All categories

2 years ago

What would be the force required to accelerate an object that weighs 35 N up at a rate of .60 m/s^2?

Physics

1 answer:

natta225 [31]2 years ago

3 0

Answer: $2.14\ N$

Explanation:

Given

Force required to accelerate the object at the rate of $0.6\ m/s^2$

The weight of the object is $W=35\ N$

Mass of the object can be identified by dividing the weight by acceleration due to gravity

$\Rightarrow m=\dfrac{W}{g}\\\\\Rightarrow m=\dfrac{35}{9.8}\\\\\Rightarrow m=3.57\ kg$

$\therefore \text{Force F=}m\times a\\\Rightarrow F=3.57\times 0.6\\\Rightarrow F=2.14\ N$

You might be interested in

Anyone who wants to shw booooobs than u can see my body/cqg-fhfh-ptx

zlopas [31]

The answer yes, she gladly will

7 0

2 years ago

What does the relationship between the digestive system and other body system tell you about the body's need for homeostasis, th

Nesterboy [21]

The body continuously tries to maintain homeostasis. Therefore, the digestive system eliminate waste from your body along as providing the body with the nutrients needed. It balances out your every meal.

5 0

2 years ago

Vector A has y-component Ay= +15.0 m . A makes an angle of 32.0 counterclockwise from the +y-axis. What is the x component of A?

Gemiola [76]

Answer:

x-component=-9.3 m

Magnitude of A=17.7m

Explanation:

We are given that

$A_y=+15 m$

$\theta=32^{\circ}$

We have to find the x-component of A and magnitude of A.

According to question

$A_y=\mid A\mid cos\theta$

Substitute the values then we get

$15=\mid A\mid cos32$

$\mid A\mid =\frac{15}{cos32}=\frac{15}{0.848}$

$\mid A\mid=17.7$ m

$tan\theta=\frac{perpendicular\;side}{Base}$

$tan32=\frac{A_x}{A_y}=\frac{A_x}{15}$

$0.62\times 15=A_x$

$A_x=9.3$

The value of x-component of A is negative because the vector A lie in second quadrant.

Hence, the x- component of A=-9.3 m

6 0

2 years ago

Read 2 more answers

What kind of energy is in a rock at the edge of a cliff.

soldi70 [24.7K]

Answer:

kinetic energy

Explanation:

8 0

1 year ago

Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o

polet [3.4K]

Answer:

a. charge C experiences the greatest net force, and charge B receives the smallest net force

b. ratio=9

Explanation:

<u>Electrostatic Force</u>

Two point-charges $q_1$ and $q_2$ separated a distance d will exert a force on each other of a magnitude given by the Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{r^2}$

Where k is the proportional constant of value

$k=9*10^9\ N.m^2/c^2$

The diagram provided in the question shows four identical charges (let's assume their value is Q) separated by identical distance (of value d). The force between the charges next to others is

$\displaystyle F_1=\frac{k\ Q\ Q}{d^2}$

$\displaystyle F_1=\frac{k\ Q^2}{d^2}$

The force between charges separated 2d is

$\displaystyle F_2=\frac{k\ Q^2}{(2d)^2}$

$\displaystyle F_2=\frac{k\ Q^2}{4d^2}$

And the force between the charges A and D is

$\displaystyle F_3=\frac{k\ Q^2}{(3d)^2}$

$\displaystyle F_3=\frac{k\ Q^2}{9d^2}$

Now, let's analyze each charge and the force applied to them by the others

Let's recall equally signed charges repel each other and differently signed charges attrach each other

Charge A. It receives force to the left from B and C and to the right from D

$\displaystyle F_A=-F_1-F_2+F_3=-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}$

$\displaystyle F_A=\frac{k\ Q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

$\displaystyle F_A=-\frac{41}{36}F_1$

Charge B. It receives force to the right from A and D and to the left from C

$\displaystyle F_B=F_1-F_1+F_2=\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{4d^2}$

$\displaystyle F_B=\frac{1}{4}F_1$

Charge C. It receives forces to the right from all charges.

$\displaystyle F_C=F_2+F_1+F_1=\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{d^2}$

$\displaystyle F_C=\frac{9}{4}F_1$

Charge D. It receives forces to the left from all charges

$\displaystyle F_D=-F_3-F_2-F_1=-\frac{k\ Q^2}{9d^2}-\frac{k\ Q^2}{4d^2}-\frac{k\ Q^2}{d^2}$

$\displaystyle F_D=-\frac{49}{36}F_1$

Comparing the magnitudes of each force is just a matter of computing the fractions

$\displaystyle \frac{41}{36}=1.13,\ \frac{1}{4}=0.25,\ \frac{9}{4}=2.25,\ \frac{49}{36}=1.36$

We can see the charge C experiences the greatest net force, and charge B receives the smallest net force

The ratio of the greatest to the smallest net force is

$\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}=9$

The greatest force is 9 times the smallest net force

7 0

3 years ago