A cell phone is released from the top with the speed of 10ms what is the speed 3s after?

Air bags are designed to deploy in 10 ms. Given that the air bags expand 20 cm as they deploy, estimate the acceleration of the

joja [24]

As it is given that the air bag deploy in time

$t = 10 ms = 0.010 s$

total distance moved by the front face of the bag

$d = 20 cm = 0.20 m$

Now we will use kinematics to find the acceleration

$d = v_i*t + \frac{1}{2}at^2$

$0.20 = 0 + \frac{1}{2}a*0.010^2$

$0.20 = 5 * 10^{-5}* a$

$a = 4000 m/s^2$

now as we know that

$g = 10 m/s^2$

so we have

$a = 400g$

so the acceleration is 400g for the front surface of balloon

3 0

3 years ago

Consider the hypothesis below and answer the question that follows. Adding salt to water increases the water’s boiling point. If

irakobra [83]

The hypothesis because its very hard to make and it confounds me

8 0

3 years ago

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Consider the image above. Vi = the initial velocity and Vf = the final velocity. Is there acceleration? Explain your answer.

lutik1710 [3]

<span><span>Velocity is a vector, and the initial and final ones are in opposite directions.
There must have been acceleration in order to change the direction of motion.</span>

A) No. The initial and final velocities are the same.
This is all wrong, and not the correct choice.
It's "Yes", and the initial and final velocities are NOT the same.

B) Yes. The ball had to slow down in order to change direction.
This is poor, and not the correct choice.
The "Yes" is correct, but the explanation is bad.
Acceleration does NOT require any change in speed.

C) No. Acceleration is the change in velocity. The ball's velocity is constant.
This is all wrong, and not the correct choice.
It's "Yes", there IS acceleration, and the ball's velocity is NOT constant.

D) Yes. Even though the initial and final velocities are the same, there is a change in direction for the ball.
This choice is misleading too.
The "Yes" is correct ... there IS acceleration.
The change in direction is the reason.
The initial and final velocities are NOT the same. Only the speeds are.
</span>

3 0

3 years ago

Read 2 more answers

A severe storm has an average peak wave height of 16.4 feet for waves hitting the shore. suppose that a storm is in progress wit

Agata [3.3K]

Before we answer this question, let us first understand what alternate hypothesis is.

The alternative hypothesis is the hypothesis which is used in the hypothesis testing and this is opposite to the null hypothesis. This is the test hypothesis which is usually taken to be that the observations are the result of a real effect in an experiment.

In this case since what we want to set up is the statistical test to see if the waves are dying down, then this means we are trying to determine if the wave height are decreasing, so lesser than 16.4 feet. Therefore:

The alternative hypothesis would state (ANSWER)

Ha: μ less than 16.4 feet and P-value area is on the left of the mean.

While the null hypothesis is the opposite and would state

H0: mu equals 16.4 feet

4 0

3 years ago

Three point charges are placed on the x−y plane: a + 50.0-nC charge at the origin, a −50.0-nC charge on the x axis at 10.0 cm, a

butalik [34]

Answer:

(a) F = 0.00322i - 0.00793j with magnitude |F| = 0.00856N

(b) E = -42846.7 N/C

Explanation:

The diagram attached below explains some parameters.

Parameters given:

Charge Q1 = +50 nC at point (0, 0)

Charge Q2 = -50 nC at point (0.1, 0)

Charge Q3 = +150 nC at point (0.1, 0.08)

* The distances are in meters.

(a) The total electric force on the charge Q3 due to Q1 and Q2 is the vector sum of the forces due to Q1 and Q2. Mathematically,

F = F1 + F2

FORCE DUE TO Q1 i.e. F(Q1, Q3)

We have to find the x and y components.

From the diagram, we can find θ using SOHCAHTOA:

θ = tan⁻¹ (0.08/0.1)

θ = 38.66⁰

The distance between Q1 and Q3 can be found using Pythagoras theorem:

x² = 0.08² + 0.1²

x = 0.128 m

F1 = Fx(Q1, Q3)i + Fy(Q1, Q3)j

F1 = iF(Q1, Q3)cosθ + jF(Q1, Q3)sinθ

F(Q1, Q3) = (k * Q1 * Q3) / r²

k = Coulombs constant

F(Q1, Q3) = (9 * 10⁹ * 50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.128)²

F(Q1, Q3) = 0.00412N

F1 = i0.00412 * cos38.66 + j0. 00412 * sin38.66

F1 = 0.00322i + 0.00257j N

FORCE DUE TO Q2 i.e. F(Q2, Q3)

We have to find the x and y components.

F2 = Fx(Q2, Q3)i + Fy(Q2, Q3)j

F2 = iF(Q2, Q3)cos90 + jF(Q2, Q3)cos0

F(Q2, Q3) = (k * Q2 * Q3) / r²

F(Q2, Q3) = (9 * 10⁹ * -50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.08)²

F(Q2, Q3) = -0.0105N

F2 = -i0.0105 * cos90 - j0.0105 * cos0

F2 = - 0.0105j N

Hence, the total force will be

F = F1 + F2

F = 0.00322i + 0.00257j - 0.0105j

F = 0.00322i - 0.00793j N

The magnitude of this force is:

|F| = √(0.00322² + (-0.00793²)

|F| = 0.00856N

(b) The electric field at charge Q3 is the sum of the electric fields due to Q1 and Q2:

E = E1 + E2

E1, electric field due to Q1 = kQ1/r²

E1 = (9 * 10⁹ * 50 * 10⁻⁹) / (0.128²)

E1 = 27465.8 N/C

E2, electric field due to Q2 = (9 * 10⁹ * -50 * 10⁻⁹) / (0.08²)

E1 = -70312.5N/C

The total electric field:

E = E1 + E2

E = 27465.8 - 70312.5

E = -42846.7 N/C

3 0

2 years ago