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A cannonball is fired horizontally at the same time a ball being dropped from the same height How do the times it takes them to

ipn [44]

Answer:

The cannonball and the ball will both take the same amount of time before they hit the ground.

Explanation:

For a ball fired horizontally from a given height, there is only a vertical acceleration on it towards the ground. This acceleration is equal to the acceleration due to gravity (g = 9.81 m/s^2). A ball dropped from a height will also only experience the same vertical acceleration downwards which is also equal to g = 9.81 m/s^2.

Therefore both the cannonball and the ball will take the same amount of time to hit the ground if they are released/fired from the same height.

3 0

3 years ago

Objects with masses of 135 kg and a 435 kg are separated by 0.500 m. Find the net gravitational force exerted by these objects o

IrinaVladis [17]

Answer:

The net gravitational force on the mass is $1.27\times 10^{-5}$

Explanation:

We have by Newton's law of gravity the force of attraction between masses $m_{1},m_{2}$

$F_{att}=G\frac{m_{1}m_{2}}{r^{2}}$

Applying vales we get

Force of attraction between 135 kg mass and 38 kg mass is

$F_{1}=6.67\times 10^{-11}\frac{135\times 38}{(0.25)^{2}}\\\\F_{1}=5.47\times 10^{-6}N$

Force of attraction between 435 kg mass and 38 kg mass is

$F_{2}=6.67\times 10^{-11}\frac{435\times 38}{(0.25)^{2}}\\\\F_{2}=1.76\times 10^{-5}N$

Thus the net force on mass 38.0 kg is $F_{2}-F_{1}=1.76\times 10^{-5}-5.47\times 10^{-6}\\\\F_{2}-F_{1}=1.27\times 10^{-5}$

8 0

3 years ago

A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 33

Anuta_ua [19.1K]

Answer:

$\frac{v_{2}}{v_{1}}=2$ .

Explanation:

The average kinetic energy per molecule of a ideal gas is given by:

$\bar{K}=\frac{3k_{B}T}{2}$

Now, we know that $\bar{K} = (1/2)m\bar{v}^{2}$

Before the absorption we have:

$(1/2)m\bar{v_{1}}^{2}=\frac{3k_{B}T_{1}}{2}$ (1)

After the absorption,

$(1/2)m\bar{v_{2}}^{2}=\frac{3k_{B}T_{2}}{2}$ (2)

If we want the ratio of v2/v1, let's divide the equation (2) by the equation (1)

$\frac{v_{2}^{2}}{v_{1}^{2}}=\frac{T_{2}}{T_{1}}$

$\frac{v_{2}}{v_{1}}=\sqrt{\frac{T_{2}}{T_{1}}}$

$\frac{v_{2}}{v_{1}}=\sqrt{\frac{1340}{335}}$

$\frac{v_{2}}{v_{1}}=\sqrt{4}$

Therefore the ratio will be $\frac{v_{2}}{v_{1}}=2$

I hope it helps you!

4 0

3 years ago

Read 2 more answers

A gas, behaving ideally, fills a fixed volume container at a pressure P₁ and at a temperature T₁. The temperature of the contain

Lynna [10]

Answer:

c. P₁/T₁=P₂/T₂

Explanation:

neither Avogadro’s, Charles’, or Boyle’s law formula can be used, since some parameters like volume is not given,

to find P₂, given P₁, T₁, and T₂ we will therefore use Gay-lussac's law.

gay lussacs law state that, provided volume is kept constant, pressure is directly proportional to temperature.

the volume volume is said to be filled, i.e its is kept constants when temperature is change

4 0

3 years ago

Due to historical difficulty in delivering supplies by plane, one of your colleagues has suggested you develop a catapult for sl

ikadub [295]

Answer:

Please see below as the answer is self-explanatory.

Explanation:

We can take the initial velocity vector, which magnitude is a given (67 m/s) and project it along two directions perpendicular each other, which we choose horizontal (coincident with x-axis, positive to the right), and vertical (coincident with y-axis, positive upward).
Both movements are independent each other, due to they are perpendicular.
In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.
Applying the definition of cosine of an angle, we can find the horizontal component of the initial velocity vector, as follows:

$v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)$

Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:

$t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s (2)$

In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.
Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:

$\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (3)$

In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:

$v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)$

Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:

$\Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)$

Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.

4 0

3 years ago