Can someone help me with this? (The answer marked in red text is a incorrect answer)

Why is it more difficult to lean over and push a heavy box across the floor than it is to attach a rope and pull the box at the

melisa1 [442]

Answer:

Case 1: Pushing Diagram 1

Leaning over and Pushing the heavy box from the floor, the push will be divided in to two parts, one is horizontal that can help the box move, and one is vertically downwards, which increases the downward force of the heavy object (an addition to the gravity) and thus increases friction, making it very hard to push. When you push at certain angle, you are exhibiting two forces as shown in diagram 1.

Horizontal force acting along the plane.
Vertical force downward perpendicular to the surface.

Case 2: Pulling Diagram 2

Pulling on a rope similar object at the same angle, the pull can be divided into two parts, one is horizontal that can help the box move, and one is vertically upwards, which decreases the downwards force of the box (a subtraction in the gravity) and thus decreases friction, making it very easy to pull. When you pull at a certain angle, you are exhibiting two forces as shown in diagram 2.

Horizontal force acting along the plane.
Vertical force upward perpendicular to the surface.

So, in the case of pushing, it adds an extra weight on the object, which results in difficulty to push that object at the same angle. In case of pulling, the upward perpendicular force, it tries to lift the object upward and divided the weight partially. Thus making it easier to move the object at same angle.

8 0

3 years ago

John throws a ball with a velocity of 30 m/s at an angle of 60 degrees. What is the horizontal component of the velocity?

umka21 [38]

The horizontal component of the velocity is equal to: D. 15 m/s.

Given the following data:

Velocity = 30 m/s

Angle = 60°

To determine the horizontal component of the velocity:

The horizontal component of the velocity represents the influence of velocity in displacing an object or projectile in the horizontal direction.

Mathematically, the horizontal component of velocity is given by the formula:

$V_x = Vcos(\theta)$

Substituting the given parameters into the formula, we have;

$\\\\V_x = 30cos(60)\\\\V_x = 30 \times 0.5$

Horizontal component, Vx = 15 m/s

Read more on horizontal component here: brainly.com/question/24681896

6 0

3 years ago

a body initially at rest, starts moving with a constant acceleration of 2ms-2 .calculate the velocity acquired and the distance

Marta_Voda [28]

a) 10 m/s

b) 25 m

Explanation:

a)

The body is moving with a constant acceleration, therefore we can solve the problem by using the following suvat equation:

$v=u+at$

where

u is the initial velocity

v is the final velocity

a is the acceleration

t is the time

For the body in this problem:

u = 0 (the body starts from rest)

$a=2 m/s^2$ is the acceleration

t = 5 s is the time

So, the final velocity is

$v=0+(2)(5)=10 m/s$

b)

In this second part, we want to calculate the distance travelled by the body.

We can do it by using another suvat equation:

$v^2-u^2=2as$

where

u is the initial velocity

v is the final velocity

a is the acceleration

s is the distance travelled

Here we have

u = 0 (the body starts from rest)

$a=2 m/s^2$ is the acceleration

v = 10 m/s is the final velocity

Solving for s,

$s=\frac{10^2-0^2}{2(2)}=25 m$

3 0

3 years ago

The current through a heater is 12 A when it is plugged into a

jasenka [17]

Answer:

A. 10Ω

Explanation:

7 0

3 years ago

Unlike the news media, which may be biased, science reporting can be depended on to be objective and not influenced by governmen

sveta [45]

The answer to the given statement above would be TRUE. Yes it is true that unlike the news media which may be biased, science reporting can be depended on to be objective and not influenced by government policy and the interests of big corporations. Hope this answers your question.

4 0

3 years ago