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erik [133]
3 years ago
15

A bullet is fired horizontally with initial velocity of 800 m/s as a target located 200 mfrom the rifle.(a).How much time is req

uired for the bullet to reach the target
Physics
1 answer:
Aneli [31]3 years ago
3 0

Answer:

t = 0.25 seconds

Explanation:

Given that,

Initial speed of a bullet, v = 800 m/s

Distance from the target is 200 m

We need to find the time required for the bullet to reach the target. Time is simply calculated by the definition of velocity i.e.

t=\dfrac{d}{v}\\\\t=\dfrac{200\ m}{800\ m/s}\\\\t=0.25\ s

So, it will take 0.25 seconds to reach the target.

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It is difficult for astronomers to find object like planets and asteroids because it takes a lot of time to verify the objects locations and what surrounds a certain object in order to prove and be precise of its location
3 0
3 years ago
The counter is 1.5 m high, and the bowl has a mass of 0.5 kg. How much gravitational energy is stored in the bowl-earth system?
alina1380 [7]

Answer:

7.35 J

Im assuming, upon answering the question, that the gravity in this scenario is 9.8? As 9.8 is the gravitational force upon the earth.

5 0
3 years ago
A ball thrown horizontally at vi = 30.0 m/s travels a horizontal distance of d = 55.0 m before hitting the ground. from what hei
tekilochka [14]
Assume no air resistance, and g = 9.8 m/s².

Let
x =  angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity

The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x)  s

With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
 55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0

Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°

The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s

If t = 5.8096 s,
  u*t = 9.467*5.8096 = 55 m (Correct)
or
 u*t = 28.469*15.8096 = 165.4 m (Incorrect)

Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s

The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m

Answer: h = 110.4 m

7 0
3 years ago
Which of the following describe circumstances under which an exculpatory clause is generally unenforceable? Select all that appl
Sveta_85 [38]

Answer:

a. When the parties have unequal bargaining power  

c. When it covers public transportation  

Explanation:

Exculpatory clauses are often found in agreements between a company and a consumer when the activity is in danger, such as at a fitness center or ski resort. The company wants the consumer to understand the risk involved and to avoid lawsuits, so it includes a disclaimer in its contract. These clauses are used to limit liability, so they are not enforced when the parties have unequal bargaining power and when covering public transport.

6 0
3 years ago
(d) Suppose you use a spring to launch a payload horizontally from the asteroid so that the payload ends up far from the asteroi
nydimaria [60]

Answer:

ks= 133.2 N/m

Explanation:

  • Assuming that we can neglect the gravitational potential energy of the mass, and that no other forces acting on the payload, total mechanical energy must be conserved.
  • This energy, at any time, is part elastic potential energy (stored in the spring) and part kinetic energy.
  • When the spring is initially compressed, the payload is at rest, so all energy is elastic potential.
  • Once the spring has returned to its natural state, all this elastic potential energy must have been turned into kinetic energy.
  • If the payload is launched horizontally, and no gravity is present,this means that its final speed will be horizontal only also, according to Newton's First Law.
  • So, we can write the following equation:

       \Delta U + \Delta K = 0 (1)

  • where ΔU = -1/2*k*(Δx)²  (2)
  • and ΔK = 1/2*m*v² (3)
  • Replacing in (2) and (3) by the givens, and simplifying, we can find the stiffness ks as follows:

       k_{s} =\frac{m*v^{2}}{\Delta x^{2}} = \frac{29 kg*(3m/s)^{2}}{(1.4m)^{2}} = 133.2 N/M (4)

5 0
2 years ago
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