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4 years ago

In Motion studies, the position in space and time at which motion began is selected as the ______.

Physics

2 answers:

Sonbull [250]4 years ago

6 0

I am pretty sure that the answer is B.

nydimaria [60]4 years ago

4 0

Answer:

b) origin

Explanation:

Motion is when a body changes position in space with respect to time. There are four types of motion namely

1. circular motion: this is the motion of an object along a circular path eg. the motion of a car tyre.

2. random motion: it is the motion of a body in zigzag manner e.g the motion of smoke from the exhaust pipe of a vehicle as it rises into the atmosphere

3. oscillatory : also known as the to and fro movement of a body e.g a pendulum swing

4. Linear motion: The motion of a body on a straight line. also known as translational motion e.g the motion of a car

Parameters/ terminologies used in motion are

velocity

displacement

distance

acceleration,

time,

speed

Motion in space and time starts from a point known as Origin

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Can a vector have a component greater than its magnitude

galina1969 [7]

Answer:

Explanation:

The components of a vector can not have a magnitude greater than the vector itself.The magnitude of a component is always lesser than vector because the magnitude is a product of the vector and cos Ф where the value of cos Ф , x, is -1<x<1. Additionally, applying the Pythagorean relationship, you expect the sum of squares of the components to equal the square of the magnitude of the vectors.However, there is a situation where a component of a vector has a magnitude which equals the magnitude of the vector.

6 0

3 years ago

A revolutionary war cannon, with a mass of 2260 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 109 m/s aft

Anton [14]

Answer:

The gain in velocity is 0.37m/s

Explanation:

We need solve this problem though the conservation of momentum. That is,

$m_1 v_1 = m_2 v_2$

$m_1=2260Kg\\m_2=15.5Kg\\v_2= 109m/s$

Using the equation to find $v_1$ ,

$v_1=\frac{m_2 v_2}{m_1}\\v_1=\frac{15.5*109}{2260}\\v_1= 0.7475$

Using the conservation of energy equation, we have,

$KE= \frac{1}{2}m*v^2$

$KE_{cball}=\frac{1}{2}(15.5)(109)^2=92077.75J$

$KE_{cannon}=\frac{1}{2}(2260)(0.7475)^2=631.39J$

$Total KE= 92077.75+13425530=92708.9J$

Now this energy over the cannonball

$KE=\frac{1}{2}m*v_2^2$

$92708.9=\frac{1}{2}15.5v_2^2$

$V_2 = 109.37m/s$

The gain in velocity is 0.37m/s

4 0

4 years ago

In a closed system, the loss of momentum of one object ________ the gain in momentum of another object.

Virty [35]

In a closed system, the loss of momentum of one object is same as________ the gain in momentum of another object

according to law of conservation of momentum, total momentum before and after collision in a closed system in absence of any net external force, remains conserved . that is

total momentum before collision = total momentum after collision

P₁ + P₂ = P'₁ + P'₂

where P₁ and P₂ are momentum before collision for object 1 and object 2 respectively.

P'₁ - P₁ = - (P'₂ - P₂)

so clearly gain in momentum of one object is same as the loss of momentum of other object

8 0

3 years ago

A 15kg dog jumps out of a 40kg canoe. If the dogs velocity is 1.2m/s, what is the velocity of the canoe?

vivado [14]

Answer:

v = -0.45 m/s

Explanation: