When neither air mass displaces the adjacent one, their boundary is called a(n) _______

If a box is pulled with a force of 100 N at an angle of 25

love history [14]

The X and Y components of the force are 90.63 Newton and 42.26 Newton respectively.

Given the following data:

Force = 100 Newton.

Angle of inclination = 25°

To determine the X and Y components of the force:

<h3>The horizontal component (X) of a force:</h3>

Mathematically, the horizontal component of a force is given by this formula:

$F_x = Fcos \theta\\\\F_x =100 \times cos25\\\\F_x =100 \times 0.9063$

Fx = 90.63 Newton.

<h3>The vertical component (Y) of tensional force:</h3>

Mathematically, the vertical component of a force is given by this formula:

$F_y = Fsin \theta\\\\F_y =100 \times sin25\\\\F_y =100 \times 0.4226$

Fy = 42.26 Newton.

Read more on horizontal component here: brainly.com/question/4080400

6 0

2 years ago

Without heat from the Sun, the water cycle would A) reverse. B) not work. C) slow down. D) not be affected.

Tanzania [10]

B) not work ,because the water would freeze

4 0

3 years ago

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Consider a block on frictionless ice. Starting from rest, the block travels a distance din

sweet [91]

Answer:

The distance is now 4d

Explanation:

Mechanical Force

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object.

The acceleration can be calculated by solving for a:

$\displaystyle a=\frac{F}{m}$

Once we know the acceleration, we can calculate the distance traveled by the block as follows:

$\displaystyle d = vo.t+\frac{at^2}{2}$

If the block starts from rest, vo=0:

$\displaystyle d = \frac{at^2}{2}$

Substituting the value of the acceleration:

$\displaystyle d = \frac{\frac{F}{m}t^2}{2}$

Simplifying:

$\displaystyle d = \frac{Ft^2}{2m}$

When a force F'=4F is applied and assuming the mass is the same, the new acceleration is:

$\displaystyle a'=\frac{4F}{m}$

And the distance is now:

$\displaystyle d' = \frac{4Ft^2}{2m}$

Dividing d'/d:

$\displaystyle \frac{d' }{d}=\frac{\frac{4Ft^2}{2m}}{\frac{Ft^2}{2m}}$

Simplifying:

$\displaystyle \frac{d' }{d}=4$

Thus:

d' = 4d

The distance is now 4d

3 0

2 years ago

A spring stretches 0.2 cm per newton of applied force. An object is suspended from the spring and a deflection of 3 cm is observ

siniylev [52]

Answer:

$m=1.53kg$

Explanation:

To solve this problem we use the Hooke's Law:

$F=k*\Delta x$ (1)

F is the Force needed to expand or compress the spring by a distance Δx.

The spring stretches 0.2cm per Newton, in other words:

1N=k*0.2cm ⇒ k=1N/0.2cm=5N/cm

The force applied is due to the weight

$F=mg$

We replace in (1):

$mg=k*\Delta x$

We solve the equation for m:

$m=k*\Delta x/g=5*3/9.81=1.53kg$

7 0

3 years ago

Help needed Can a body be accelerating if it is moving in circle

belka [17]

Answer:

When a body is moving on a circle it is accelerating because centripetal acceleration is always acting on it towards the center.

Please see the attached picture...

From the above diagram,we can say the acceleration is always acting on the body when it moves in a circle.

Hope this helps.......

Good luck on your assignment.......

5 0

3 years ago

Read 2 more answers

When neither air mass displaces the adjacent one, their boundary is called a(n) ________ front?