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Svetach [21]
2 years ago
7

The launching velocity of a missile is 20.0 m/s, and it is shot at 53º above the horizontal. Which of the velocity components (n

eglecting air resistance) remains constant throughout the flight?
Physics
2 answers:
DochEvi [55]2 years ago
6 0

Answer:

Horizontal component of velocity will remains conserved which is v = 12 m/s

Explanation:

Two components of velocity of the projectile is given as

\vec v = 20 cos53 \hat i + 20 sin53 \hat j

\vec v = 12\hat i + 16\hat j

now we know that the acceleration is only due to gravity as we are considering the force is only vertically downwards due to gravity and there is no air resistance during whole motion.

So we will have acceleration as

\vec a = 0\hat i - g\hat j

now we can say that here we do not have any acceleration along horizontal direction so this component of velocity will remain conserved.

So the constant component of velocity is along horizontal component and it is given by v = 12 m/s

Gekata [30.6K]2 years ago
5 0
Neglecting air resistance, the horizontal component remains constant. The angle doesn't matter.
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In the produce section of a supermarket, five pears are placed on a spring scale. The placement of the pears stretches the sprin
tankabanditka [31]

Answer:

The displacement of the spring due to weight is 0.043 m

Explanation:

Given :

Mass m = 2 Kg

Spring constant k = 450 \frac{N}{m}

According to the hooke's law,

  F = -kx

Where F = force, x = displacement

Here,

F = mg         ( g = 9.8 \frac{m}{s^{2} } )

F = 2 \times 9.8 = 19.6 N

Now for finding displacement,

  x = \frac{F}{k}

Here minus sign only represent the direction so we take magnitude of it.

  x = \frac{19.6}{450}

  x = 0.043 m

Therefore, the displacement of the spring due to weight is 0.043 m

8 0
2 years ago
How does end A of the rod react when the charged ball approaches it after a great many previous contacts with end A? Assume that
sleet_krkn [62]

Answer: IT IS STRONGLY REPELLED

Explanation: The laws of guiding magnetic attraction or repulsion of Magnetic materials,states that when like poles are brought together they repel each other, but when unlike poles are brought together they are attracted.

The rod will be strongly repelled because the forces on the rod is greater and has the same Polarity as the charged ball.

6 0
1 year ago
An object weighs 200 newtons at a distance of 100 kilometers above the center of a small uniform planet. how much will the objec
disa [49]

Since the law of gravitation is an inverse square law if you quadruple the radius the f will drop by a factor of 16 SO the object would weigh 200/16 = 12.5N

In other words, as the distance, or radius, quadruples the weight becomes 1/16 of the original weight. Just plug in 4 for r and when you square it you get 16. The numerator is 1 so that is how the weight becomes 1/16.

7 0
2 years ago
A muon formed high in the Earth's atmosphere is measured by an observer on the Earth's surface to travel at speed V - 0.983c for
Alex_Xolod [135]

Answer:

The moun lives 2.198*10^-6 s as measured by its own frame of reference

The Earth moved 648 m as measured by the moun's frame of reference

Explanation:

From the point of view of the observer on Earth the muon traveled 3.53 km at 0.983c

0.983 * 3*10^8 = 2.949*10^8 m/s

Δt = d/v = 3530 / 2.949*10^8 = 1.197*10^-5 s

The muon lived 1.197*10^-5 s from the point of view of the observer.

The equation for time dilation is:

\Delta t' = \Delta t * \sqrt{1 - \frac{v^2}{c^2}}

Then:

\Delta t' = 1.197*10^-5 * \sqrt{1 - \frac{(0.983c)^2}{c^2}} = 2.198*10^-6 s

From the point of view of the moun Earth moved at 0.983c (2.949*10^8 m/s) during a time of 2.198*10^-6, so it moved

d = v*t = 2.949*10^8 * 2.198*10^-6 = 648 m

7 0
1 year ago
A 2200 kg truck has put its front bumper against the rear bumper of a 2400 kg SUV to give it a push. With the engine at full pow
leonid [27]

Answer:

a) The maximum possible acceleration the truck can give the SUV is 7.5 meters per second squared

b) The force of the SUV's bumper on the truck's bumper is 18000 newtons

Explanation:

a) By Newton's second law we can find the relation between force and acceleration of the SUV:

F=ma

With F the maximum force the truck applies to the SUV, m the mass of the SUV and a the acceleration of the SUV; solving for a:

a=\frac{F}{m}=\frac{18000}{2400}\approx7.5\,\frac{m}{s^{2}}

b) Because at this acceleration the truck's bumper makes a force of 18000 N on the SUV’s bumper by Third Newton’s law the force of the SUV’s bumper on the truck’s bumper is 18000 N too because they are action-reaction force pairs.

7 0
1 year ago
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