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Natasha_Volkova [10]
3 years ago
13

How much time would Simpson save by increasing his average velocity to 26m/s East?

Physics
1 answer:
igor_vitrenko [27]3 years ago
7 0

Answer:

8 minutes

Explanation:

Part of the question is missing. Complete question is:

<em>Simpson drives his car 144 km with an average velocity of 24 m/s toward the east. How much time would Simpson save by increasing his average velocity to 26m/s East?</em>

Solution:

First of all, we have to calculate the time it takes for Simpson to complete the drive moving at the initial velocity, 24 m/s. We can use the formula:

t=\frac{d}{v}

where

d=144 km = 1.44\cdot 10^5 m is the displacement

v=24 m/s is the velocity

Substituting,

t=\frac{1.44\cdot 10^5}{24}=6000 s = 100 min

While driving at 26 m/s, the time taken will be

t=\frac{1.44\cdot 10^5}{26}=5538 s=92 min

So, Simpson will save 8 minutes.

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Three masses are located in the x-y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m
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<h2>Answer:</h2>

D. (1m, 0.5m)

<h2>Explanation:</h2>

The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;

x = (m₁x₁ + m₂x₂ + m₃x₃) / M       ----------------(i)

y = (m₁y₁ + m₂y₂ + m₃y₃) / M       ----------------(ii)

Where;

M = sum of the masses

m₁ and x₁ = mass and position of first mass in the x direction.

m₂ and x₂ = mass and position of second mass in the x direction.

m₃ and x₃ = mass and position of third mass in the x direction.

y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.

From the question;

m₁ = 6kg

m₂ = 4kg

m₃ = 2kg

x₁ = 0m

x₂ = 3m

x₃ = 0m

y₁ = 0m

y₂ = 0m

y₃ = 3m

M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg

Substitute these values into equations (i) and (ii) as follows;

x = ((6x0) + (4x3) + (2x0)) / 12

x = 12 / 12

x = 1 m  

y = (6x0) + (4x0) + (2x3)) / 12

y = 6 / 12

y = 0.5m

Therefore, the center of mass of the system is at (1m, 0.5m)

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Is it accurate to describe the physical universe as composed of only matter and energy?
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3 In a television tube, an electron starting from rest experiences a force of 4.0 × 10−15 N over a distance of 50 cm. The final
MAXImum [283]

Answer:

The final speed of the electron = 2.095×10⁸ m/s

Explanation:

From newton's fundamental equation of dynamics,

F = ma ........................Equation 1

Where F = force, m = mass of the electron, a = acceleration of the electron.

making a the subject of the equation,

a = F/m.................... Equation 2

Given: F = 4.0×10⁻¹⁵ N,

Constant: m =  9.109×10⁻³¹ kg.

Substituting into equation 2

a = 4.0×10⁻¹⁵/9.109×10⁻³¹

a = 4.39×10¹⁶ m/s².

Using newton's equation of motion,

v² = u²+2as .......................... Equation 3

Where v = final velocity of the electron, u = initial velocity of the electron, a = acceleration of the electron, s = distance covered by the electron.

Given: u = 0 m/s(at rest), s = 50 cm = 0.5 m, a = 4.39×10¹⁶ m/s²

Substituting into equation 3

v² = 0² + 2(0.5)(4.39×10¹⁶)

v = √(4.39×10¹⁶)

v = 2.095×10⁸ m/s

Thus the final speed of the electron = 2.095×10⁸ m/s

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3 years ago
On a roller coaster ride the total mass of a cart - with two passengers included - is 319 kg. Peak K is at 43.6 m above the grou
Dafna1 [17]

The mechanical energy is lost due to friction between the two peak is 78,458.688 J

<h3>What is mechanical energy?</h3>

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.

M.E = KE +PE

Given is a roller coaster ride the total mass of a cart - with two passengers included - is 319 kg. Peak K is at 43.6 m above the ground and peak L is at 24.4 m. At location K the speed of the cart is 16.4 m/s, and at location L it is 12.4 m/s.

Total energy at peak K,

TE₁ = 1/2 mv₁² +mgh₁

Substitute the values, we get

TE₁ =  1/2 x319 x 16.4² +319 x 9.81 x 43.6

TE₁ = 179,340. 524 J

Total energy at peak L,

TE₂ = 1/2 mv₂² +mgh₂

Substitute the values, we get

TE₂ =  1/2 x319 x 12.4² +319 x 9.81 x 24.4

TE₂ = 100,881. 836 J

The mechanical energy lost is

M.E =  TE₁ -TE₂

M.E =  179,340. 524 J -  100,881. 836 J

M.E = 78,458.688 J

Thus, the mechanical energy is lost due to friction between the two peak i

Learn more about mechanical energy.

brainly.com/question/13552918

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