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EleoNora [17]
2 years ago
15

A shell is fired from the ground with an initial speed of 1.60 × 103 m/s (approximately five times the speed of sound) at an ini

tial angle of 51.0 ◦ to the horizontal. The acceleration of gravity is 9.81 m/s 2 . a) Neglecting air resistance, find the shell’s horizontal range. Answer in units of m.
Physics
1 answer:
Volgvan2 years ago
6 0

Answer:

The horizontal range will be 2.55\times 10^5m

Explanation:

We have given initial speed of the shell u = 1.6\times 10^3m/sec

Angle of projection = 51°

Acceleration due to gravity g=9.8m/sec^2

We have to find maximum range

Horizontal range in projectile motion is given by

R=\frac{u^2sin2\Theta }{g}=\frac{(1.60\times 10^3)^2sin(2\times 51^{\circ})}{9.81}=2.55\times 10^5m

So the horizontal range will be 2.55\times 10^5m

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Which action might lead scientists to develop new explanations about the
sweet [91]

Answer:

The correct option is;

B. Designing experiments to replicate the conditions in which life may have first evolved on Earth

Explanation:

The proof to the hypothesis that life originated from inanimate inorganic, or non-living  molecules which is an explanation for the origin of life on Earth was provided by an experiment designed and performed in 1953 by Stanley L. Miller and Harold C. Urey which consisted of using chemicals proposed in the hypothesis and combining them through a specific design process to replicate expected atmospheric condition before the life began on Earth.

With such successful design of experiments to replicate the conditions in which life may have first evolved on Earth, it was possible to better explain the hypothesis that life originated from inorganic molecule.

5 0
3 years ago
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P6: An object of mass m sits on a spring of constant k in an elevator that is accelerating upwards with acceleration a. a) In te
tankabanditka [31]

Answer:

(a). The spring compressed is \dfrac{ma+mg}{k}.

(b). The acceleration is 1.5 g.

Explanation:

Given that,

Acceleration = a

mass = m

spring constant = k

(a). We need to calculate the spring compressed

Using balance equation

kx-mg=ma

x=\dfrac{ma+mg}{k}....(I)

The spring compressed is \dfrac{ma+mg}{k}.

(b). If the compression is 2.5 times larger than it is when the mass sits in a still elevator,

The compression is given by

x=2.5\times x_{0}

Here, acceleration is zero

So, x=2.5\times\dfrac{mg}{k}

We need to calculate the acceleration

Put the value of x in equation (I)

2.5\times \dfrac{mg}{k}=\dfrac{ma+mg}{k}

2.5\times\dfrac{mg}{k}=\dfrac{m}{k}(a+g)

a=2.5g-g

a=1.5g

Hence, (a). The spring compressed is \dfrac{ma+mg}{k}.

(b). The acceleration is 1.5 g.

8 0
2 years ago
In the graph, which region shows nonuniform positive acceleration?
cricket20 [7]

Answer: A.AB

Explanation:

This Velocity vs Time graph shows the acceleration of a body or object, since acceleration is the variation of velocity in time.

As we can see in the attached image, the graph can be divided in four segments:

OA: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a positive slope, hence we are dealing with a positive uniform acceleration.

AB: In this segment the acceleration is changing at a nonuniform rate, since  in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be positive. This means the <u>acceleration is nonuniform and positive.</u>

BC: In this segment the acceleration is changing at a nonuniform rate, since  in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be negative. This means the acceleration is nonuniform and negative.

CD: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a negative slope, hence we are dealing with a negative uniform acceleration.

From all these segments, the only one that fulfils the nonuniform positive acceleration condition is option A:

Segment AB

3 0
2 years ago
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ohms Long describes the relationship between electronic current voltage and ___
levacccp [35]

Answer:

<em>voltage</em><em> </em><em>,</em><em>current</em><em> </em><em>and</em><em> </em><em>resistance</em><em> </em>

5 0
3 years ago
If a weightlifter lifts a 286 kg mass 0.25 meters above his head, how much PEg does the mass have?
vazorg [7]

Answer:

Change in potential energy is 700.7 J.

Explanation:

Given:

Mass of the object is, m=286 kg

Height to which the object is raised above his head, h=0.25 m

Acceleration due to gravity is, g=9.8 m/s²

We know that for a mass m raised to a height h, the change in potential energy is given as:

\Delta PE=mgh, where, \Delta PE\rightarrow \textrm{Change in Potential Energy}

Now, plug in the given values and solve.

\Delta PE=286\times 9.8\times 0.25=700.7\textrm{ J}

Therefore, the change in the potential energy is 700.7 J.

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3 years ago
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