Ask question

Ask question

All categories

4 years ago

15

when an object moves with constant velocity, does its average velocity during any time interval differ from its instataneous vel

ocity at any instant?

1 answer:

BigorU [14]4 years ago

8 0

For a constant-velocity object, the average and instantaneous are the same. So the answer is no. It's like taking a running average of a string of numbers that are all the same number. The average is always the sum of the numbers divided by how many have accumulated, which will always equate to the repeated number.

You might be interested in

Describe how the rider exerts a force on the motorcycle

LUCKY_DIMON [66]

Well, one example is that the weight of the rider puts downward force on the motorcycle, which is absorbed by the suspension or shocks or something.

4 0

3 years ago

A 0.5 m diameter wagon wheel consists of a thin rim having a mass of 7 kg and six spokes, each with a mass of 1.2 kg. 1.2 kg 7 k

Arte-miy333 [17]

Explanation:

It is given that,

Mass of the rim of wheel, m₁ = 7 kg

Mass of one spoke, m₂ = 1.2 kg

Diameter of the wagon, d = 0.5 m

Radius of the wagon, r = 0.25 m

Let I is the the moment of inertia of the wagon wheel for rotation about its axis.

We know that the moment of inertia of the ring is given by :

$I_1=m_1r^2$

$I_1=7\times (0.25)^2=0.437\ kgm^2$

The moment of inertia of the rod about one end is given by :

$I_2=\dfrac{m_2l^2}{3}$

l = r

$I_2=\dfrac{m_2r^2}{3}$

$I_2=\dfrac{1.2\times (0.25)^2}{3}=0.025\ kgm^2$

For 6 spokes, $I_2=0.025\times 6=0.15\ kgm^2$

So, the net moment of inertia of the wagon is :

$I=I_1+I_2$

$I=0.437+0.15=0.587\ kgm^2$

So, the moment of inertia of the wagon wheel for rotation about its axis is $0.587\ kgm^2$ . Hence, this is the required solution.

4 0

4 years ago

For a top player, a tennis ball may leave the racket on the serve with a speed of 55 m/s (about 120 mi/h). If the ball has a mas

inn [45]

Answer:

Yes is large enough

Explanation:

We need to apply the second Newton's Law to find the solution.

We know that,

$F= ma$

And we know as well that

$a= \frac{v}{t}$

Replacing the aceleration value in the equation force we have,

$F= \frac{mv}{t}$

Substituting our values we have,

$F= \frac{(0.060)(55)}{4*10^{-3}}$

$F=825N$

The weight of the person is then,

$W = mg$

$W = (60)(9.8) = 558N$

<em>We can conclude that force on the ball is large to lift the ball</em>

6 0

3 years ago

A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected

gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta$

Put the value into the fomrula

$5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45$

$15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}$

$15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}$ ....(I)

Along y -axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta$

Put the value into the formula

$0+0=5\times v_{1}\sin30-3\times v_{2}\sin45$

$\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0$ ...(II)

From equation (I) and (II)

$v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}$

$v_{1}=2.19\ m/s$

Put the value of v₁ in equation (I)

$\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0$

$v_{2}=\dfrac{5.475\times\sqrt{2}}{3}$

$v_{2}=2.58\ m/s$

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0

3 years ago

The gun, mount, and train car of a railway had a total mass of 1.22 x 10^6 kg. The gun fired a projectile that was 80 cm in diam

kakasveta [241]

1) According to the law of conservation of momentum ..
<span>Horiz recoil mom of gun (M x v) = horiz. mon acquired by shell (m x Vh) </span>

<span>1.22^6kg x 5.0 m/s = 7502kg x Vh </span>
<span>Vh = 1.22^6 x 5 / 7502 .. .. Vh = 813 m/s </span>

<span>Barrel velocity V .. .. cos20 = Vh / V .. ..V = 813 /cos20 .. .. ►V = 865 m/s </span>

<span>2) Using the standard range equation .. R = u² sin2θ /g </span>

<span>R = 865² x sin40 / 9.80 .. .. ►R = 49077 m .. (49 km)</span>

5 0

3 years ago