Ask question

Ask question

All categories

tekilochka [14]

3 years ago

14

What is the relationship between the angle of an incline and the acceleration of an object moving down the incline? How would yo

u determine the acceleration using the angle?

1 answer:

iren2701 [21]3 years ago

5 0

Answer:

See Explanation

Explanation:

The relationship between angle of an incline and the acceleration of an object moving down the incline.

As the angle of an incline increases, so does the acceleration of the body moving down the incline increases, resolving the force acting on an inclined object

Parallel force = mgsin, perpendicular = mgcosΘ

With th weigh component 'mg' of the parallel force accounting for the acceleration of the body down the incline.

mgsinΘ = ma

Fnet = ma

B.) From Fnet = ma

Fnet = ma

a = Fnet / m

Where Fnet = Net force = mgsinΘ, a = acceleration

You might be interested in

Two identical satellites orbit the earth in stable orbits. one satellite orbits with a speed v at a distance r from the center o

wariber [46]

The available options are: (found the complete text on internet)
A- at a distance less than r
B- at a distance equal to r
C- at a distance greater than r

Solution:
The correct answer is C) at a distance greater than r.

In fact, the gravitational attraction between the satellite and the Earth provides the centripetal force that keeps the satellite in circular orbit, so we can write
$G \frac{Mm}{r^2}=m \frac{v^2}{r}$
where the term on the left is the gravitational force, while the term on the right is the centripetal force, and where
G is the gravitational constant
M is the Earth mass
m is the satellite mass
r is the distance of the satellite from the Earth's center
v is the satellite speed

Re-arranging the equation, we get
$r= \frac{GM}{v^2}$
and we see from this formula that, if the second satellite has a speed less than the speed v of the first satellite, it means that the denominator of the fraction is smaller, and so r is larger for the second satellite.

7 0

3 years ago

Two massless bags contain identical bricks, each brick having a mass M. Initially, each bag contains four bricks, and the bags m

stepladder [879]

Answer: $F_{2}=\frac{3}{4}F_{1}$

Explanation:

According to Newton's law of universal gravitation:

$F=G\frac{m_{1}m_{2}}{r^2}$

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

In this case we have two situations:

1) Two bags with masses $4M$ and $4M$ mutually exerting a gravitational attraction $F_{1}$ on each other:

$F_{1}=G\frac{(4M)(4M)}{r^2}$ (1)

$F_{1}=G\frac{16M^2}{r^2}$ (2)

$F_{1}=16\frac{GM^2}{r^2}$ (3)

2) Two bags with masses $2M$ and $6M$ mutually exerting a gravitational attraction $F_{2}$ on each other (assuming the distance between both bags is the same as situation 1):

$F_{2}=G\frac{(2M)(6M)}{r^2}$ (4)

$F_{2}=G\frac{12M^2}{r^2}$ (5)

$F_{2}=12\frac{GM^2}{r^2}$ (6)

Now, if we isolate $\frac{GM^2}{r^2}$ from (3):

$\frac{F_{1}}{16}=\frac{GM^2}{r^2}$ (7)

Substituting $\frac{GM^2}{r^2}$ found in (7) in (6):

$F_{2}=12(\frac{F_{1}}{16})$ (8)

$F_{2}=\frac{12}{16}F_{1}$ (9)

Simplifying, we finally get the expression for $F_{2}$ in terms of $F_{1}$ :

$F_{2}=\frac{3}{4}F_{1}$

5 0

3 years ago

A 5. 3 ft -ft-tall girl stands on level ground. The sun is 30 ∘ above the horizon. How long is her shadow?

slamgirl [31]

The trigonometric ratios are sine, cosine and tangent. We can use cosine to find the length of the shadow from the calculation as 1.7 ft.

<h3>What is trigonometric ratio?</h3>

The term trigonometric ratio has to do with sine, cosine and tangent which are used to solve problems that involve the right angled triangle.

Using the geometry of the right angle triangle, we can find the tangent of the angle 30 in order to obtain the length of the shadow.

Tan 30° = opp/3

opp = 3Tan 30°

opp = 1.7 ft

The length of the girl's shadow from the calculation is 1.7 ft.

Learn more about tangent: brainly.com/question/14022348

8 0

2 years ago

A toy gun uses a spring to project a 4.5-g soft rubber sphere horizontally. The spring constant is 8.0 N/m, the barrel of the gu

marishachu [46]

Answer:

1.93 m/s

Explanation:

Parameters given:

Mass = 4.5g = 0.0045kg

Spring constant = 8.0 N/m

Length of barrel = 13 cm = 0.013m

Frictional force = 0.035N

Compression = 5.8 cm = 0.058m

First, we find the P. E. stored in the spring:

P. E. = ½*k*x²

P. E. = ½ * 8 * 0.058² = 0.013J

Then, we find the work done by the frictional force while the sphere is leaving the barrel of the gun:

Work = Force * distance

The distance here is the length of the barrel.

Work = 0.035 * 0.13 = 0.0046 J

The kinetic energy of the sphere can now be found:

K. E. = P. E. - Work done

K. E. = 0.013 - 0.0046 = 0.0084J

We can now find the speed using the formula for K. E.:

K. E. = ½*m*v²

0.0084 = ½ * 0.0045 * v²

v² = 0.0084/0.00255 = 3.733

=> v = 1.93 m/s

4 0

3 years ago

Read 2 more answers

C. A helium atom has a diameter of approximately 9.8 • 10-11 meters. What is the diameter of a helium atom in nanometers? Given

Studentka2010 [4]

$1\ nm = 10^{-9} m\\\\1\ m =\frac{1\ nm}{10^{-9}}$

Since the diameter of helium atom is approximately $9.8\times 10^{-11}\ m$ , therefore the diameter of helium atom in nano meter,

$9.8\times 10^{-11}\ m=9.8\times 10^{-11} \times\frac{1\ nm}{10^{-9}}=0.098\ nm$

4 0

3 years ago