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3 years ago

9

A particular grocery bag can withstand 40 N of tension force before it rips. If a shopper places 3 kg of food in the bag, what i

s the maximum acceleration the bag can withstand before ripping?

2 answers:

snow_lady [41]3 years ago

8 0

Answer;

= 13.33 m/s²

Solution;

From the second Newton's law of motion;

F= ma

F is the force = 40 N

M is the mass = 3 kg

a is the acceleration;

40 = 3 × a

a = 40/3

= 13.33 m/s²

Hence; the maximum acceleration the bag can withstand before ripping is 13.33 m/s²

Lesechka [4]3 years ago

3 0

Answer:

3.3 is right

Explanation:

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In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue

agasfer [191]

Answer:

a) p₀ = 1.2 kg m / s, b) p_f = 1.2 kg m / s, c) θ = 12.36, d) v_{2f} = 1.278 m/s

Explanation:

a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved

a) the initial impulse is

p₀ = m v₁₀ + 0

p₀ = 0.6 2

p₀ = 1.2 kg m / s

b) as the system is isolated, the moment is conserved so

p_f = 1.2 kg m / s

we define a reference system where the x-axis coincides with the initial movement of the cue ball

we write the final moment for each axis

X axis

p₀ₓ = 1.2 kg m / s

p_{fx} = m v1f cos 20 + m v2f cos θ

p₀ = p_f

1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ

1.2482 = v_{2f} cos θ

Y axis

p_{oy} = 0

p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ

0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ

0.2736 = v_{2f} sin θ

we write our system of equations

0.2736 = v_{2f} sin θ

1.2482 = v_{2f} cos θ

divide to solve

0.219 = tan θ

θ = tan⁻¹ 0.21919

θ = 12.36

let's look for speed

0.2736 = v_{2f} sin θ

v_{2f} = 0.2736 / sin 12.36

v_{2f} = 1.278 m / s

7 0

2 years ago

AleksAgata [21]

Answer:

B.C. D. G.

Explanation:

A vector quantity, has both magnitude and direction. A tip to remember is if you can add a direction to it! You wouldnt say 30 pounds north, but you would say 30 mph north.

<em>I hope this helped! Comment if you have any questions! :)</em>

3 0

3 years ago

"A ball with a mass of 0.1 kg is rolling at a velocity of 5 m/s. What is its

Mumz [18]

Answer:

velocity

Explanation:

because the si unit of mass is kg, velocity is m/s, acceleration is m/S2 , moment is kgm2/s . so 5 is given as velocity.

3 0

2 years ago

In a historical movie, two knights on horseback start from rest 86 m apart and ride directly toward each other to do battle. Sir

Harlamova29_29 [7]

Answer:

Relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

Explanation:

Let the distance covered by Sir George be $S_{1}$

and the distance covered by Sir Alfred be $S_{2}$

Since the knights collide, hence they must have traveled for the same amount of time just before collision

From one of the equations of motion for linear motion

$S = ut + \frac{1}{2}at^{2}$

Where $S$ is the distance traveled

$u$ is the initial velocity

$a$ is the acceleration

and $t$ is the time

For Sir George,

$S = S_{1}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.21 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{1} = (0)t + \frac{1}{2}(0.21)t^{2}\\S_{1} = 0.105 t^{2}\\$

$t^{2} = \frac{S_{1}}{0.105}$

Now, for Sir Alfred

$S = S_{2}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.26 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{2} = (0)t + \frac{1}{2}(0.26)t^{2}\\S_{2} = 0.13 t^{2}\\$

$t^{2} = \frac{S_{2}}{0.13}$

Since, they traveled for the same time, $t$ just before collision, we can write

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$

Since, the two nights are 86 m apart, that is, the sum of the distances covered by the knights just before collision is 86 m. Then we can write that

$S_{1} + S_{2} = 86$ m

Then, $S_{2} = 86 - S_{1}$

Then,

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$ becomes

$\frac{S_{1}}{0.105}= \frac{86 -S_{1}}{0.13}$

$0.13{S_{1}}= 0.105({86 -S_{1}})\\0.13{S_{1}}= 9.03 - 0.105S_{1}}\\0.13{S_{1}} + 0.105S_{1}}= 9.03 \\0.235{S_{1}} = 9.03\\{S_{1}} =\frac{9.03}{0.235}$

$S_{1} = 38.43$ m

∴ Sir George covered a distance of 38.43 m just before collision.

Hence, relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

6 0

3 years ago

A velocity selector has a magnetic field of magnitude 0.22 T perpendicular to an electric field of magnitude 0.51 MV/m.

ohaa [14]

Answer:

speed of a particle = 2.31 × $10^{6}$ m/s

energy of proton required = 27.77 KeV

energy of electron required = 15.171 eV

Explanation:

given data

magnetic field of magnitude = 0.22 T

electric field of magnitude = 0.51 MV/m

to find out

speed of a particle and energy must protons have to pass through undeflected and energy must electrons have to pass through undeflected

solution

we know that force due to magnetic and electric field is express as

force due to magnetic field B = qvB ..............1

and force due to electric field E = qE .....................2

so without deflection force due to magnetic field = force due to electric field

so here qvB = qE

and V = $\frac{E}{B}$ ...................3

put here value

V = $\frac{0.51*10^6}{0.22}$

speed of a particle = 2.31 × $10^{6}$ m/s

and

now energy of proton required will be here as

energy of proton required = mass of proton × $\frac{V^2}{2}$

put here value

energy of proton required = 1.67 × $10^{-27}$ × $\frac{(2.31*10^6)^2}{2}$

energy of proton required = 4.45 × $10^{-15}$ J

energy of proton required = 4.45 × $10^{-15}$ J ÷ (1.602 × $10^{-19}$

energy of proton required = 27777.777 eV

energy of proton required = 27.77 KeV

and

now we get here energy of electron required that is

energy of electron required = mass of electron × $\frac{V^2}{2}$

put here value

energy of electron required = 9.11 × $10^{-31}$ × $\frac{(2.31*10^6)^2}{2}$

energy of electron required = 24.305× $10^{-19}$ J

energy of electron required = 24.305 × $10^{-19}$ J ÷ (1.602 × $10^{-19}$

energy of electron required = 15.171 eV

5 0

3 years ago

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