1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
marta [7]
2 years ago
14

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]2 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

You might be interested in
If it requires 3.0 J of work to stretch a particular spring by 2.1 cm from its equilibrium length, how much more work will be re
-BARSIC- [3]

Answer:

Work done = 13605.44

Explanation:

Data provided in the question:

For elongation of 2.1 cm (0.021 m) work done by the spring is 3.0 J

The relation between Energy (U) and the elongation (s) is given as:

U = \frac{1}{2}kx^2   ................(1)

where,

k is the spring constant

on substituting the valeus in the above equation, we get

3.0 = \frac{1}{2}k\times0.021^2

or

k = 13605.44 N/m

now

for the elongation x = 2.1 + 4.1 = 6.2 cm = 0.062 m

using the equation 1, we have

U = \frac{1}{2}\times13605.44\times (0.062)^2

or

U = 26.149 J

Also,

Work done = change in energy

or

W = 26.149 - 3.0 = 23.149 J

4 0
2 years ago
A laser emits two wavelengths (λ1 = 420 nm; λ2 = 630 nm). When these two wavelengths strike a grating with 450 lines/mm, they pr
Westkost [7]

A) Order of the first laser: 3, order of the second laser: 2

B) The overlap occurs at an angle of 34.9^{\circ}

Explanation:

A)

The formula that gives the position of the maxima (bright fringes) for a diffraction grating is

d sin \theta = m \lambda

where

d is spacing between the lines in the grating

\theta is the angle of the maximum

m is the order of diffraction

\lambda is the wavelength of the light

For laser 1,

d sin \theta = m_1 \lambda_1

For laser 2,

d sin \theta = m_2 \lambda_2

where

\lambda_1 = 420 nm\\\lambda_2 = 630 nm

Since the position of the maxima in the two cases overlaps, then the term d sin \theta on the left is the same for the two cases, therefore we can write:

m_1 \lambda_1 = m_2 \lambda_2\\\frac{m_1}{m_2}=\frac{\lambda_2}{\lambda_1}=\frac{630}{420}=\frac{3}{2}

Therefore:

m_1 = 3

m_2 = 2

B)

In order to find the angle at which the overlap occurs, we use the 1st laser situation:

d sin \theta = m_1 \lambda_1

where:

N = 450 lines/mm = 450,000 lines/m is the number of lines per unit length, so the spacing between the lines is

d=\frac{1}{N}=\frac{1}{450,000}=2.2\cdot 10^{-6} m

m_1 = 3 is the order of the maximum

\lambda_1 = 420 nm = 420\cdot 10^{-9} m is the wavelength of the laser light

Solving for \theta, we find the angle of the maximum:

sin \theta = \frac{m_1 \lambda_1}{d}=\frac{(3)(420\cdot 10^{-9})}{2.2\cdot 10^{-6}}=0.572

So the angle is

\theta=sin^{-1}(0.572)=34.9^{\circ}

Learn more about diffraction:

brainly.com/question/3183125

#LearnwithBrainly

5 0
3 years ago
Rjejdndjoskdnrksjdjsjsne
Ilia_Sergeevich [38]
Rjejdndjoskdnrksjdjsjsne
4 0
3 years ago
A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providi
OleMash [197]

Answer:

The sled required 9.96 s to travel down the slope.

Explanation:

Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.

Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).

The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.

The magnitude of the friction force is calculated as follows:

Ff = μ · Fn

where :

μ = coefficient of kinetic friction

Fn =  normal force

The normal force has the same magnitude as the y-component of the gravity force:

Fgy = Fg · cos 30º = m · g · cos 30º

Where

m = mass

g = acceleration due to gravity

Then:

Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º

Fgy = 744 N

Then, the magnitude of Fn is also 744 N and the friction force will be:

Ff = μ · Fn = 0.151 · 744 N = 112 N

The x-component of Fg, Fgx, is calculated as follows:

Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N

The resulting force, Fr, will be the sum of all these forces:

Fw + Fgx - Ff = Fr

(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).

Fr = 161 N + 430 N - 112 N = 479 N

With this resulting force, we can calculate the acceleration of the sled:

F = m·a

where:

F = force

m = mass of the object

a = acceleration

Then:

F/m = a

a = 479N/87.7 kg = 5.46 m/s²

The equation for the position of an accelerated object moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.

x = 1/2· a ·t²

Let´s find the time at which the position of the sled is 271 m:

271 m = 1/2 · 5.46 m/s² · t²

2 · 271 m / 5.46 m/s² = t²

<u>t = 9.96 s </u>

The sled required almost 10 s to travel down the slope.

8 0
3 years ago
The net force acting on a 5 kg discus is 50 n. what is the acceleration of the discus
Lyrx [107]
Answer: 10 m/s^2

Explanation:

1) The second law of Newton gives the definition and formula to calculate the net force:

Net force acting on an object = mass * acceleration.

2) From that, when you know the net force acting of the object and its mass, you can solve for the acceleration:

acceleration = Net force / mass

acceleration = 50 N / 5 kg = 10 m/s^2, which is the answer.
8 0
3 years ago
Other questions:
  • Interactive LearningWare 8.1 reviews the approach that is necessary for solving problems such as this one. A motorcyclist is tra
    12·1 answer
  • Beginning with earth, summarize the structure of the universe. Also Andromeda galaxy is located appromaxily 2.5 million light-ye
    11·1 answer
  • Which phrase best describes sir isaac newton's contributions to modern science and therefore to the industrial revolution?
    14·1 answer
  • Edit question A 37-cm-long wire of linear density 18 g/m vibrating at its second mode, excites the third vibrational mode of a t
    11·1 answer
  • Why does a small pebble sin in water?
    7·1 answer
  • Whats evidence do scientists use to support the theory of plate tectonics
    7·1 answer
  • How are mass and inertia related
    8·1 answer
  • Which of the following are true about electrons? a.) They are positively charged b.) They are negatively charged c.) They are fo
    13·2 answers
  • How was the scientific revolution important to the enlightenment
    9·1 answer
  • As a light wave strikes a shiny surface like a piece of polished metal, it will be
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!