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Nookie1986 [14]
3 years ago
12

Answers the questions 1. At what distance did the object start to move ?

Physics
2 answers:
tester [92]3 years ago
4 0

Answer:

at one second :p

Explanation:

Citrus2011 [14]3 years ago
3 0

Answer:

it started to move a 1 second

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At high speeds, a particular automobile is capable of an acceleration of about 0.540 m/s^2. At this rate, how long (in seconds)
Mama L [17]

Answer:

t = 6.68 seconds

Explanation:

The acceleration of the automobile, a=0.54\ m/s^2

Initial speed of the automobile, u = 91 km/hr = 25.27 m/s

Final speed of the automobile, v = 104 km/hr = 28.88 m/s

Let t is the time taken to accelerate from u to v. It can be calculated as the following formula as :

t=\dfrac{v-u}{a}

t=\dfrac{28.88-25.27}{0.54}

t = 6.68 seconds

So, the time taken by the automobile to accelerate from u to v is 6.68 seconds. Hence, this is the required solution.

5 0
3 years ago
A positively charged particle of mass 7.2 x 10-8 kg is traveling due east with a speed of 88 m/s and enters a 0.6-T uniform magn
Marianna [84]

Answer:

q = 8.57 10⁻⁵ mC

Explanation:

For this exercise let's use Newton's second law

         F = ma

where force is magnetic force

        F = q v x B

the bold are vectors, if we write the module of this expression we have

         F = qv B sin θ

as the particle moves perpendicular to the field, the angle is θ= 90º

        F = q vB

the acceleration of the particle is centripetal

        a = v² / r

we substitute

        qvB = m v² / r

         qBr = m v

          q =\frac{m\  v}{B\  r}

The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use

          v = d / t

the distance is ¼ of the circle,

          d = \frac{1}{4} \  2\pi  r

           d =\frac{\pi }{2r}

we substitute

           v =  \frac{\pi  r}{2t}

           r = \frac{2 \ t  \ v}{\pi }

           

let's calculate

           r =\frac{2 \ 2.2  \ 10^{-3} \ 88}{\pi } 2 2.2 10-3 88 /πpi

           r = 123.25 m

         

let's substitute the values

           q = \frac{ 7.2 \ 10^{-8} \ 88}{ 0.6 \ 123.25}7.2 10-8 88 / 0.6 123.25

            q = 8.57 10⁻⁸ C

Let's reduce to mC

           q = 8.57 10⁻⁸ C (10³ mC / 1C)

           q = 8.57 10⁻⁵ mC

4 0
2 years ago
. A student claims that if lighting strikes a metal flagpole, the force exerted by the Earth’s magnetic field on the current in
Ratling [72]

Answer:

Explanation:

Given that, current generated from lightning range from

10⁴ A < I < 10^5 A

We know that,

The magnetic force is given as

F = iLB

The magnetic field on the earth surface is

B = 10^-5 T

So, let assume the worst case of a 15m flag pole

L = 15m

Then,

F = iLB

F = 10^5 × 10 × 10^-5

F = 15 N

Therefore, 15N is fairly strong so it will come to the material that was use for the material of the flag pole.

Therefore, it is possible that the student is right depending on the material of the flag pole.

7 0
3 years ago
hi, Need help with triangle law of vector addition worksheet and a verifying Newtons second law worksheet?
Yuri [45]

Use Newton's second law and the free body diagram to determine the net force and acceleration of an object. In this unit, the forces acting on the object were always directed in one dimension.

The object may have been subjected to both horizontal and vertical forces but there was no single force directed both horizontally and vertically. Moreover, when free-body diagram analysis was performed, the net force was either horizontal or vertical, never both horizontal and vertical.

Times have changed and we are ready for situations involving two-dimensional forces. In this unit, we explore the effects of forces acting at an angle to the horizontal. This makes the force act in two dimensions, horizontal and vertical. In such situations, as always in situations involving one-dimensional network forces, Newton's second law applies.

Learn more about Newton's second law here:-brainly.com/question/25545050

#SPJ9

5 0
11 months ago
The Lamborghini Huracan has an initial acceleration of 0.75g. Its mass, with a driver, is 1510 kg.
Cerrena [4.2K]

Answer:

11109.825 N

Explanation:

Given Data:

total mass =m=1510 kg

initial acceleration (a) =0.75g                  ( g=9.81 m/s² )

F=ma

  = (1510)*( 0.75*9.81)

  = 11109.825 N

4 0
3 years ago
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