speed, which is the magnitude of v. Note the value of a is just the acceleration due to gravity, which is what makes the rocket orbit and not fly off into space.

$9.8 \frac{m}{s^2}=\frac{v^2}{6.381 \times 10^6m} \\ \\ 9.8 \frac{m}{s^2} \times 6.381 \times 10^6m=v^2 \\ \\ v= \sqrt{9.8 \frac{m}{s^2} \times 6.381 \times 10^6m} =24755 \frac{m}{s}$

8 0

3 years ago

A 37 kg object has an applied force of 85N [R] acting on it. The coefficient of

Sliva [168]

Answer:

Explanation:

This is quite tricky! You need to do 2 different equations to solve all the parts of this problem. First is finding the acceleration in one dimension, which has an equation of

F - f = ma

where F is the applied Friction,

f is the frictional force acting against F,

m is the mass of the object, and

a is the acceleration of the object (NOT the velocity!)

This is Newton's Second Law expanded on a bit. The sum of the forces working on an object is equal to the object's mass times its acceleration. We have F, but we need f which is found in the equation

f = μ $F_n$ which is the coefficient of kinetic friction times the weight of the object. Weight is found in the equation

w = mg where m is mass and g is the pull of gravity. Let's start there and work backwards:

w = 37(9.8) to 2 sig figs so

w = 360N. Now fill that in to find f:

f = (.17)(360) to 2 sig figs so

f = 61. Now for the final answer in the original equation way back up at the top:

85 - 61 = 37a and do the subtraction on the left side first:

24 = 37a and then we divide to 2 sig figs to get

a = .65 m/s/s

Since we are moving in a straight line (as opposed to on an angle) the displacement is found in

d = rt which simply says that the distance an object moves is equal to its rate times the time. Therefore,

d = 2.2(3.4) to 2 sig figs so

d = 7.5 m

6 0

3 years ago