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4 years ago

13

When does a cold front develop? A. when a cold air mass invades a warm air mass B. when two masses of cold air meet C. when two

masses of warm air meet

2 answers:

ella [17]4 years ago

7 0

It should be A since a cold front develops when a cold air mass moves into an area with warmer air.

mr_godi [17]4 years ago

6 0

Answer:

when a cold air mass invades a warm air mass

Explanation:

:)

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A bucket of water with mass 5 kg sits on the ground with a coefficient of static friction of 0.35. What is the maximum force of

allochka39001 [22]

Answer:

The force of static friction is 17.15 N

Explanation:

It is given that,

Mass of the bucket, m = 5 kg

The coefficient of static friction is, $\mu=0.35$

We need to find the maximum force of static friction. It is given by :

$F=\mu mg$

$F=0.35\times 5\ kg\times 9.8\ m/s^2$

F = 17.15 N

So, the force of static friction is 17.15 N. Hence, this is the required solution.

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4 years ago

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Answer:

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Explanation:

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3 years ago

A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 14 m/s. A passenger accidentally drops his camera fr

fgiga [73]

Answer:

<em>The balloon is 66.62 m high</em>

Explanation:

<u>Combined Motion </u>

The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

$\displaystyle y=y_o+v_o\ t-\frac{g\ t^2}{2}$

The values are

$\displaystyle y_o=15\ m$

$\displaystyle v_o=14\ m/s$

We must find the values of t such that the height of the camera is 0 (when it hits the ground)

$\displaystyle y=0$

$\displaystyle y_o+v_o\ t-\frac{g\ t^2}{2}=0$

Multiplying by 2

$\displaystyle 2y_o+2v_ot-gt^2=0$

Clearing the coefficient of $t^2$

$\displaystyle t^2-\frac{2\ V_o}{g}\ t-\frac{2\ y_o}{g}=0$

Plugging in the given values, we reach to a second-degree equation

$\displaystyle t^2-2.857t-3.061=0$

The equation has two roots, but we only keep the positive root

$\displaystyle \boxed {t=3.69\ s}$

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is

$\displaystyle Y_r=15+V_r.t$

$\displaystyle Y_r=15+14\times3.69$

$\displaystyle \boxed{Y_r=66.62\ m}$

3 0

3 years ago

The only force acting on a 2.5 kg canister that is moving in an xy plane has a magnitude of 3.0 N. The canister initially has a

Effectus [21]

Answer:44.58 J

Explanation:

mass of block $\left ( m\right )=2.5 kg$

Force magnitude=3 N

Initial velocity = $2.3\hat{i} m/s$

Final velocity= $6.4\hat{j} m/s$

Initial Kinetic Energy= $\frac{1}{2}mv^2$

= $\frac{1}{2}\times 2.5\times 2.3^2=6.612 J$

Final Kinetic Energy= $\frac{1}{2}mv^2$

= $\frac{1}{2}\times 2.5\times 6.4^2=51.2 J$

Work Done =Final -Initial Kinetic energy=51.2-6.612=44.58 J

4 0

3 years ago

A moon orbits a planet every 42 hours with a mean orbital radius of .002819 AU. The mass of the moon is 8.932 x 1022 kg. Using N

Pepsi [2]

Answer:

The mass of the planet is $1.9407\times10^{27}\ kg$

Explanation:

Given that,

Time period = 42 hours = 151200 sec

Orbital radius = 0.002819 AU = 421716397.5 m

Mass of moon $m=8.932\times10^{22}\ kg$

We need to calculate the mass of the planet

Using Kepler’s third law

$T^2\propto a^3$

$T^2=\dfrac{4\pi^2}{G(M+m)}\times a^3$

Where, a = orbital radius

T = time period

G = gravitational constant

M = mass of moon

m = mass of planet

Put the value into the formula

$(151200)^2=\dfrac{4\pi^2}{6.673\times10^{-11}(8.932\times10^{22}+m)}\times(421716397.5)^3$

$(8.932\times10^{22}+m)=\dfrac{4\pi^2}{6.673\times10^{-11}}\times\dfrac{(421716397.5)^3}{(151200)^2}$

$(8.932\times10^{22}+m)=1.94087\times10^{27}$

$m=1.94087\times10^{27}-8.932\times10^{22}$

$m=1.9407\times10^{27}\ kg$

Hence, The mass of the planet is $1.9407\times10^{27}\ kg$

8 0

4 years ago