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3 years ago

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An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time

?
But use this formula with the steps: x = xo + vo t + ½ a t 2

1 answer:

Tju [1.3M]3 years ago

5 0

Answer:

675m

Explanation:

Given parameters:

Initial velocity = 0m/s

Acceleration = 6m/s²

Time = 15s

Unknown:

Distance traveled by the body = ?

Solution:

To solve this problem; we use the expression;

S = ut + $\frac{1}{2}$ at²

Where u is the initial velocity

t is the time

a is the temperature

Insert the parameters and solve;

S = 0 x 15 + $\frac{1}{2}$ x 6 x 15²

S = 675m

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A 0.37-kg object connected to a light spring with a force constant of 23.2 N/m oscillates on a frictionless horizontal surface.

mars1129 [50]

Answer:

a) v = 31.67 cm / s , b) v = -29.36 cm / s , c) v= 29.36 cm/s, d) x = 3.46 cm

Explanation:

The angular velocity in a simple harmonic movement is

w = √ K / m

w = √ 23.2 / 0.37

w = 7,918 rad / s

a) the expression against the movement is

x = A cos (wt + Ф)

Speed is

v = dx / dt = - A w sin (wt + Ф)

The maximum speed occurs for cos = ± 1

v = A w

v = 4.0 7,918

v = 31.67 cm / s

b) as the object is released from rest

0 = -A w sin (0+ Фi)

sin Ф = 0

Ф = 0

The equation is

x = 4.0 cos (7,918 t)

v = -4.0 7,918 sin (7,918 t)

v = - 31.67 sin (7.918t)

Let's look for the time for a displacement of x = 1.5 cm, remember that the angles must be in radians

7,918 t = cos⁻¹ 1.5 / 4.0

t = 1,186 / 7,918

t = 0.1498 s

We look for speed

v = -31.67 sin (7,918 0.1498)

v = -29.36 cm / s

c) if the object passes the equilibrium equilibrium position again at this point the velocity has the same module, but the opposite sign

v = 29.36 cm / s

d) let's look for the time for the condition v = v_max / 2

31.67 / 2 = 31.67 sin ( 7,918 t)

7.918t = sin⁻¹ 0.5

t = 0.5236 / 7.918

t = 0.06613

With this time let's look for displacement

x = 4.0 cos (7,918 0.06613)

x = 3.46 cm

6 0

3 years ago

A force causes a mass of 4 kg to have an acceleration of 8 m/s2. Suppose something causes the mass to be one-quarter of its orig

solniwko [45]

Explanation:

From Newton's second law:

F = ma

Given that m = 4 kg and a = 8 m/s²:

F = (4 kg) (8 m/s²)

F = 32 N

If m is reduced to 1 kg and F stays at 32 N:

32 N = (1 kg) a

a = 32 m/s²

So the acceleration increases by a factor of 4.

5 0

3 years ago

1) You slam on the brakes of your car in a panic, and skid a certain distance on a straight level road. If you had been travelin

aleksandr82 [10.1K]

Answer:

d = 4 d₀o

Explanation:

We can solve this exercise using the relationship between work and the variation of kinetic energy

W = ΔK

In that case as the car stops v_f = 0

the work is

W = -fr d

we substitute

- fr d₀ = 0 - ½ m v₀²

d₀ = ½ m v₀² / fr

now they indicate that the vehicle is coming at twice the speed

v = 2 v₀

using the same expressions we find

d = ½ m (2v₀)² / fr

d = 4 (½ m v₀² / fr)

d = 4 d₀o

3 0

3 years ago

Part of one row of the periodic table is shown. 4 blue boxes in a row. The first has S n at the center and 50 above with tin and

nydimaria [60]

Answer:

Atoms of tellurium (Te) have the greatest average number of neutrons equal to 76.

Explanation:

In the periodic table, Elements are represented with their respected symbols. Above the symbol is the elements atomic number which is equal to the number of protons in each atom. Below the symbol is the mass number of that element which is roughly equal to the sum of neutrons and protons of that atom.

To calculate the number of neutrons we can take the difference of Atomic number and mass number:

Number of neutrons = mass number - atomic number

<u>- Tin:</u>

Atomic number = 50

Mass number = 119

Number of neutrons = mass number - atomic number = 119 - 50

Number of neutrons = 69

<u>- Antimony(Sb):</u>

Atomic number = 51

Mass number = 122

Number of neutrons = mass number - atomic number = 122 - 51

Number of neutrons = 71

<u>- Tellurium(Te):</u>

Atomic number = 52

Mass number = 128

Number of neutrons = mass number - atomic number = 128 - 52

Number of neutrons = <u>76</u>

<u>- Iodine(I):</u>

Atomic number = 53

Mass number = 127

Number of neutrons = mass number - atomic number = 127 - 53

Number of neutrons = 74

Here, the greatest number of neutrons is for the atoms of Tellurium(Te).

7 0

4 years ago

Read 2 more answers

PLEASE HELP ME 45 POINTS

sergij07 [2.7K]

Answer:

a) We kindly invite you to see the explanation and the image attached below.

b) The acceleration of the masses is 4.203 meters per square second.

c) The tension force in the cord is 28.02 newtons.

d) The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is 3.551 meters per second.

Explanation:

a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.

b) Both masses are connected to each other by the same cord, the direction of acceleration will be dominated by the mass of greater mass (mass A) and both masses have the same magnitude of acceleration. By the 2nd Newton's Law, we create the following equation of equilibrium:

Mass A

$\Sigma F = T - m_{A}\cdot g = -m_{A}\cdot a$ (1)

Mass B

$\Sigma F = T - m_{B}\cdot g = m_{B}\cdot a$ (2)

Where:

$T$ - Tension force in the cord, measured in newtons.

$m_{A}$ , $m_{B}$ - Masses of blocks A and B, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$a$ - Net acceleration of the each block, measured in meters per square second.

By subtracting (2) by (1), we get an expression for the acceleration of each mass:

$m_{B}\cdot a +m_{A}\cdot a = T-m_{B}\cdot g -T + m_{A}\cdot g$

$(m_{B}+m_{A})\cdot a = (m_{A}-m_{B})\cdot g$

$a = \frac{m_{A}-m_{B}}{m_{B}+m_{A}} \cdot g$

If we know that $m_{A} = 5\,kg$ , $m_{B} = 2\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the acceleration of the masses is:

$a = \left(\frac{5\,kg-2\,kg}{5\,kg+2\,kg}\right) \cdot\left(9.807\,\frac{m}{s^{2}} \right)$

$a = 4.203\,\frac{m}{s^{2}}$

The acceleration of the masses is 4.203 meters per square second.

c) From (2) we get the following expression for the tension force in the cord:

$T = m_{B}\cdot (a+g)$

If we know that $m_{B} = 2\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $a = 4.203\,\frac{m}{s^{2}}$ , then the tension force in the cord:

$T = (2\,kg)\cdot \left(4.203\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}} \right)$

$T = 28.02\,N$

The tension force in the cord is 28.02 newtons.

d) Given that system starts from rest and net acceleration is constant, we determine the time taken by the block to cover a distance of 1.5 meters through the following kinematic formula:

$\Delta y = \frac{1}{2}\cdot a\cdot t^{2}$ (3)

Where:

$a$ - Net acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

$\Delta y$ - Covered distance, measured in meters.

If we know that $a = 4.203\,\frac{m}{s^{2}}$ and $\Delta y = 1.5\,m$ , then the time taken by the system is:

$t = \sqrt{\frac{2\cdot \Delta y}{a} }$

$t = \sqrt{\frac{2\cdot (1.5\,m)}{4.203\,\frac{m}{s^{2}} } }$

$t \approx 0.845\,s$

The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is calculated by the following formula:

$v = a\cdot t$ (4)

Where $v$ is the final speed of the system, measured in meters per second.

If we know that $a = 4.203\,\frac{m}{s^{2}}$ and $t \approx 0.845\,s$ , then the final speed of the system is:

$v = \left(4.203\,\frac{m}{s^{2}} \right)\cdot (0.845\,s)$

$v = 3.551\,\frac{m}{s}$

The final speed of the system is 3.551 meters per second.

8 0

3 years ago