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3 years ago

Hector drew the diagram below to illustrate projectile motion.

Physics

2 answers:

Serjik [45]3 years ago

6 0

Answer

He should make the arrow for “Path” curve downward.

Explanation

The force of gravity is usually directly downward. So from the diagram, it is correctly labelled.

If the object was given a horizontal force, the direction of the inertial is also correct. Inertial is the force the resist the change of state of motion.

What Hector should change is the path followed by the object. It will be a curve not a straight line as it is drawn.

The correct answer is He should make the arrow for “Path” curve downward.

Eduardwww [97]3 years ago

5 0

Let us review the answers given.
1. He should switch the labels "Gravity" and "Inertia".
These labels are correct, so no change is required.

2. He should make the arrow for "Path" curve downward.
This statement is correct.

3. He should switch the labels "Inertia" and "Path".
No change is required.

4. He should make the arrow for "Inertia" curve downward.
This statement is incorrect.

Answer:
Hector should make the arrow for "Path" curve downward as shown in the figure.

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Two electrons are at rest and separated by a distance of 4.32 × 10-10 m. When they are released they accelerate away from each o

sasho [114]

Answer:

Speed of electron when their separation increased by a factor of 4.10 is 9.41 x 10⁵ m/s .

Explanation:

The electric potential energy is given by the relation :

$U = \frac{kq_{1}q_{2} }{r}$

Here q₁ and q₂ are the two charge particles and r is the distance between them and k is electric constant.

In this case, there are two electrons which are separated by the distance 4.32 x 10⁻¹⁰ m.

Let e be the electron charge and r₁ be the distance between them. Then, the initial electric potential energy is :

$U_{1} = \frac{ke^{2} }{r_{1} }$

Now, the distance between the electrons increases by the factor of 4.10. Let r₂ be the new distance between them i.e. r₂ = 4.10 r₁.

Thus, the new electric potential energy is :

$U_{2} = \frac{ke^{2} }{r_{2} }=\frac{ke^{2} }{4.10r_{1} }$

Applying law of conservation of energy :

ΔU = ΔK

Here ΔU is change in electric potential energy and ΔK is change in kinetic energy.

( U₁ - U₂ ) = ( K₂ - K₁ )

Here K₂ and K₁ are initial and final kinetic energy of electron.

Since, the electron initially is at rest, so its initial kinetic energy is zero. Thus, the above equation becomes:

K₂ = U₁ - U₂

$\frac{1}{2}mv^{2}=\frac{ke^{2} }{r_{1} }- \frac{ke^{2} }{4.10r_{1} }$

Here m and v are the mass and final speed of electron respectively.

$v^{2}=\frac{2}{m} \frac{ke^{2} }{r_{1} }(1- \frac{1 }{4.10 })$

Substitute 9.1 x 10⁻³¹ kg for m, 9 x 10⁹ N m² C⁻² for k, 1.6 x 10⁻¹⁹ C for e and 4.32 x 10⁻¹⁰ m for r₁ in the above equation.

$v^{2}=\frac{2}{9.1\times10^{-31} } \frac{9\times10^{9}\times(1.6\times10^{-19})^{2} }{4.32\times10^{-10} }(1- \frac{1 }{4.10 })$

$v^{2}=8.86\times10^{11}$

v = 9.41 x 10⁵ m/s

5 0

2 years ago

A man 6 feet tall walks at a rate of 6 feet per second away from a light that is 15 feet above the ground.

Tems11 [23]

Answer:(a)10 ft/s

(b)4 ft/s

Explanation:

Given

height of light $=15 feet$

height of man $=6 feet$

$\frac{\mathrm{d} x}{\mathrm{d} t}=6 ft/s$

From diagram

$\frac{15}{y}=\frac{6}{y-x}$

$5(y-x)=2y$

$3y=5x$

differentiate both sides

$3\times \frac{\mathrm{d} y}{\mathrm{d} t}=5\times \frac{\mathrm{d} x}{\mathrm{d} t}$

Tip of shadow is moving at the rate of

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{5}{3}\times 6=10 ft/s$

(b)rate at which length of his shadow is changing

Length of shadow is $y-x$

differentiating w.r.t time

$\frac{\mathrm{d} (y-x)}{\mathrm{d} t}=\frac{\mathrm{d} y}{\mathrm{d} t}-\frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{\mathrm{d} (y-x)}{\mathrm{d} t}=10-6=4 ft/s$

7 0

3 years ago

In projectile mtion, what is the y-component of the initial velocity? if V= Vi = 100 m/s and the angle with horizontal axis Θ =

Damm [24]

Answer:

hence the y - component of the velocity is 50m/s

4 0

2 years ago

a city government received a $1 million grant to build swimming pools and skating rinks for youth. based on the data provided in

ivolga24 [154]

We have that for the Question "A city government received a $1 million grant to build swimming pools and skating rinks for youth. based on the data provided in the graph, what is the opportunity cost of building one swimming pool?" it can be said that

OC=2

From the question we are told

A city government received a $1 million grant to build swimming pools and skating rinks for youth. based on the data provided in the graph, what is the opportunity cost of building one swimming pool?

Generally the equation for the Opportunity cost is mathematically given as

$OC=\frac{y_1-y_2}{x_1-x_2}\\\\OC=\frac{2-0}{1-0}\\\\OC=2$

Therefore

From the graph of the question we can ascertain that Opportunity cost

OC=2

For more information on this visit

brainly.com/question/18670421

Graph is attached below

6 0

2 years ago

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0 °C for the following reacti

Yakvenalex [24]

The equilibrium constant of the reaction at 25⁰c will be 426827.5.

Theory-

Equilibrium constant :The equilibrium constant comes from the chemical equilibrium law. For the chemical equilibrium state, at a fixed constant temperature, the ratio of the product of the reaction's multiplication to the concentration of its reactants' multiplication, and each is raised to the power to the corresponding coefficients of the elements in the reaction.

The chemical equilibrium is given by for a general chemical reaction.

a. A+ b. B ⇌ c. C+ d. D,.

Kc =[C]c [D]d/[A]a [B]b.

Gibb's free energy :The second law of thermodynamics can be arranged in such a way that it gives a new expression when a chemical reaction happens at a constant temperature and constant pressure.

G=H-TS