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Dmitriy789 [7]
3 years ago
11

An object of mass m is oscillating with a period T. The position x of the object as a function of time is given by the equation

x(t)=Acosωt . The maximum net force exerted on the object while it is oscillating has a magnitude F. Which of the following expressions is correct for the maximum speed of the object during its motion?
Physics
1 answer:
Scrat [10]3 years ago
3 0

What the equation given?

  • x(t)=Acos\omegat

Maximum velocity occurs at the equilibrium position

So

  • x=0

Now

  • x(0)=Acos0[/tex]
  • x(0)=A

Now

As we know the formula

\\ \rm\rightarrowtail V_max=A\omega

These expressions can be used

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An RV is traveling 60 km/h along a highway. A boy sitting near the driver of the RV throws a ball to another boy at the back end
alina1380 [7]

Answer:

Speed of the ball relative to the boys: 25 km/h

Speed of the ball relative to a stationary observer: 35 km/h

Explanation:

The RV is travelling at a velocity of

v_{RV}=+60 km/h

Here we have taken the direction of motion of the RV as positive direction.

The boy sitting near the driver throws the ball back with speed of 25 km/h, so the velocity of the ball in the reference frame of the RV is

v_B = -25 km/h

with negative sign since it is travelling in the opposite direction relative to the RV. Therefore, this is the velocity measured by every observer in the reference frame of the RV: so the speed measured by the boys is

v = 25 km/h

Instead, a stationary observer outside the RV measures a velocity of the ball given by the algebraic sum of the two velocities:

v = +60 km/h + (-25 km/h) = +35 km/h

So, he/she measures a speed of 35 km/h.

5 0
3 years ago
You can think of the work-kinetic energy theorem as the second theory of motion, parallel to Newton's laws in describing how out
kiruha [24]

Answer:

a) 4 289.8 J

b) 4 289.8 J

c) 6 620.1 N

d) 411 186.3 m/s^2

e) 6 620.1 N

Explanation:

Hi:

a)

The kinetic energy of the bullet is given by the following formula:

K = (1/2) m * v^2

With

    m = 16.1 g = 1.61 x 10^-2 kg

     v = 730 m/s

K = 4 289.8 J

b)

the work-kinetic energy theorem states that the work done on a system is the same as the differnce in kinetic energy of the same. Since the initial state of the bullet was at zero velocity (it was at rest)  Ki = 0, therefore:

W = ΔK = Kf - Ki  = 4 289.8 J

c)

The work done by a force is given by the line intergarl of the force along the trayectory of the system (in this case the bullet).

If we consider a constant force (and average net force) directed along the trayectory of the bullet, the work and the force will be realted by:

W = F * L

Where F is the net force and L is the length of the barrel, that is:

F = (4 289.8 J) / (64.8 cm) = (4 289.8 Nm) / (0.648 m) = 6620.1 N

d)

The acceleration can be found dividing the force by the mass:

a = F/m = (6620.1 N) /(16.1 g) = 411 186.3 m/s^2

e)

The force will have a magnitude equal to c) and direction along the barrel towards the exit

5 0
3 years ago
The average velocity of blood flowing in a certain 4-mm-diameter artery in the human body is 0.28 m/s. The viscosity and density
OLga [1]

Answer:

V = 3.5 x 10⁻⁶ m³/s = 3.5 cm³/s

Explanation:

The volume flow rate of the blood in the artery can be given by the following formula:

V = Av

where,

V = Volume flow rate = ?

A = cross-sectional area of artery = πd²/4 = π(0.004 m)²/4 = 1.26 x 10⁻⁵ m²

v = velcoity = 0.28 m/s

Therefore,

V = (1.26\ x\ 10^{-5}\ m^2)(0.28\ m/s)

<u>V = 3.5 x 10⁻⁶ m³/s = 3.5 cm³/s</u>

4 0
3 years ago
How do liquids affect friction?
Sophie [7]

Although liquids offer resistance to objects moving through them, they also smooth surfaces and reduce friction. Liquids tend to get thinner (less viscous) as they are heated. ... The rubbing produces friction and the result is heat.
6 0
3 years ago
Everyone's favorite flying sport disk can be approximated as the combination of a thin outer hoop and a uniform disk, both of di
Umnica [9.8K]

Answer:

The length of the boomerang is 0.364 m

Explanation:

The moment of inertia is:

I=\frac{1}{2} m_{d} r^{2} +m_{h} r^{2}

Where

md = 0.05 kg

mh = 0.12 kg

r = d/2 = 0.273/2 = 0.1365 m

I=\frac{0.05*(0.1365)^{2} }{2} +(0.12)*(0.1365)^{2} =2.7x10^{-3} kgm^{2}

The length of the boomerang is:

L_{b} =\sqrt{\frac{12I}{m} } =\sqrt{\frac{12*2.7x10^{-3} }{0.245} } =0.364m

5 0
3 years ago
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