11. Consider the velocity vs time graph below. Which object is not moving at time = 0?

A car of mass 2400kg moving at 20 m/s slams into a cement wall and comes to a halt?

Brut [27]

Impulse = change in momentum

The car's momentum was (mass) x (speed)

Momentum = (2400 kg) x (20 m/s)

Momentum = 48,000 km-m/s

To completely stop the car, the impulse = -48,000 km-m/s .

7 0

3 years ago

The gnaphosid spider Drassodes cupreus has evolved a pair of lensless eyes for detecting polarized light. Each eye is sensitive

BartSMP [9]

Answer:

Part A

The intensity is $I = 618 W/m^2$

Part B

The intensity is $I_1= 81.884 W/m^2$

Explanation:

From the question we are told that

The intensity of the light detected by first eye is $I = 700 W/m^2$

Now at initial state according the question the light ray is perpendicular to the eye so it means that it is at 90° the eye

Now the first question is to obtain the intensity the first eye (the first in this case is the one focused on the light )would detect when the head is rotated by 20° its previous orientation

This is mathematically evaluated as

$I = I_i cos^2 ( 20^o)$

$I = 700\ cos^2 (20)$

$I = 618 W/m^2$

Now the second question is to obtain the intensity the first eye (the first eye in this case is the one that is not focused on the light )would detect when the head is rotated by 20° its previous orientation

Now in this case the angle between the eye and the light is 90-20 = 70°

So

$I_1 = 700 \ cos^2 (70)$

$I_1= 81.884 W/m^2$

5 0

3 years ago

Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma

Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

Mass of inertia is given by

$I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2$

Angular acceleration is given by

$\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454$

The coefficient of friction is 0.54454

At r = 0.25 m

$\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N$

The force needed to stop the wheel is 104.00902 N

5 0

2 years ago

On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const

IceJOKER [234]

Answer:

a)

Explanation:

Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

$x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} (1)$

Since the car starts from rest, v₀ =0.
We know the value of t = 5 sec., but we need to find the value of a.
Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

$a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2 (2)$

Replacing a and t in (1):

$x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m. (3)$

Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

$a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2 (4)$