A bicycle with an initial velocity of +6 m/s accelerates at a

Which of the following statements about galaxies is true? A. Galaxies typically contain millions of stars. B. Galaxies actually

Sloan [31]

the answer is D i took the quiz so it is definitely D

6 0

3 years ago

Read 2 more answers

Help!!!!

BARSIC [14]

Can we see the diagram? Thanks.

6 0

3 years ago

A 2000 kg truck traveling north at 34 km/h turns east and accelerates to 58 km/h. (a) What is the change in the truck's kinetic

barxatty [35]

Explanation:

It is given that,

Mass of the truck, m = 2000 kg

Initial velocity of the truck, u = 34 km/h = 9.44 m/s

Final velocity of the truck, v = 58 km/h = 16.11 m/s

(a) Change in truck's kinetic energy, $\Delta E=\dfrac{1}{2}m(v^2-u^2)$

$\Delta E=\dfrac{1}{2}\times 2000\ kg\times (16.11^2-9.44^2)$

$\Delta E=170418.5\ J$

$\Delta E=1.7\times 10^5\ J$

(b) Change in momentum of the truck, $\Delta p=m(v-u)$

$\Delta p=2000\ kg\times (16.11-9.44)$

$\Delta p=13340\ kg-m/s$

Hence, this is the required solution.

6 0

4 years ago

How much work would a child do while puling a 12-kg wagon a distance of 3m with a 22 N force directed 30 degrees with respect to

Alika [10]

Answer:

The work done will be 57.15 J

Explanation:

Given that,

Mass = 12 kg

Distance = 3 m

Force = 22 N

Angle = 30°

We need to calculate the work done

The work done is defined as,

$W = Fd\cos\theta$

Where, F = force

d = displacement

Put the value into the formula

$W=22\times3\times\cos30^{\circ}$

$W=22\times3\times\dfrac{\sqrt{3}}{2}$

$W = 57.15\ J$

Hence, The work done will be 57.15 J

6 0

3 years ago

You charge an initially uncharged 89.9-mf capacitor through a 30.5-ω resistor by means of a 9.00-v battery having negligible int

blsea [12.9K]

1) The differential equation that models the RC circuit is :

(d/dt)V_capacitor + (V_capacitor/RC) = (V_source/RC)

Where the time constant of the circuit is defined by the product of R*C

Time constant = T = R*C = (30.5 ohms) * (89.9-mf) = 2.742 s

2) Charge of the capacitor 1.57 time constants

1.57*(2.742) = 4.3048 s

The solution of the differential equation is

V_capac (t) = (V_capac(0) - V_capac(∞))e ^(-t /T) + V_capac(∞)

Since the capacitor is initially uncharged V_capac(0) = 0

And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery V_capac(∞) = 9V

This means,

V_capac (t) = (-9V)e ^(-t /T) + 9V

The charge in a capacitor is defined as Q = C*V

Where C is the capacitance and V is the Voltage across

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /T) + 9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V = -1.87V +9V

V_capac (4.3048 s) = 7.1275 V

Q (4.3048 s) = 89.9mF*(7.1275V) = 0.6407 C

3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V

Q (∞) = 89.9mF*(9V) = 0.8091 C

7 0

3 years ago