A train accelerates at -1.5 m/s2 for 10 seconds. If the train had an initial speed of 32 m/s, what is its new speed?

Physics

1 answer:

Nata [24]3 years ago

7 0

B should be the answer

Look at this example to help you

You might be interested in

What is oxygenated blood? De-oxygenated blood?

WITCHER [35]

Oxygenated blood that has oxygen in them while de-oxygenated blood has carbon dioxide. in which the oxygenated blood carries the oxygen throughout the body since that cells need oxygen to function. called "gas exchange." once the cells got their required oxygen. the carbon dioxide needs somewhere to go, thus having deoxygenated blood. and that carbon dioxide needs to get out of the body

6 0

3 years ago

Read 2 more answers

The potential energy of a catapult was completely converted into kinetic energy by releasing a small stone with a mass of 20 gra

DerKrebs [107]

1,000 grams = 1 kilogram
20 grams = 0.02 kilogram

Kinetic energy = (1/2) (mass) x (speed)²

(1/2) (0.02) x (15)² =

(0.01) x (225) = 2.25 joules

4 0

3 years ago

Read 2 more answers

A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca

svetoff [14.1K]

Answer:

$q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Explanation:

Given that $L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V$ , we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

$E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000 \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}$

#To find the particular solution:

$Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$