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3 years ago

When a 9.9 kg object is in free fall, it feels a force of 97.02 N. what is its acceleration?

Physics

1 answer:

Free_Kalibri [48]3 years ago

6 0

The acceleration of an object in free fall is $9.8 m/s^2$

Explanation:

We can solve the problem by using Newton's second law of motion, which states that:

$F=ma$

where

F is the net force on an object

m is the mass of the object

a is its acceleration

For the object in this problem, we have:

m = 9.9 kg is its mass

F = 97.02 N is the net force on the object

So, we can rearrange the equation to find its acceleration:

$a=\frac{F}{m}=\frac{97.02}{9.9}=9.8 m/s^2$

This value is actually known as acceleration of gravity (g), and it is the acceleration that any object in free fall has near the Earth's surface, regardless of its mass.

Learn more about Newton's second law here:

brainly.com/question/3820012

And about free fall:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

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a net force of 219 N is exterted on a rock. the rock has an acceleration of 3m/s^2 due to this force. what is the mass of the ro

Sonbull [250]

Answer:

Explanation:

The mass of the object can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{219}{3} \\$

We have the final answer as

Hope this helps you

6 0

3 years ago

Which force is represented by the arrow at

shutvik [7]

A is pulling the block straight down toward the center of the Earth, no matter what the slope of the plane may be. A is the force of gravity.

The directions of B and C both depend on the slope of the plane.

B is a force that's parallel to the plane, pulling the block UP the plane. B is the force of friction.

C is a force perpendicular to the plane, preventing the block from falling down through the plane. C is the normal force.

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2 years ago

Read 2 more answers

510 g squirrel with a surface area of 935 cm2 falls from a 4.8-m tree to the ground. Estimate its terminal velocity. (Use the dr

Yuki888 [10]

Answer:

The terminal velocity is $v_t =17.5 \ m/s$

Explanation:

From the question we are told that

The mass of the squirrel is $m_s = 50\ g = \frac{50}{1000} = 0.05 \ kg$

The surface area is $A_s = 935 cm^2 = \frac{935}{10000} = 0.0935 \ m^2$

The height of fall is h =4.8 m

The length of the prism is $l = 23.2 = 0.232 \ m$

The width of the prism is $w = 11.6 = 0.116 \ m$

The terminal velocity is mathematically represented as

$v_t = \sqrt{\frac{2 * m_s * g }{\dho_s * C * A } }$

Where $\rho$ is the density of a rectangular prism with a constant values of $\rho = 1.21 \ kg/m^3$

$C$ is the drag coefficient for a horizontal skydiver with a value = 1

A is the area of the prism the squirrel is assumed to be which is mathematically represented as

$A = 0.116 * 0.232$

$A = 0.026912 \ m^2$

substituting values

$v_t = \sqrt{\frac{2 * 0.510 * 9.8 }{1.21 * 1 * 0.026912 } }$

$v_t =17.5 \ m/s$

7 0

3 years ago

Which statement is true about effective nuclear charge?a. Effective nuclear charge decreases as we move to the right across a ro

MissTica

Answer:

Option b. Effective nuclear charge increases as we move to the right across a row in the periodic table

Explanation:

The <em>effective nuclear charge </em>is a measure of how strong the protons in the nucleus of an atom attract the outermost electrons of such atom.

The <em>effective nuclear charge</em> is the net positive charge experienced by valence electrons and is calculated (as an approximation) by the equation: Zeff = Z – S, where Z is the atomic number and S is the number of shielding electrons.

The shielding electrons are those electrons in between the interesting electrons and the nucleus of the atom.

Since the shielding electrons are closer to the nucleus, they repel the outermost electrons and so cancel some of the attraction exerted by the positive charge of the nucleus, meaning that the outermost electrons feel less the efect of attraction of the protons. That is why in the equation of Zeff, the shielding electrons (S) subtract the total from the atomic number Z.

The <em>effective nuclear charge</em>, then, is responsible for some properties and trends in the periodic table. Here, you can see how this explains the trend of the atomic radius (size of the atom) accross a row in the periodic table.

As the<em> effective nuclear charge</em> is larger, in a same row of the periodic table, the shielding effect is lower, the outermost electrons are more strongly attracted by the nucleus, and the size of the atoms decrease. That is why as we move to the right in the periodic table, the size of the atoms decrease.

3 0

3 years ago

To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel

alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0

3 years ago