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2 years ago

A 15-kilogram cart is at rest on a horizontal

1 answer:

6 0

The most likely answer to this problem would be (1) more mass and more inertia.

A 15-kilogram cart at rest and a 5-kilogram box would make up a 20-kilogram cart and box that is at rest on a horizontal surface. The mass changed into something more, of course, as a result of combining the two object into one and by combining the two objects' mass, the inertia that it previously possessed as a cart by itself was increased when the inertia of the box was also combined to the cart.

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Mariulka [41]

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6 0

3 years ago

What are four drawbacks to using fossil fuels in Alberta thermal power stations?

puteri [66]

Answer:

Fossil fuels have been used for centuries to generate power, but there are many disadvantages associated with their use:

Fossil fuels pollute the environment.

Fossil fuels are non-renewable and unsustainable.

Drilling for fossil fuels is a dangerous process.

Explanation:

can you help me please? brainly.com/question/21745542

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3 years ago

Assuming both graduated cylinders are holding water at room temperature, which cylinder has more thermal energy?

natta225 [31]

Answer:

The correct option is;

The graduate cylinder with more water has more thermal energy because it is holding more water molecules

Explanation:

Given that the thermal energy of the system is the energy possessed by the system by virtue of the increased motion of the particles by virtue of a transfer of heat, when the content of the system is heated

The thermal energy, Q is given by the following equation;

Q = Mass, m × The specific heat capacity, C × The change in temperature, ΔT

Given that the graduated cylinder with more water has more mass and therefore, more water molecules, than the cylinder with less water, the cylinder with more water has more thermal energy.

3 0

3 years ago

A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s

Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

<u>Given:</u>

Time of flight = t = 20 s
Angle of the initial velocity of projectile with the horizontal = $\theta = 30^\circ$

<u>Assume:</u>

Initial velocity of the projectile = u
R = Range of the projectile during the time of flight
H = maximum height of the projectile
D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

Initial horizontal velocity of the projectile = $u\cos \theta$
Initial horizontal velocity of the projectile = $u\sin \theta$

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

$\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s$

Hence, the initial speed of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

$\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m$

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

$R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m$

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

$\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m$