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2 years ago

Step gently onto a bathroom scale, read the dial, and then jump from a chair onto the same scale.

Physics

1 answer:

disa [49]2 years ago

6 0

Answer:

1)The second is correct. The scale shows a higher value initially and after a short time it shows the same value of rising gently

2) The correct answer is the first does not change

Explanation:

1) Let's look for the solution of the problem before seeing the statements given

We use Newton's second law

Fb - w = ma

Fb = ma + mg

Fb = m (a + g)

When you go up smoothly, the acceleration a is zero, but when you go up sharply it is different from zero, after a short time it reaches zero and the strength of the bastecía is the same as at the beginning.

Affirmations

The second is correct. The scale shows a higher value initially and after a short time it shows the same value of rising gently

2) changes the value of g

The value of the acceleration of gravity depends on the mass and radius of the Earth only, not the conditions or movements of the person. The value of g does not change

The correct answer is the first does not change

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Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s

frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

y₁ = v₀₁ t + ½ g t²

y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

t ’= (t-t₀)

y₂ = v₀₂ t ’+ ½ g t’²

y₂ = 0 + ½ g (t-t₀)²

y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

S = y₁ -y₂

S = ½ g t²– ½ g (t-t₀)²

S = ½ g [t² - (t²- 2 t to + to²)]

S = ½ g (2 t t₀ - t₀²)

S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to. The rock y has not left and the distance increases

For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂

3 0

2 years ago

A 2 000-kg sailboat experiences an eastward force of 3 000 N by the ocean tide and a wind force against its sails with a magnitu

Vesnalui [34]

Answer:

The magnitude of the resultant acceleration is 2.2 $m/s^2$

Explanation:

Mass (m) of the sailboat = 2000 kg

Force acting on the sailboat due to ocean tide is $F_1$ = 3000N

Eastwards means takes place along the positive x direction

Then $F_{1x}$ = 3000N and $F_{1y}$ = 0

Wind Force acting on the Sailboat is $F_2$ = 6000N directed towards the northwest that means at an angle 45 degree above the negative x axis

Then

$F_{2x}$ = -(6000N) cos 45 degree = -4242.6 N

$F_{2y}$ = (6000N) cos 45 degree = 4242.6 N

Hence , the net force acting on the sailboat in x direction is

$F_x = F_{1x}+ F_{2x}$

= - 3000 N + 4242.6 N

= - 3000 N +4242.6 N

= 1242.6N

Net Force acting on the sailboat in y direction is

$F_y = F_{1y}+ F_{2y}$

= 0+ 4242.6N

= 4242.6N

The magnitude of the resultant force =

Using pythagorean theorm of 1243 N and 4243 N

$\sqrt{(1242.6)^2 + (4242.6)^2$

$\sqrt{(1544054.76) + (17999654.8)}$

$\sqrt{(19543709.5)^2}$

4420.8 N

F = ma

$a = \frac{F}{m}$

$a =\frac{4420.8}{ 2000}$

=2.2 $m/s^2$

4 0

2 years ago

A square block of steel with volume 10 cm3 and mass of 75 g is cut precisely in half. The density of the two smaller pieces is n

dem82 [27]

A. Density only depends on the substance. It doesn't matter whether you have a little chip of it or a supertanker full of it ... the density doesn't change.

4 0

2 years ago

Read 2 more answers

Two speakers, A and B, produce identical sound waves. A listener is 3.2 m away from speaker A. The listener finds the lowest fre

earnstyle [38]

Answer:

0.83 m or 5.57 m

Explanation:

Destructive interference will occur when the distances from the speakers differ by 1/2 wavelength.

The length of 1 cycle of 72.4 Hz is ...

λ = v/f = (343 m/s)/(72.4 Hz) ≈ 4.738 m

So, the distance of the listener from speaker B is ...

3.2 m ± (4.738 m)/2 = {0.83 m, 5.57 m} . . . either of these distances

_____

The location could be at additional multiples of 4.738 m, but we think not. The sound intensity drops off with the square of the distance from the speaker, so identical sound waves from the speakers will sound quite different at different distances from the speakers. For best interference, the distances need to be as close to the same as possible. That will be at 3.2 m and 5.57 m.

_____

<em>Comment on the speed of sound</em>

We don't know what speed you are to use for the speed of sound. We have used 343 m/s. Some sources use 340 m/s, which will give a result different by 2 or 3 cm.

8 0

2 years ago

two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0

2 years ago