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skad [1K]
3 years ago
6

3. A worker pushes a box across the floor to the right at a constant speed with a force of 25N. What

Physics
1 answer:
WITCHER [35]3 years ago
6 0

Answer:

Friction between the box and the floor is 25N to the left.

Explanation:

According to Newton's second law of motion, the net force acting on an object is equal to the produce between the object's mass and its acceleration:

F_{net}=ma

where

m is the mass of the object

a is its acceleration

In this problem, we have two forces acting on the object:

- The applied force, F = 25 N, to the right

- The force of friction F_f, opposing the motion of the box, so to the left

So we can write the net force as

F_{net}=F-F_f

Also, we know that the box is moving at constant speed: this means its acceleration is zero, so

a=0

Therefore

F_{net}=0

WHich means:

F-F_f=0

And therefore,

F_f=F=25 N

which means that the force of friction is also 25 N.

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6 0
3 years ago
The action of two forces. One is a forward force of 1157 N provided by traction between the wheels and the road. The other is a
Pie

Complete question

A 2700 kg car accelerates from rest under the action of two forces. one is a forward force of 1157 newtons provided by traction between the wheels and the road. the other is a 902 newton resistive force due to various frictional forces. how far must the car travel for its speed to reach 3.6 meters per second? answer in units of meters.

Answer:

The car must travel 68.94 meters.

Explanation:

First, we are going to find the acceleration of the car using Newton's second Law:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

with m the mass , a the acceleration and \sum\overrightarrow{F} the net force forces that is:

(F-f) (2)

with F the force provided by traction and f the resistive force:

(2) on (1):

(F-f)=ma

solving for a:

a=\frac{F-f}{m} =\frac{1157N-902N}{2700kg} =0.094\frac{m}{s^{2}}

Now let's use the Galileo’s kinematic equation

Vf^{2}=Vo^{2}+2a\varDelta x (3)

With Vo te initial velocity that's zero because it started from rest, Vf the final velocity (3.6) and \varDelta x the time took to achieve that velocity, solving (3) for \varDelta x:

\varDelta x= \frac{Vf^{2}}{2a} = t= \frac{(3.6\frac{m}{s})^2}{2*0.094\frac{m}{s^{2}}}

t=68.94 m

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3 years ago
How are metals identified i the periodic table??
zaharov [31]

Answer:

okay here is a thing I learned when I was younger in my middle school:

Explanation:

my teacher would tell me that metals are considered a weak metals are on the left side and the good metals are located on the right side because the only way I remembered was the right means it is really strong and the left is weak and not that supportive. but I think that's how I still think it is or other people may have their own opinions. but hope this helped out with your question!

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3 years ago
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A short-wave radio antenna is supported by two guy wires, 155 ft and 175 ft long. Each wire is attached to the top of the antenn
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Answer:

163.8 ft

Explanation:

In triangle ABD

AB = 155 ft

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Using Pythagorean theorem in triangle ADC

AC^{2} = AD^{2} + DC^{2} \\175^{2} = 148^{2} + DC^{2} \\DC = 93.4 ft

d = distance between the anchor points

distance between the anchor points is given as

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There are two atoms of potassium bonded to one atom of sulfur.
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