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2 years ago

13

What is the acceleration of a train that takes 22 seconds to go from a speed of 73 kilometers per hour to a speed of 45 kilomete

rs per hour?

1 answer:

gladu [14]2 years ago

4 0

Answer:

.354 m/s^2

Explanation:

Acceleration is change in velocity divided by change in time

(73 - 45 ) = 28 km / hr velocity change this is 7.778 m/s

7.778 m/s / 22 s = .354 m/s^2

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A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10 m/s. The instan

gizmo_the_mogwai [7]

Answer:

The velocity just before hitting the ground is $v_f = 30 m/s$

Explanation:

From the question we are told that

The initial speed is $u = 10 m/s$

The final speed is $v = 30 \ m/s$

From the equations of motion we have that

$v^2 =u^2 + 2as$

Where s is the distance travelled which is the height of the cliff

So making it the subject of the the formula we have that

$s = \frac{v^2 - u^2 }{2a}$

Where a is the acceleration due to gravity with a value $a = 9.8m/s^2$

So

$s = \frac{30^2 - 10^2 }{2 * 9.8 }$

$s = 40.8 \ m$

Now we are told that was through horizontally with a speed of

$v_x =10 m/s$

Which implies that this would be its velocity horizontally through out the motion

Now it final velocity vertically can be mathematically evaluated as

$v_y = \sqrt{2as}$

Substituting values

$v_y = \sqrt{(2 * 9.8 * 40.8)}$

$v_y = 28.3 \ m/s$

The resultant final velocity is mathematically evaluated as

$v_f = \sqrt{v_x^2 + v_y^2}$

Substituting values

$v_f = \sqrt{10^2 + 28.3^2}$

$v_f = 30 m/s$

5 0

3 years ago

The smallest unit of an element that has all of the properties of the element is a/an

AysviL [449]

Answer:B

Explanation: an atom is the smallest particle of an element that can take part in a chemical reaction.

7 0

3 years ago

Which would require the greater energy; slowing down of the orbital speed of the Earth so it crashes into the sun, or speeding u

tresset_1 [31]

Answer: Speeding up the orbital speed of earth so it escapes the sun require the greater energy.

Explanation: To find the answer, we need to know more about the Orbital and escape velocities.

<h3>What is Orbital and Escape velocity?</h3>

Orbital velocity can be defined as the minimum velocity required to put the satellite in its orbit around the earth.
The expression for orbital velocity near to the surface of earth will be,

$V_o=\sqrt{gR}$

Escape velocity can be defined as the minimum velocity with which a body must be projected from the surface of earth, so that it escapes from the gravitational field of earth.
The expression for orbital velocity will be,

$V_e=\sqrt{2gR}$

If we want to get into the sun, we want to slow down almost completely, so that your speed relative to the sun became almost zero.
We need about twice the raw speed to go to the sun than to leave the sun.

Thus, we can conclude that, the speeding up the orbital speed of earth so it escapes the sun require the greater energy.

Learn more about orbital and escape velocity here:

brainly.com/question/28045208

#SPJ4

3 0

2 years ago

A rod has a radius of 10 mm is subjected to an axial load of 15 N such that the axial strain in the rod is ????௫ = 2.75*10-6, de

EleoNora [17]

Answer:

Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension

ex = бx/E

бx = Fx/A = Fx/π $r^{2}$

Using both equation and solving for the modulus of elasticity E

E = бx/ex = Fx / π $r^{2}$ ex

E = $\frac{15}{pi (10 * 10^{-3})^{2} * 2.75 * 10^{-6} } = 17.368 * 10^{9} Pa = 17.4 GPa$

Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius

ey = $\frac{1}{E}$ (бy - v (бx + бz)) = $-\frac{v}{E}$ бx

= $\frac{vFx}{Epir^{2} }$ = $\frac{0.23 * 15}{pi (10 * 10^{-3)^{2} } * 17.362 * 10^{9} } = -0.63 *10^{-6}$

Finally

ey = Δr / r

Δr = ey * r = 10 * $-0.63* 10^{-6} mm = -6.3 * 10^{-6} mm$

Δd = 2Δr = $-12.6 * 10^{-6} mm$

Explanation:

5 0

3 years ago

Two equal, but oppositely charged particles are attracted to each other electrically. The size of the force of attraction is 48.

galina1969 [7]

Given:

The force of attraction is F = 48.1 N

The separation between the charges is

$\begin{gathered} r=\text{ 60.9 cm} \\ =\text{ 60.9}\times10^{-2}\text{ m} \end{gathered}$

Also, the magnitude of charge q1 = q2 = q.

To find the magnitude of charge.

Explanation:

The magnitude of charge can be calculated by the formula

$\begin{gathered} F=\frac{k(2q)}{r^2} \\ q=\frac{Fr^2}{2k} \end{gathered}$

Here, k is the Coulomb's constant whose value is

$k\text{ = 9}\times10^9\text{ N m}^2\text{ /C}^2$

On substituting the values, the magnitude of charge will be

$\begin{gathered} q=\frac{48.1\times(60.9^\times10^{-2})^2}{2\times9\times10^{^9}} \\ =9.91\times10^{-10}\text{ C} \\ =9.91\times10^{-4}\text{ }\mu C \end{gathered}$

Thus, the magnitude of each charge is 9.91 x 10^(-4) micro Coulombs.

6 0

1 year ago