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Reptile [31]
2 years ago
12

As an object moves along the x axis, many measurements are made of its position, enough to generate a smooth, accurate graph of

x versus t. Which of the following quantities for the object cannot be obtained from this graph alone? (Select all that apply.)
the displacement during some time interval
the speed at any instant
the velocity at any instant
the average velocity during some time interval
the acceleration at any instant
Physics
2 answers:
Bas_tet [7]2 years ago
6 0
The best and most correct answers among the choices provided by your question are he second and third choices. 
<span>The velocity at any instant the average velocity during some time interval cannot be obtained from the graph alone.</span>


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
vladimir1956 [14]2 years ago
3 0

Answer:

the acceleration at any instant

Explanation:

1). the displacement during some time interval

We can determine the displacement of the object from the graph because we know that displacement is defined as the change in position of the object and from graph change in position can be obtained easily

2). the speed at any instant

3). the velocity at any instant

speed at any instant of time and velocity at any instant of time must be same in the magnitude and we can determine it by the slope of the graph

As we know that instantaneous velocity is defined as

v = \frac{dx}{dt}

so slope of the graph at any instant will give an idea about instantaneous speed of velocity

4). the average velocity during some time interval

To find average velocity we need the displacement of object at a given interval of time and ratio of this displacement and time interval is known as average velocity

So we can find it from graph

5). the acceleration at any instant

To find the instantaneous acceleration we need to find

a = \frac{dv}{dt}

this is not possible to find from x-t graph because in order to find this slope we need to find velocity time graph.

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Grace [21]

Complete question is;

A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:

(a) the work done by the force of gravity on the projectile,

(b) the change in kinetic energy of the projectile since it was fired, and

(c) the final kinetic energy of the projectile.

(d) Are any of the answers changed if the initial angle is changed?

Answer:

A) W = mgh

B) ΔKE = mgh

C) K2 = mgh + ½mv_o²

D) No they wouldn't change

Explanation:

We are expressing in terms of m, v0​, h, and g. They are;

m is mass

v0 is initial velocity

h is height of projectile fired

g is acceleration due to gravity

A) Now, the formula for workdone by force of gravity on projectile is;

W = F × h

Now, Force(F) can be expressed as mg since it is force of gravity.

Thus; W = mgh

Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.

Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.

B) Change in kinetic energy is simply;

ΔKE = K2 - K1

Where K2 is final kinetic energy and K1 is initial kinetic energy.

However, from conservation of energy, we now that change in kinetic energy = change in potential energy.

Thus;

ΔKE = ΔPE

ΔPE = U2 - U1

U2 is final potential energy = mgh

U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.

Thus;

ΔKE = ΔPE = mgh

Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.

C) As seen in B above,

ΔKE = ΔPE

Thus;

½mv² - ½mv_o² = mgh

Where final kinetic energy, K2 = ½mv²

And initial kinetic energy = ½mv_o²

Thus;

K2 = mgh + ½mv_o²

Similar to a and B above, this will not change even if initial angle is changed

D) All of the answers wouldn't change because their equations don't depend on the angle.

5 0
2 years ago
A majorette in a parade is performing some acrobatic twirlings of her baton. Assume that the baton is a uniform rod of mass 0.12
den301095 [7]

Question

Initially, the baton is spinning about a line through its center at angular velocity 3.00 rad/s.  What is its angular momentum? Express your answer in kilogram meters squared per second.

Answer:

0.0192 kgm^{2}/s

Explanation:

The angular momentum L of the baton moving about an axis perpendicular to it, passing through the center of the baton is,

L = \frac{1}{{12}}m{l^2}\omega

Here, l is the length of the baton.

Substitute 0.120 kg for m, 3 rads/s for \omega[\tex] and 0.8 m for l [tex]\begin{array}{c}\\L = \frac{1}{{12}}m{l^2}\omega \\\\ = \frac{1}{{12}}\left( {0.120{\rm{ kg}}} \right){\left( {{\rm{80}}{\rm{.0 cm}}} \right)^2}{\left( {\frac{{1 \times {{10}^{ - 2}}{\rm{m}}}}{{1{\rm{ cm}}}}} \right)^2}\left( {{\rm{3}}{\rm{.00 rad/s}}} \right)\\\\ = 0.0192{\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\\\end{array}

5 0
2 years ago
Read 2 more answers
A 0.45 Caliber bullet (m = 0.162 kg) leaving the muzzle of a gun at 860
Charra [1.4K]

Answer:139.32kgm/s

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As the external magnetic field decreases, an induced current flows in the coil. what is the direction of the induced magnetic fi
GrogVix [38]
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Mrac [35]

Answer: The final volume V₂ of the container is  0.039 m³.

Explanation:

Since the temperature is constant, the gas would expand isothermally.

For isothermal expansion,

P₁V₁=P₂V₂

Where, P₁ and P₂ are the initial and final pressure and V₁ and V₂ are initial and final volume.

It is given that:

V₁ = 0.0250 m³

P₁ = 1.5 × 10⁶ Pa

P₂ = 0.950 × 10⁶ Pa

V₂ = ?

⇒ 1.5 × 10⁶ Pa × 0.0250 m³ = 0.950 × 10⁶ Pa × V₂

⇒V₂ = 0.039 m³

Hence, the final volume V₂ of the container is  0.039 m³.

4 0
1 year ago
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