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labwork [276]
3 years ago
10

A rocket-driven sled Sonic Wind No. 2, is used for investigating the physiological effects of large accelerations on people. It

runs on a straight, level track. Starting from rest, it can reach a speed of 250 m/s in a time 2s with approximately constant acceleration.What is the best single equation?
Physics
1 answer:
rewona [7]3 years ago
7 0

Answer:

The best single equation in order to fin the acceleration is

a=v_f/t=250/2=125m/s^2

Explanation:

The best single equation in order to find the acceleration is the following kinematics equation:

v_f=v_o+at

As the rocket start from the rest, the initial velocity value is zero. Then the acceleration value is:

a=v_f/t=250/2=125m/s^2

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100 POINTS!! PLEASE DONT ANSWER UNLESS YOU KNOW BOTH ANSWERS PLEASE.
JulijaS [17]

Answer:

1. is B and 2. is A

Explanation:

6 0
3 years ago
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A small boat sailed straight north out of a harbor in strong east wind (blowing from west to east). After sailing for 120 minute
Amiraneli [1.4K]

it can be said that  the speed of the east wind is

v=0.3608m/s

From the question we are told

A small boat sailed <u>straight </u>north out of a harbor in <em>strong </em>east wind (blowing from west to east).

After sailing for 120 minutes, it ended up hitting a buoy 60^\circ60 ∘ to the north-east of the harbor. If the straight-line distance between the buoy and the harbor is 3 km,

  • what is the speed of the east wind?.

<h3> the speed of the east wind</h3>

Generally the equation for the distance  is mathematically given as

BA=3000sin60

BA=2598.07m

Therefore

the speed of the east wind

V_w=\frac{BA}{120*60}\\\\V_w=\frac{2598.07}{120*60}

v=0.3608

For more information on this visit

brainly.com/question/22568180

7 0
2 years ago
Estimate the magnitude of the electric field due to the proton in a hydrogen atom at a distance of 5.29 x 10-11 m, the expected
V125BC [204]
Would be A 1012 N/C because The magnitude of the electric field at distance r from a point charge q is E=k
e
​
q/r
2
, so
E=
(5.11×10
−11
m)
2

(8.99×10
9
N.m
2
/C
2
)(1.60×10
−19
C)
​

=5.51×10
11
N/C∼10
1
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making (e) the best choice for this question.
3 0
3 years ago
If the distance between the first order maximum and the tenth order maximum of a double-slit pattern is 18 mm and the slits are
Setler79 [48]

Answer:

Wavelength of light is 600 nm

Explanation:

Given

Distance between the first order maximum and the tenth order maximum of a double-slit pattern = 18 mm

Separation between the slits = 0.15 mm

Distance of screen from the slits = 50 cm

Wavelength

= \frac{18*10^{-3} * 0.15 *10^{-3}}{0.50*9} \\= 6 *10^{-7}\\= 600nm

4 0
3 years ago
You measure the magnetic field of a long, current-carrying wire to be B is equal to start fraction 50 times mu naught over pi en
FromTheMoon [43]

Answer:

  I = 1.875 A

Explanation:

For this exercise we use Ampere's law

             ∫ B . ds = μ₀  I

We use a circular path around the wire whereby B and ds are parallel, whereby the dot product is reduced to the algebraic product

               ds = 2π dr

               B (2πr) = μ₀  I

              I = B 2π R /μ₀  

              r= 7.5 cm = 0.075 m

calculate

             I = (50 μ₀ /π) 2π 0.075 /μ₀  

             I = 1.875 A

           

5 0
3 years ago
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