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Leto [7]
4 years ago
15

Plz answer them I need answers ASAP

Physics
1 answer:
vaieri [72.5K]4 years ago
5 0

Answer:

4. 13.04m/s²

5. 4x10^11m/s²

6. –3m/s²

Explanation:

Acceleration is defined as rate of change of velocity with time. Mathematically it is represented as:

a = (v – u)/t

a is the acceleration.

v is the final velocity.

u is the initial velocity.

t is the time

With the above formula, we can obtain the acceleration of each question.

4. Determination of the acceleration of the spacecraft.

Data obtained from the question include the following:

Initial velocity (u) = 0

Final velocity (v) = 67.8m/s

Time (t) = 5.2secs

Acceleration (a) =..?

a = (v – u)/t

a = (67.8 – 0)/5.2

a = 13.04m/s²

Therefore, the acceleration of the spacecraft is 13.04m/s².

5. Determination of the acceleration of the electron.

Data obtained from the question include the following:

Initial velocity (u) = 0

Final velocity (v) = 2.6x10^5m/s

Time (t) = 6.5x10^-7secs

Acceleration (a) =..?

a = (v – u)/t

a = (2.6x10^5 – 0)/ 6.5x10^-7

a = 4x10^11m/s²

Therefore, the acceleration of the electron is 4x10^11m/s².

6. Determination of the acceleration of the truck.

Data obtained from the question include the following:

Initial velocity (u) = 20m/s

Final velocity (v) = 5m/s

Time (t) = 5secs

Acceleration (a) =..?

a = (v – u)/t

a = (5 – 20)/5

a = –15/5

a = –3m/s².

Therefore, the acceleration of the truck is –3m/s².

You might be interested in
An object is pulled with two forces, 10 N northward and 15 N southward. The direction of the net force is to the An object is pu
ValentinkaMS [17]

Answer:

check image

Explanation:

For any question related to newons law of motion first draw the free body diagram(FBD),

7 0
4 years ago
Lamar writes several equations trying to better understand potential energy. What conclusion is best supported by lamar's worm?
kumpel [21]

Complete question is;

Lamar writes several equations trying to better understand potential energy. What conclusion is best supported by Lamar’s work?

A) The elastic potential energy is the same for any distance from a reference point.

B) The gravitational potential energy equals the work needed to lift the object.

C) The gravitational potential energy is the same for any distance from a reference point.

D) The elastic potential energy equals the work needed to stretch the object

Answer:

B) The gravitational potential energy equals the work needed to lift the object.

Explanation:

In physics, we know that potential energy is the energy of a body at rest while the energy of a body in motion is known as kinetic energy.

However,the work required to lift a body from it's position of rest is equal to the Gravitational potential energy of that body.

Elastic potential energy is the one that is stored as a result of force applied to deform an elastic object. Thus, it is not equal to the work needed to stretch the object and it is also not the same for any distance from reference point.

Thus, looking at the options, Option B is correct

6 0
3 years ago
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.8×106N , one an angle 11degrees west of north an
sp2606 [1]

Answer: 2,9 ×10¹⁰J.

Explanation:

1. The <em><u>work </u></em>is a measure of energy and is calculated as the product of the displacement times the parallel force to such displacement.

That means that the only components of the force that contribute to work are those that result parallel to the displacement.

2. Since it is given that the <em>two tugboats "pull the tanker a distance 0.83km toward the north"</em>, that is the displacement, and you have to calculate the net force toward the north.

3. <u>Tugboat #1</u>.

a) Force magnitude: F₁ = 1.8×10⁶N

b) Angle: α = 11° West of North

c) North component of the force F₁: Fy₁ = F₁cos(α) =  1.8×10⁶N  × cos(11°) = 1.77×10⁶N

4. <u>Tugboat #2</u>:

a) Force magnitude: F₂ = 1.8×10⁶N

b) Angle:  = 11° East of North

c) North component of the force F₂: Fy₂ = F₂cos(β) =  1.8×10⁶N  × cos(11°) = 1.77×10⁶N =

5. <u>Total net force, Fn</u>:

Fn = Fy₁ + Fy₂ = 1.77×10⁶N + 1.77×10⁶N = 3.54×10⁶N

6. <u>Work, W</u>:

Displacement, d = 0.83 km = 8,300 m

W = Fn×d = 3.54×10⁶N×8,300m = 29,000 ×10⁶J = 2,9 ×10¹⁰J

The answer is rounded to two significant figures because both data, Force and displacement, have two significant figures.

7 0
3 years ago
Does anyone know delta and star math ?
Goshia [24]
Google it and click on the first website. 
7 0
4 years ago
Assuming perfect optics, if the smallest feature you can resolve when observing around 660 nm is of angular size 0.04 arcsec, wh
Dmitriy789 [7]

Answer:

The value is \theta _2 =  0.08 \ arcsec

Explanation:

From the question we are told that

   The first wavelength is  \lambda_1 =  660 \ nm  = 660 *10^{-9 }

   The  first angular size is  \theta_1 =  0.04 \  arcsec

    The second wavelength is  \lambda _2 =  1320 \  nm  =  1320 *10^{-9 } \  m

Generally according to  Rayleigh Criterion we have that

          \theta=  1.22 *  \frac{\lambda }{D}

given every other thing remains constant we have that

         \theta =  k *  \lambda

Here k  represented constant so

         k  =  \frac{\theta }{\lambda}

=>      \frac{\theta_1}{ \lambda_1}  =  \frac{\theta_2}{ \lambda_2}

=>      \frac{\theta_1}{ \theta_2}  =  \frac{\lambda_1}{ \lambda_2}

So

        \frac{ 0.04}{ \theta_2}  =0.5

=>     \theta _2 =  0.08 \ arcsec

3 0
3 years ago
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