All categories

3 years ago

Please help me. i have 2 hours

Physics

1 answer:

mestny [16]3 years ago

7 0

the missing word is clockwise moment. I hope this helps good luck

You might be interested in

The Solar System is made up of eight planets, numerous comets, asteroids, and moons, and the Sun. The force that holds all of th

Nastasia [14]

Answer:

B. GRAVITY is the answer

7 0

3 years ago

Read 2 more answers

Please help!?!?

Vanyuwa [196]

Since bulb is connected in the closed circuit at the position of D

as well as switch B is also closed in that position so the current will flow through the bulb and bulb will glow in that position

So the most appropriate correct option will be

D. The light bulb will be on

5 0

4 years ago

Read 2 more answers

Consider the free-body diagram. If you want the box to move, the force applied while dragging must be greater than the

VLD [36.1K]

You would want it to be greater than D. friction force

It needs be greater than the friction applied to it.

6 0

3 years ago

The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ

prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

$\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})$

= $(30)^{2} + 2(-0.25(412 - 0))$

= 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

y = $200 e^{\frac{x}{1000}}$

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

Now, second derivative of the train is calculated as follows.

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

$\frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}$

Radius of curvature of the train is as follows.

$\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}$

= $\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}$

= 3808.96 m

Now, we will calculate the normal component of the train as follows.

$a_{n} = \frac{\nu^{2}_{B}}{\rho}$

= $\frac{(26.34)^{2}}{3808.96}$

= 0.1822 $m/s^{2}$

The magnitude of acceleration of train is calculated as follows.

a = $\sqrt{(a_{t})^{2} + (a_{n})^{2}}$

= $\sqrt{(-0.25)^{2} + (0.1822)^{2}}$

= $0.309 m/s^{2}$

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is $0.309 m/s^{2}$ .

6 0

3 years ago

What was created 300 years ago by scientist carolus linnaeus

Norma-Jean [14]

<span>the classification system</span>

6 0

3 years ago