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nordsb [41]
4 years ago
8

A crate pushed along the floor with velocity vâ i slides a distance d after the pushing force is removed. if the mass of the cra

te is doubled but the initial velocity is not changed, what distance does the crate slide before stopping?
Physics
1 answer:
Alenkinab [10]4 years ago
7 0
<span>d The mass is doubled which means that both the momentum and kinetic energy are also doubled. Also the normal force that's acting along with the coefficient of kinetic friction is also doubled. So the friction that's working to slow down the crate is doubled. So the crate will have double the kinetic energy that needs to be dissipated, but the rate of dissipation is also doubled, so the total time required to dissipate the kinetic energy is the same. And since both crates start out with the same velocity and since they'll lose energy (and velocity) at the same proportional rate, they'll take the same distance to slide to a stop.</span>
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A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand
enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

v=gt

And the distance traveled downwards is:

\displaystyle y=\frac{gt^2}{2}

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

\displaystyle t=\sqrt{\frac{2y}{g}}

Replacing into the first equation:

\displaystyle v=g\sqrt{\frac{2y}{g}}

Rationalizing:

\displaystyle v=\sqrt{2gy}

Let's call v1 the final speed of the package dropped from a height H. Thus:

\displaystyle v_1=\sqrt{2gH}

Let v2 be the final speed of the package dropped from a height 4H. Thus:

\displaystyle v_2=\sqrt{2g(4H)}

Taking out the square root of 4:

\displaystyle v_2=2\sqrt{2gH}

Dividing v2/v1 we can compare the final speeds:

\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}

Simplifying:

\displaystyle v_2/v_1=2

The final speed of the second package is twice as much as the final speed of the first package.

5 0
3 years ago
Charge g is distributed in a spherically symmetric ball of radius a. (a) Evaluate the average volume charge density p. (b) Now a
nasty-shy [4]

Answer:

Explanation:

The volume of a sphere is:

V = 4/3 * π * a^3

The volume charge density would then be:

p = Q/V

p = 3*Q/(4 * π * a^3)

If the charge density depends on the radius:

p = f(r) = k * r

I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.

Q = \int\limits^{2*\pi}_0\int\limits^\pi_0  \int\limits^r_0 {k * r} \, dr * r*d\theta* r*d\phi

Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0  \int\limits^r_0 {r^3} \, dr * d\theta* d\phi

Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 {\frac{r^4}{4}} \, d\theta* d\phi

Q = k *\int\limits^{2*\pi}_0 {\frac{\pi r^4}{4}} \,  d\phi

Q = \frac{\pi^2 r^4}{2}}

Since p = k*r

Q = p*π^2*r^3 / 2

Then:

p(r) = 2*Q / (π^2*r^3)

3 0
3 years ago
The mass of a hot-air balloon and its cargo (not including the air inside) is 170 kg. The air outside is at 10.0°C and 101 kPa.
scoundrel [369]

Answer:

108.37°C

Explanation:

P₁ = Initial pressure = 101 kPa

V₁ = Initial volume = 530 m³

T₁ = Initial temperature = 10°C = 10+273.15 =283.15 K

P₂ = Final pressure = 101 kPa (because it is open to atmosphere)

V₂ = Final volume = 530 m³

P₁V₁ = n₁RT₁

⇒101×530 = n₁RT₁

⇒53530 J = n₁RT₁

P₂V₂ = n₂RT₂

⇒53530 J = n₂RT₂

\frac{m_1}{m_2}=\frac{\rho V_1}{\rho V_1-170}\\\Rightarrow \frac{m_1}{m_2}=\frac{1.244\times 530}{1.244\times 530-170}=1.347\\\Rightarrow \frac{m_1}{m_2}=1.347\\\Rightarrow \frac{n_1}{n_2}=1.347

Dividing the first two equations we get

1=\frac{n_1}{n_2}\frac{T_1}{T_2}\\\Rightarrow 1=1.347\frac{283.15}{T_2}\\\Rightarrow T_2=1.347\times 283.15= 381.52\ K

∴Temperature must the air in the balloon be warmed before the balloon will lift off is 381.25-273.15 = 108.37°C

8 0
3 years ago
Consider the 65 N light fixture supported as in the figure. Find the tension in the supporting wires.
ASHA 777 [7]

By using Lami's theorem formula, the tension in the supporting wires is 48.6 Newtons

TENSION

  • Tension is also a force having Newton as S.I unit.
  • The tension in the wire will be the same.

This question can be solved by using either vector diagram or by using  Lami's theorem.

The sum of two given angles  = 42 + 42 = 84 degrees

The third angle = 180 - 84 = 96 degrees.

Below is the Lami's theorem formula

\frac{T}{sin\alpha } = \frac{T}{sin\beta } = \frac{W}{sinY}

Where

\alpha  = \beta = 42 + 90 = 132 degrees

Y = 96 degrees

W = 65 N

By using the formula, we have

\frac{T}{sin\alpha } =  \frac{W}{sinY}

T/sin 132 = 65/sin96

Cross multiply

T = 0.743 x 65.57

T = 48.56 N

Therefore, the tension in the supporting wires is 48.6 Newtons approximately.

Learn more about Tension here: brainly.com/question/24994188

3 0
3 years ago
I need help me with my question
lisov135 [29]

The tilt of the moon's axis does not allow for monthly alignment, so the lunar and solar eclipse do not happen every month.

<h3>How do the lunar and solar eclipse occur?</h3>
  • For the occurrence of lunar and solar eclipse, the sun, moon and the earth must remain in a plan and along a straight line.
  • When the earth appears in between the sun and the moon, lunar eclipse occurs.
  • When the moon appears in between the sun and the earth, solar eclipse occurs.
  • The moon and earth are rotating not only around the sun, but also around the black hole of Milky way galaxy.
  • So they are not present in a plan as well as in a straight line in every full moon and new moon time.

Thus, we can conclude that the option D is correct.

Learn more about the lunar eclipse and solar eclipse here:

brainly.com/question/8643

#SPJ1

4 0
2 years ago
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