How to fix this? I need to save both

7.4 s

Explanation:

Given:

[tex]v_0 = 73\:\text{m/s}[/tex]

[tex]v = 0[/tex]

[tex]a = -9.8\:\text{m/s}^2[/tex]

[tex]t = ?[/tex]

To solve the time it takes for the object to come to a stop, we are going to use the equation below:

[tex]v = v_0 + at \Rightarrow t = \dfrac{v - v_0}{a}[/tex]

Using the given values above, we get

[tex]t = \dfrac{0 - 73\:\text{m/s}}{-9.8\:\text{m/s}^2}[/tex]

[tex]\;\;\;\;= 7.4\:\text{s}[/tex]

Answer:

It releases some of the energy into the atmosphere as hot steam.

Explanation:

The force will be reduced to 1/4 of the original

Explanation:

According to Coulomb's law, the force between two charges [tex]q_a\:\text{and}\:q_b,[/tex] separated by a distance r is given by

[tex]F = k\dfrac{q_aq_b}{r^2}[/tex]

where k is the Coulomb constant.

Now let F' be the force between the two charges when their separation distance is doubled. We can write this force as

[tex]F' = k\dfrac{q_aq_b}{(2r)^2} = k\dfrac{q_aq_b}{4r^2}[/tex]

[tex]\;\;\;\;\;= \frac{1}{4}\left(k\dfrac{q_aq_b}{r^2}\right) = \frac{1}{4}F[/tex]

Therefore, the force will be reduced to a quarter of its original value.

Hope you could understand.

If you have any query, feel free to ask.

Answer:

P = 2 pi (L / g)^1/2

P2 / P1 = (8 / 2)^1/2 = 2

The period would be twice as long or 5.6 sec.

(a) The final velocity of the two vehicles if the collision was inelastic is 1.1 m/s.

(b)  For the elastic collision, the final velocity of the car is 9.81 m/s backwards and the final velocity of the truck is 2.19 m/s forward.

(c) The time taken to exert the given force is 0.00625 m (s).

The given parameters;

(a) The final velocity of the two vehicles if the collision was inelastic is calculated as follows;

[tex]m_1 u_1 + m_2u_2 = v(m_1+ m_2)\\\\12m + 10m(0) = v(m + 10m)\\\\12m = v(11m)\\\\v = \frac{12m}{11m} \\\\v = 1.1 \ m/s[/tex]

(b) The final velocity of the two vehicles if the collision was elastic is calculated as follows;

[tex]m_1 u_1 + m_2u_2 = m_1v_1 + m_2v_2\\\\\12m \ + \ 10m(0) = mv_1 + 10mv_2\\\\12m = m(v_1 + 10v_2)\\\\12 = v_1 + 10 v_2\ \ - --(1)[/tex]

Apply one-directional velocity equation:

[tex]u_1 +v_1 = u_2 + v_2\\\\12 + v_1 = 0 + v_2\\\\12+ v_1 = v_2 \ \ --- (2)[/tex]

Substitute the value of [tex]v_2[/tex] into equation (1);

[tex]12 = v_1 + 10(12 + v_1)\\\\12= v_1 + 120 + 10v_1\\\\12- 120 = 11v_1\\\\-108 = 11v_1\\\\v_1 = \frac{-108}{11} \\\\v_1 = -9.81 \ m/s\\\\[/tex]

Solve for [tex]v_2[/tex];

[tex]v_2 = 12 + v_1\\\\v_2 = 12 - 9.81\\\\v_2 = 2.19 \ m/s[/tex]

Thus, for the elastic collision, the final velocity of the car is 9.81 m/s backwards and the final velocity of the truck is 2.19 m/s forward.

(c)

The change in the momentum of the truck is calculated as;

[tex]\Delta P = m_2(v_2 - u_2)\\\\\Delta P = 10m(2.19)\\\\\Delta P = 21.9m[/tex]

The time taken to exert the given force is calculated as follows;

[tex]Ft = \Delta P\\\\t = \frac{\Delta P}{F} \\\\t = \frac{21.9 \ m}{3500} \\\\t = 0.00625 \ m (seconds)[/tex]

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Answer:

One force will be gravity & inertia.

Explanation:

Bioth are combine to keep Earth in orbit around the sun, and the moon in orbit around Earth

The time taken for the two balls to hit each other is 8 s.

The given parameters:

The time taken for the two balls to hit each other is calculated by applying relative velocity formula as shown below;

[tex](V_1 - (-V_2) )t = s\\\\(V_1 + V_2) t = s\\\\(0.3 + 0.2) t = 4\\\\0.5t = 4\\\\t = \frac{4}{0.5} \\\\t = 8 \ s[/tex]

Thus, the time taken for the two balls to hit each other is 8 s.

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Answer:

A

Explanation:

acceleration = velocity / time

Hope that helps

Answer:

D

Explanation:

Average acceleration = final velocity- initial velocity, divided by elapsed time.

ā = V - Vo/t

= change in velocity ÷ change in time

Answer:

Explanation:

300000(9.8) = ω²(0.0160)

ω = 13555 rad/s

13555 rad/s (60 s/min/ 2π rad/rev) = 129,445 rev/min

Answer:

True;   ar = v^2 / R      Radial acceleration because it moves in a circular path

    at = α R = tangential acceleration because its speed changes

a = at + ar    total acceleration equals sum of radial and tangential

Answer:

pressure = force / area

then pressure = 25 / 5 = "5" N/m^2

The horizontal distance where the block hits the floor is 3.2 m.

The given parameters:

The final velocity of the bullet-block system after the collision is calculated by applying the principle of conservation of linear momentum;

[tex]m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\0.015(500) + 1(0) =v(0.015 + 1)\\\\7.5 = 1.015v\\\\v = \frac{7.5}{1.015} \\\\v = 7.39 \ m/s[/tex]

The time taken for the bullet-block system to fall to the floor after collision is calculated as follows;

[tex]h = v_0_yt + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 0.92}{9.8} }\\\\t = 0.43 \ s[/tex]

The horizontal distance where the block hits the floor is calculated as follows;

[tex]X = v_x t\\\\X = 7.39 \times 0.43\\\\X = 3.2 \ m[/tex]

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Answer:

Depends on the weight i would be 7.120

Answer:

[tex]a[/tex]

Explanation:

is a corect anser

Answer:

The total amount of energy, kinetic plus potential, remains the same.

Explanation:

Answer:

The right answer for this question is 85%.

(I had the same question.)

Answer:

The progressive loss of material from a surface by the mechanical action of a fluid on a surface is called erosion.

Answer:

Yes.

Explanation:

There is no movement in magnetic, chemical, electrostatic, or nuclear (potential) energy. The other options for that question can't be right. Mechanical energy is a form of kinetic, so B cannot be true. Thermal energy is also kinetic, which makes C and D incorrect as well.

Answer:

oK so  here's  what you should do is  add .09 and 0.6

Explanation:

Answer:

Velocity of the stone after 1.5 seconds has passed = 37 m/s

Explanation:

Initial velocity (u) = 22 m/s

Time (t) = 1.5 sec

Acceleration due to gravity (g) = 10 m/s²

By using kinematics equation:

v = u + gt

v = 22 + 10 × 1.5

v = 22 + 15

v = 37 m/s

Final velocity (v) = 37 m/s

Answer:

1/2 m v^2 + 1/2 I ω^2 = m g h       conservation of energy

I = 2/5 m R^2     inertia of solid sphere

1/2 m v^2 + 1/5 m ω^2 R^2 = m g h

1/2 v^2 + 1/5 v^2 = g h

v^2 = 10 g h / 7 = 1.43 * 9.80 * 19 m^2/s^2 = 266 m^2/s^2

v = 16.3 m/s

v = R ω

ω = 16.3 / .6 = 27.2 / sec

The density of the liquid will be equal to   [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]

The density of an object is defined as the ratio of the mass of an object to the volume of the object.

Volume of tube = 2^2 * pi * 30 = 377 cm^3

Volume of tube submerged = 25* 377 / 30 = 314 cm^3

Buoyancy = weight of liquid displaced

Volume of liquid displaced = 314 cm^3

Mass of tube and lead = 250 + 30 = 280 g

Now from the mass density by definition

[tex]\rho = \dfrac{m}{v}[/tex]

[tex]m=\rho \times v[/tex]

Mass of liquid displaced = Mass being supported

[tex]314 \times \rho = 280 g[/tex]

[tex]\rho= \dfrac{280}{ 314 } = .892 \frac{g}{cm^3}[/tex]

Thus the density of the liquid will be equal to   [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]

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The density of the liquid is 1.67 g/[tex]cm^3[/tex].

The volume of the tube is

30 * 4 * 3.14 * 0.25 = 94.2 [tex]cm^3[/tex].

The mass of the lead pellets and the plastic tube is

30 + 250 = 280 g.

The volume of the lead pellets is

250 / 11.34 = 22 [tex]cm^3[/tex].

The volume of the liquid that the tube displaces is

94.2 - 22 = 72 [tex]cm^3[/tex].

The density of the liquid is

280 / 72 = 1.67 g/ [tex]cm^3[/tex].

Therefore, the density of the liquid is 1.67 g/ [tex]cm^3[/tex].

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#SPJ3

Answer:

F1 = G m1 m2 / R^2        force of attraction

F2 = G m1 m2 / (R/2)^2

F2 / F1 = 4       the force of gravity will be quadrupled

Answer: Time (days) = 88, the (mass) = 0.2180

Explanation:

Time (days) = 88, the (mass) = 0.2180

[tex]a_c = 3.14\:\text{m/s}^2[/tex]

Explanation:

First, we need to convert the given angular speed [tex]\omega[/tex] from rev/s to rad/s:

[tex]2.50\:\dfrac{\text{rev}}{\text{s}}×\dfrac{2\pi\:\text{rad}}{1\:\text{rev}} = 15.7\:\text{rad/s}[/tex]

The centripetal acceleration [tex]a_c[/tex] is defined as

[tex]a_c = \dfrac{v^2}{r}[/tex]

Recall that [tex]v = r\omega[/tex] so we can write [tex]a_c[/tex] as

[tex]a_c = \dfrac{(r\omega)^2}{r} = \omega^2r[/tex]

[tex]\;\;\;\;\;=(15.7\:\text{rad/s})^2(0.200\:\text{m}) = 3.14\:\text{m/s}^2[/tex]

Answer:

Depends on what are you looking at. All of the following are valid definitions, it just depends on which way you are analizing the problem.

From a kinematics point of view:

Acceleration is, by definition, is a vector quantity that measures the rate of change of the rate of change in position (add brackets if it helps visualizing the idea). This leads to the following different definitions - which are more like means of calculating it

From a dinamics point of view

Acceleration is the effect of a force applied to a body, and measures the ratio of the force applied to a body of mass m and the mass itself (which is another formulation of Newton second law):

[tex]a = \frac Fm[/tex]

Acceleration is the rate of change of velocity

To define acceleration, We need to know more about motion.

Motion: This can be defined as the change in position of a body from one point to another. When an object accelerates, it undergoes motion.

Definition

Acceleration can be defined as the rate of change of velocity. The S.I unit of acceleration is meter-per-squared seconds. (m/s²)

The formula of acceleration is

⇒ Where:

Hence, Acceleration is the rate of change of velocity

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Answer:

3(1.5) = 4.5 V

Explanation:

The law of reflection allows to find the results for the questions about ray reflection in a plane mirror are:

    a) Attachment we see a diagram of the incident and reflected rays, incident and reflected angles are equal.

    b)  The extension of the reflected rays is what forms the image.

    c)  The image's distance  is 20 cm behind the flat mirror.

    d) The point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C  

    e) the Rays at points A, B, C cannot perceive in the point E.

    f) attachment we see the rays that come out from the pencil eraser.

    g) The image is behind the mirror at 15 cm.

The geometric interaction describes the interaction of light rays with surfaces, looking for where the rays are directed, it is described by two phenomenological laws:

             [tex]\theta_i = \theta_{r}[/tex]  

From these two general laws, geometric optics establishes a relationship for the formation of the image, called the constructor's equation.

            [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

Where f is the focal length, p and q are the distance to the object and the image, respectively.

In this exercise, the medium is a mirror, which is why it must comply with the law of reflection.

a) In the attachment we see a diagram of the incident and reflected rays for the three points.

According to the law of reflection, the incident and reflected angles are equal.

b) From the diagram we can see that the extension of the reflected rays is what forms the image, which is called virtual and is located behind the mirror.

c) In the diagram we see two rays to form the image, we see that the distance to the object is equal to the distance to the image.

From the constructor's equation a plane mirror has an infinite radius.

      p = -q

Therefore the image's distance  is 20 cm behind the flat mirror.  Therefore the distance to the object and the image are the same, the negative sign indicates that the image is behind the mirror.

d) A person located at point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C since their angle of reflection is not equal to the incident angle.

To perceive a ray it must have an angle of incidence of 25º.

e) Point E is located very far from the pencil, so the incident angle increases as does the reflected angle.

the Rays at points A, B, C cannot perceive.

f) In the attachment we see the rays that come out from the pencil eraser, they indicate that the distance to the plane mirror is 15.0 cm,

g) The image is behind the mirror at 15 cm.

In conclusion using the law of reflection we can find the results for the questions are:

     a) Attachment we see a diagram of the incident and reflected rays, incident and reflected angles are equal.

     b)  The extension of the reflected rays is what forms the image.

     c)  The image's distance  is 20 cm behind the flat mirror.

     d) The point D (normal for an angle of 50º) cannot perceive the rays coming from point A, B, C  

    e) the Rays at points A, B, C cannot perceive in the point E.

    f) attachment we see the rays that come out from the pencil eraser.

   g) The image is behind the mirror at 15 cm.

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The acceleration due to gravity on Mars is 11.81 m/s².

The given parameters:

The equation of motion to determine the acceleration due to gravity on the moon is calculated as follows;

[tex]s = vt + \frac{1}{2} gt^2[/tex]

where;

Since the time measured is two way time, the new equation for the total distance traveled is calculated as;

[tex]v = \frac{2d}{t} \\\\2d = vt\\\\d = \frac{vt}{2} \\\\d = \frac{320 \times 27.1}{2} \\\\d = 4,336 \ m[/tex]

The acceleration due to gravity is calculated as follows;

[tex]s = vt + \frac{1}{2} gt^2\\\\4,336 = 0 \ + \ \frac{1}{2} \times g \times (27.1)^2\\\\4,336 = 367.21g\\\\g = \frac{4,336}{367.21} \\\\g = 11.8 1 \ m/s^2[/tex]

Thus, the acceleration due to gravity on Mars is 11.81 m/s².

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