A train's mass is 500 kg and it's acceleration is 5 m/s. What is the net force on the train

A stone is thrown from a cliff and lands in the sea. Air resistance is negligible. Which statement is correct whilst the stone i

erik [133]

Since the stone is travelling, the displacement cannot be constant. This leaves the last 2 choices as possibilities. Now consider the acceleration that can act on the stone. Acceleration due to gravity is the only one. That means the horizontal component of the stone has to be constant since there is no acceleration in the horizontal direction.

4 0

3 years ago

A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points

mr Goodwill [35]

Answer:

Approximately $11.0\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81 \; \rm N \cdot kg^{-1}$ , and that the tabletop is level.)

Explanation:

Weight of the book:

$W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N$ .

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, $F(\text{normal force}) \approx 10.202\; \rm N$ .

As a side note, the $F_N$ and $W$ on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction $F(\text{kinetic friction})$ on it will be equal to

$\mu_{\rm k}$ , the coefficient of kinetic friction, times
$F(\text{normal force})$ , the normal force that's acting on it.

That is:

$\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}$ .

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that $15.0\; \rm N$ applied force. The net force on the book shall be:

$\begin{aligned}& F(\text{net force}) \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}$ .

Apply Newton's Second Law to find the acceleration of this book:

$\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}$ .

6 0

2 years ago

What is the angle of refraction when a ray of light passes frm ethanol to air

77julia77 [94]

Answer: Your answer is<u> 1.36.</u>

Hope this helps!

5 0

2 years ago

You apply a horizontal force of 25N to push a shopping cart across the parking lot at a constant velocity. a) what is the net fo

AlekseyPX

(a) The net force on the shopping cart is zero.

(b) The the force of friction on the shopping cart is 25 N.

(c) When same force is applied to the shopping cart on a wet surface, it will move faster.

<h3>Net force on the shopping cart</h3>

The net force on the shopping cart is calculated as follows;

F(net) = F - Ff

where;

F is the applied force
Ff is the frictional force

ma = F - Ff

where;

a is acceleration of the cart
m is mass of the cart

at a constant velocity, a = 0

0 = F - Ff

F(net) = 0

F = Ff = 25 N

Net force is zero, and frictional force is equal to applied force.

<h3>On wet surface</h3>

Coefficient of kinetic friction of solid surface is greater than that of wet surface.

Since frictional force limit motion, when the frictional force is smaller, the object tends to move faster.

Thus, the cart will move faster on a wet surface due to decrease in friction.

Learn more about frictional force here: brainly.com/question/24386803

#SPJ1

4 0

2 years ago

3. Neha travels 4 m towards the north, 3m towards the east, and then 7 m towards south, find the displacement and total distance

Digiron [165]

Answer:

Explanation:

Explanation: total displacement =3√2m. and total distance covered=14m. I hope this is right and helps u.

7 0

3 years ago