Objects have a tendency to resist changing their motion. This property is called: *

One particle has mass m and a second particle has mass 2m. The second particle is moving with speed v and the first with speed 2

choli [55]

Answer:

Explanation:

The formula for kinetic energy to be used here is 1/2mv².

If the first particle is "particle a" and the second particle is "particle n"; there kinetic energies (K.E) will be

K.Eₐ = 1/2.m2v² = mv²

K.Eₙ = 1/2.2mv² = mv²

From the above, <u>it can be said that there kinetic energies are the same</u>.

NOTE that the m and v used in the question means mass and velocity respectively.

3 0

3 years ago

An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

5 0

3 years ago

From a static hot air balloon, a 10kg projectile is launched at a speed of 10m / s upwards. If the balloon has a mass of 90kg. W

lesya692 [45]

Answer:

c. 1.11 m/s down

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Assuming the balloon and projectile are originally at rest:

(90 kg) (0 m/s) + (10 kg) (0 m/s) = (90 kg) v + (10 kg) (10 m/s)

0 kg m/s = (90 kg) v + 100 kg m/s

v = -1.11 m/s

8 0

3 years ago

object carries a charge of -8.1 ÂµC, while another carries a charge of -2.0 ÂµC. How many electrons must be transferred from the

LenKa [72]

Number of electrons transferred: $1.91\cdot 10^{13}$

Explanation:

The charge on the first object is

$Q_1 = -8.1\mu C$

while the charge on the 2nd object is

$Q_2=-2.0 \mu C$

When they are in contact, the final charge on each object will be

$Q=\frac{Q_1+Q_2}{2}=\frac{-8.1+(-2.0)}{2}=-5.05 \mu C$

So, the amount of charge (electrons) transferred from the 1st object to the 2nd object is

$\Delta Q = Q_1 - Q = -8.1 -(5.05)=-3.05 \mu C = -3.05\cdot 10^{-6}C$

The charge of one electron is

$e=-1.6\cdot 10^{-19}C$

Therefore, the number of electrons transferred is

$N=\frac{Q}{e}=\frac{-3.05\cdot 10^{-6}}{-1.6\cdot 10^{-19}}=1.91\cdot 10^{13}$

Learn more about electrons:

brainly.com/question/2757829

#LearnwithBrainly

8 0

3 years ago

How does friction affect work

GenaCL600 [577]

Answer:

Friction works against the motion and acts in the opposite direction. When one object is sliding on another it starts to slow down due to friction. ... By rubbing them together we generate friction and, therefore, heat. The force F of friction pushes back on the block

Explanation:

4 0

3 years ago