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3 years ago

Help me i suck at physics ;-;

Physics

2 answers:

Alex_Xolod [135]3 years ago

7 0

force= mass x acceleration= 0.2 x 20= 4 N

(N stands for newton btw)

Elena L [17]3 years ago

3 0

Answer is 4 and here is a trick so you can answer it by yourself. Multiply 0.2 with 20 and you will get your answer

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A train 471 m long is moving on a straight track with a speed of 75.1 km/h. The engineer applies the brakes at a crossing, and l

hram777 [196]

Answer:

$t = 37.6 s$

Explanation:

As we know that train is initially moving with the speed

$v_i = 75.1 km/h$

now we know that

$v_i = 20.86 m/s$

now the final speed of the train when it crossed the crossing

$v_f = 15 km/h$

$v_f = 4.17 m/s$

now we can use kinematics here

$v_f^2 - v_i^2 = 2 a d$

$4.17^2 - 20.86^2 = 2 a(471)$

$a = -0.44 m/s^2$

Now the time to cross that junction is given as

$v_f - v_i = at$

$4.17 - 20.86 = a(-0.44)$

$t = 37.6 s$

4 0

3 years ago

We cannot consider something to be work unless

iren [92.7K]

<span>force applied causes movement of an object in the same direction as the applied force.</span>

5 0

4 years ago

Read 2 more answers

When the voltage is at a maximum positive value, the current is at a value that is_________.

alukav5142 [94]

Answer:

When the voltage is at a maximum positive value, the the current is at a value that is maximum and positive

Explanation:

We know that the relation between the Voltage and the current is given using the Ohm's law, which states that the voltage (V) is directly proportional to the current (I)

Mathematically,

V ∝ I

Hence,

When the voltage is at a maximum positive value, the the current is at a value that is maximum and positive

4 0

3 years ago

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be: