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3 years ago

How does the vertical acceleration at point a compare to the vertical acceleration at point c?

Physics

2 answers:

krek1111 [17]3 years ago

5 0

Both accelerations are equal to free-fall acceleration.

<h3>Further explanation</h3>

This is a matter of projectile motion, i.e., an object moving in two dimensions. The shaped is called a parabola.
The only forces acting during its flight are gravity and friction. In many situations, air resistance can be ignored.
It is moving horizontally and vertically at the same time but recalls this, the horizontal dan vertical components of the motion are independent of one another. Assuming the gravitational force is constant.

<u>Horizontal component</u>

There are no forces in the horizontal direction, so there is no horizontal acceleration, i.e., a = 0 m/s². This means the horizontal velocity must be constant along the track.

<u>Vertical component</u>

There is naturally a constant vertical force acting down, so there is a constant vertical acceleration. The value of vertical acceleration is similar to gravity (g).

See the attachment for the motion diagram.

Therefore the vertical acceleration at point A and point B is the same as free-fall acceleration.

<h3>Learn more</h3>

Determine the acceleration of the stuffed bear brainly.com/question/6268248
Particle's speed and direction of motion brainly.com/question/2814900
The energy density of the stored energy brainly.com/question/9617400

Keywords: the motion diagram, showing the trajectory, projectile motion, parabola, horizontal, vertical component, gravity, how does the vertical acceleration at point A, compare, point C, equal to free-fall acceleration

Anika [276]3 years ago

4 0

The vertical accelerations of the point a and point c are equal. because the vertical acceleration is due to the acceleration due to gravity. Acceleration due to gravity<span> may refer to: Gravitational acceleration, the acceleration caused by the gravitational attraction of massive bodies in general, which is equal to 9.8 m/s^2</span>

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The difference in weight is due to the displacement of water (the buoyancy of water is acting on the athlete thus giving her smaller weight).<span>

The amount of weight displaced or the amount of buoyant force is: </span>

Fb = 690 N - 48 N

Fb = 642 N

From newtons law, F = m*g. Using this formula, we can calculate for the mass of water displaced:

m of water displaced = 642N / 9.8m/s^2

m of water displaced = 65.5 kg

Assuming a water density of 1 kg/L, and using the formula volume = mass/density:

V of water displaced = 65.5kg / 1kg/L = 65.5 L

The volume of water displaced is equal to the volume of athlete. Therefore:

V of athlete = 65.5 L

The mass of athlete can also be calculated using, F = m*g

m of athlete = 690 N/ 9.8m/s^2

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3 years ago

Read 2 more answers

Magnets are not hammered or rubbed<br> why?

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2 years ago

You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov

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Answer:

<em>a. The frequency with which the waves strike the hill is 242.61 Hz</em>

<em>b. The frequency of the reflected sound wave is 254.23 Hz</em>

<em>c. The beat frequency produced by the direct and reflected sound is </em>

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

$f_{1} = f[\frac{v_{s} }{v_{s} -v} ]$ ..............................1

where $f_{1}$ is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = $36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}$

v = 16.27 m/s

Substituting into equation 1 we have

$f_{1}$ = $231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})$

$f_{1}$ = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

$f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]$

where $f_{2}$ is the frequency of the reflected sound;

$f_{1}$ is the frequency which the wave strikes the hill = 242.61 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

$f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]$

$f_{2}$ = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound and the reflected sound. This can be obtained as follows;

Δf = $f_{2}$ - $f_{1}$

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

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3 years ago

If v =5.00 meters/second and makes an angle of 60 with the negative direction of the y-axis coalculate the possible values of vx

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An angle of 60 degrees with the negative y-axis could mean 60 degrees clockwise or counterclockwise, which translates to two possible angles (starting from the positive x-axis and moving counterclockwise) of 210 degrees or 330 degrees.

Then the horizontal component $v_x$ of a velocity vector $\mathbf v$ with magnitude $5.00\,\dfrac{\mathrm m}{\mathrm s}$ could be one of two expressions:

$v_x=\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\cos210^\circ=-4.33\,\dfrac{\mathrm m}{\mathrm s}$

$v_x=\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\cos330^\circ=4.33\,\dfrac{\mathrm m}{\mathrm s}$

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3 years ago

The correct formula for copper (I) bromide is —

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Answer:

The correct formula is the first one.

Explanation:

Copper has two valences I and II, in this example, it's mentioned that copper's valence is I.

Then we have to look in all the formulas for one formula where copper has that valence, and that valence is in formula 1 and 3. Number 4 is discarted that formula is incorrect, that formula doesn't exist.

Number 2 is also discarted because in this formula Cu has a valence of 2.

Number three is discarted because here Bromide has a valence of 2 and that is incorrect, Bromide's valence is 1.

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3 years ago