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Radda [10]
3 years ago
5

A shell if fired from the ground with an initial velocity of 1,700 m/s at an initial angle of 55 degrees to the horizontal. Negl

ecting air resistance, what is the shell's horizontal range?
Physics
1 answer:
Triss [41]3 years ago
8 0

Answer:

Therefore the horizontal range = 294897.96 m.

Explanation:

Range of a projectile: The range is defined as the horizontal distance from the point of projection to the point where the projectile hit the projection plane again. The S.I unit of range is Meter (m).

It can be expressed mathematically as

R = u²sin2∅/g............................. Equation 1

Where R = Horizontal range, ∅ = angle of projection, u = initial velocity, g = acceleration due to gravity.

<em>Given: u = 1700 m/s, </em>∅ = 55°,

Constant: g = 9.8 m/s²

Substituting these values into equation 1

R = (1700²sin55)/9.8

R = 2890000/9.8

R = 294897.96 m.

Therefore the horizontal range = 294897.96 m.

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Why is no image formed when an object is at the focal point of a converging lens?
Temka [501]

Answer:

the refracted rays neither converge nor diverge. After refracting, the light rays are traveling parallel to each other and cannot produce an image.

Explanation:

6 0
3 years ago
The volume of a gas decreases from 15.7 mºto 11.2 m3 while the pressure changes from 1.12 atm to 1.67 atm. If the
Svetlanka [38]

Answer:

Approximately 261\; \rm K, if this gas is an ideal gas, and that the quantity of this gas stayed constant during these changes.

Explanation:

Let P_1 and P_2 denote the pressure of this gas before and after the changes.

Let V_1 and V_2 denote the volume of this gas before and after the changes.

Let T_1 and T_2 denote the temperature (in degrees Kelvins) of this gas before and after the changes.

Let n_1 and n_2 denote the quantity (number of moles of gas particles) in this gas before and after the changes.

Assume that this gas is an ideal gas. By the ideal gas law, the ratios \displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1} and \displaystyle \frac{P_2 \cdot V_2}{n_2 \cdot T_2} should both be equal to the ideal gas constant, R.

In other words:

R = \displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1}.

R =\displaystyle \frac{P_2 \cdot V_2}{n_2 \cdot T_2}.

Combine the two equations (equate the right-hand side) to obtain:

\displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1} = \frac{P_2 \cdot V_2}{n_2 \cdot T_2}.

Rearrange this equation for an expression for T_2, the temperature of this gas after the changes:

\displaystyle T_2 = \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2} \cdot T_1.

Assume that the container of this gas was sealed, such that the quantity of this gas stayed the same during these changes. Hence: n_2 = n_1, (n_2 / n_1) = 1.

\begin{aligned} T_2 &= \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2}\cdot T_1 \\[0.5em] &= \frac{1.67\; \rm atm}{1.12\; \rm atm} \times \frac{11.2\; \rm m^{3}}{15.7\; \rm m^{3}} \times 1 \times 245\; \rm K \\[0.5em] &\approx 261\; \rm K\end{aligned}.

4 0
3 years ago
When a fish expands its air bladder, the density of the fish
Mazyrski [523]
<h2>Answer: decreases</h2>

Every substance, body or material has mass and volume, however the mass of different substances occupy different volumes.

This is where density  D  appears as a  physical characteristic property of matter that establishes a relationship between the mass  m of a body or substance and the volume  V it occupies.

D=\frac{m}{V}

So, according to this equation, the density is inversely ptoportional to the volume:

If the volume increases, the density decreases.

This is what a fish does to have buoyancy, since the density of a body is related to its buoyancy:

A body will float on another fluid if its density is lower.

<h2>This is what the fish does when it expands its air bladder, incrementing its volume, hence decreasing its density.</h2>

8 0
3 years ago
A ball with a mass of 0.25 kg hits a gym ceiling with a force of 78.0 n. what happens next?
QveST [7]
<span>a. The ball accelerates downward with a force of 80.5 N. This is a rather badly worded question since the answer depends upon whether or not the impact with the gym ceiling was elastic or non-elastic. With an elastic collision, the ball will accelerate downward with it's original force plus the acceleration due to gravity. With a non-elastic collision (the energy in the ball being used to damage the ceiling of the gym), then the initial energy the ball has would be expended while causing damage to the gym ceiling and then the ball would accelerate downward solely due to the force of gravity. In either case, we need to take into consideration the force of gravity. So multiply the mass of the ball by the gravitational acceleration, giving F = 0.25 kg * 9.8 m/s^2 = 2.45 kg*m/s^2 = 2.45 N Since the initial force is 78.0 newtons, let's add them 78.0 N + 2.45 N = 80.45 N and after rounding to 3 figures, gives 80.5 N So we have a possible answer of 2.45N or 80.5N depending upon if the collision is elastic or not. And unfortunately, both possible answers are available. Since no mention of the ceiling being damaged is made in the question, and to be honest a 100% non-elastic collision is highly unlikely, I will assume the collision is elastic, so the answer is "a".</span>
6 0
3 years ago
Which of the following numbers is equal to an object's acceleration?
Elenna [48]
<span>C. the slope of that objects velocity-time graph

Hope it helps!
</span>
5 0
3 years ago
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