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12345 [234]
3 years ago
11

A woman pulls on a 6.00-kg crate, which in turn is connected to a 4.00-kg

Physics
2 answers:
djverab [1.8K]3 years ago
7 0

Answer:

B. is subject to a smaller net force but same acceleration.

Explanation:

F = m*a

So because our force applied is constant from the women pulling on the rope which means the acceleration is the same on both the 4kg create and the 6kg create. The only thing that changes here is the mass of the creates, so there is more tension force between the women and the 6kg create then there is between the 4kg create and the 6kg. It takes less force to move the 4kg create therefore the tension force is less between the two creates.

Anastasy [175]3 years ago
7 0

Answer:

Explanation:

We know that when force (F) is applied then both the crate will move together so both have same acceleration.

And from Newton's 2 nd law we have:

F= ma , where m= mass of crate and a is acceleration.

Now for crate of mass 4 kg we have:

F = ma = 4a Newton.    

For the crate of mass 6 kg we have:

F = ma = 6a Newton.    

Hence the crate having less mass is subjected with smaller net force .

Hence from the above we can conclude that the correct answer is option 3 which is : is subjected to a smaller net force and has the same acceleration.

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Which of the following is an ionic compound ? A. Gold
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Among the given choices, the ionic compound is D. Magnesium Chloride (<span>MgCl2</span>) where magnesium has a +2 charge while chloride has a -1 charge. Ionic compounds are chemical compounds comprising of ions held together by electrostatic forces named as ionic bonding.
7 0
3 years ago
A bolt of lightning discharges 9.7 C in 8.9 x 10^-5 s. What is the average current during the discharge?
Anastaziya [24]

Answer: 1.089\times 10^5\ A

Explanation:

Given

Charge discharged Q=9.7\C

time taken t=8.9\times 10^{-5}\ s

Current is given as rate of change of discharge i.e.

\Rightarrow I=\dfrac{Q}{t}\\\\\Rightarrow I=\dfrac{9.7}{8.9\times 10^{-5}}\\\\\Rightarrow I=1.089\times 10^5\ A

Therefore, the average current is 1.089\times 10^5\ A

3 0
2 years ago
A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.
GarryVolchara [31]

a) See free-body diagram in attachment

b) The book is stationary in the vertical direction

c) The net horizontal force is 35 N in the forward direction

d) The net force on the book is 35 N in the forward horizontal direction

e) The acceleration is 8.75 m/s^2 in the forward direction

Explanation:

a)

The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.

From the diagram of this book, we see there are 4 forces acting on the book:

- The applied force, F = 50 N, pushing forward in the horizontal direction

- The frictional force, F_f = 15 N, pulling backward in the horizontal direction (the frictional force always acts in the direction opposite to the motion)

- The weight of the book, W=mg, where m is the mass of the book and g=9.8 m/s^2 is the acceleration of gravity, acting downward. We can calculate its magnitude using the mass of the book, m = 4 kg:

W=(4)(9.8)=39.2 N

- The normal reaction exerted by the desk on the book, N, acting upward, and balancing the weight of the book

b)

The book is in equilibrium in the vertical direction, therefore there is no motion.

In fact, the magnitude of the normal reaction (N) exerted by the desk on the book is exactly equal to the weight of the book (W), so the equation of motion along the vertical direction is

N-W=ma

where a is the acceleration; however, since N = W, this becomes

a=0

And since the book is initially at rest on the desk, this means that there is no motion.

c)

We said there are two forces acting in the horizontal direction:

- The applied force, F = 50 N, forward

- The frictional force, F_f = 15 N, backward

Since they act along the same line, we can calculate their resultant as

\sum F = F - F_f = 50 - 15 = 35 N

and therefore the net force is 35 N in the forward direction.

d)

The net force is obtained as the resultant  of the net forces in the horizontal and vertical direction. However, we have:

- The net force in the horizontal direction is 35 N

- The net force in the vertical direction is zero, because the weight is balanced by the normal reaction

Therefore, this means that the total net force acting on the book is just the net force acting on the horizontal direction, so 35 N forward.

e)

The acceleration of the book can be calculated by using Newton's second law:

\sum F = ma

where

\sum F is the net force

m is the mass

a is the acceleration

Here we have:

\sum F = 35 N (in the forward direction)

m = 4 kg

Therefore, the acceleration is

a=\frac{\sum F}{m}=\frac{35}{4}=8.75 m/s^2 (forward)

Learn more about forces, weight and Newton's second law:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

8 0
3 years ago
An eagle carries a fish up 50 m into the sky using 90 N of force. How much work did the eagle do on the fish? (Work: W = Fd)
Leno4ka [110]
<span>Work: W = Fd. 50(distance) multiplied by 90(force) would equal 4500 J or, answer D</span>
4 0
3 years ago
Read 2 more answers
A certain field line diagram illustrates the electric field due to three particles that carry charges 5.0 μC, -3.0 μC, and -2.0
4vir4ik [10]

Answer:

6

Explanation:

Number of lines emanate from + 5 micro coulomb is 15 .

They terminates at negative charges that means at - 3 micro coulomb and - 2 micro Coulomb.

the electric field lines terminates at - 3 micro Coulomb and - 2 micro Coulomb is in the ratio of 3 : 2.

So the lines terminating at - 3 micro coulomb

                                    = \frac{3}{5}\times 15 = 9

So the lines terminating at - 2 micro coulomb

                                    = \frac{2}{5}\times 15 = 6

So, the number of filed lines terminates at - 2 micro Coulomb are 6.

4 0
3 years ago
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