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3 years ago

Motion is

Physics

1 answer:

IgorC [24]3 years ago

4 0

Explanation:

When something is moving, we call it is in motion. It can be also defined as a change of position in relation to another object.

When an object is not moving at all, we call that it was at rest.

To be in motion, there must be a change in position.

Hence, the correct options are "a change of position in relation to another object" and "something is moving".

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Man A (70kg) and Man B (90kg) are hanging motionless from a platform at rest. What is the tension TA in the top rope if the plat

Damm [24]

Answer:

The tension in the upper rope (top rope), T1 = 1,888 N

Explanation:

The Parameters that were given:

Mass A, M1 = 70kg

Mass B. M2 = 90kg

acceleration, a = 2 m/s2

Assume the rope doesn't have mass, acceleration due to gravity, g

= 9.8 m/s2

The tension, T in a platform = m (a + g)

Then the tension, T1 in the upper rope = m1 (a + g) + T2

Where T2 = Tension in the lower rope

First, we calculate T2

Since the platform accelerates upward the acceleration would be positive

T2 = m2 (a + g)

T2 = 90kg ( 2 m/s2 + 9.8 m/s2)

T2 = 1,062N

To calculate the tension T1,

T1 = m1 (a + g) + T2

= 70kg (2 m/s2 + 9.8 m/s2) + 1062N

T1 = 1,888 N

3 0

3 years ago

A student fires a cannonball diagonally

rodikova [14]

Answer:The cannons total flight time is 2.23seconds

Explanation:

A ball fired diagonally is fired at an angle of 45° to the horizontal. The motion of the fired ball is a projectile motion. A projectile is a motion in which an object fired into space with an initial velocity U is allowed to fall freely under the influence of gravitational force.

To total time of flight T of the cannon ball can be expressed as;

T = 2Usin(theta)/g where;

U is the initial velocity or speed of the ball = 31m/s

theta is the angle that the ball make with the horizontal = 45°

g is the acceleration due to gravity = 9.81m/s²

Substituting the given datas into the formula we have;

T = 31sin45°/9.81

T = 31×0.7071/9.81

T = 21.92/9.81

T = 2.23seconds

8 0

3 years ago

A heat engine receives heat of 400kW from a heat source of 1100 K, and rejects the waste heat into a heat sink at 320K. It gener

GuDViN [60]

Answer:

Explanation:

efficiency of carnot engine operating between 1100 K and 320 K

= 1100 - 320 / 1100

= .709

efficiency of heat engine = 120 / 400 = .30

2 nd law thermal efficiency of heat engine = .30 x 100 / .709

= 42.31 % .

4 0

3 years ago

A projectile is fired from a height of 80 M above sea level, horizontally with a speed of 360 M / S, calculate: The time it take

Maslowich

Answer:

(a) The projectile takes approximately 4.420 seconds to reach the water, (b) The horizontal scope of the projectile is 1591.2 meters, (c) The remaining height to descend after 2 seconds of being launched is 63.624 meters.

Explanation:

The projectile experiments a parabolic motion, where horizontal speed remains constant and accelerates vertically due to the gravity effect. Let consider that drag can be neglected, so that kinematic equation are described below:

$x = x_{o}+v_{o,x} \cdot t$

$y = y_{o} + v_{o,y}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}$

Where:

$x_{o}$ , $y_{o}$ - Initial horizontal and vertical position of the projectile, measured in meters.

$v_{o,x}$ , $v_{o,y}$ - Initial horizontal and vertical speed of the projectile, measured in meters per second.

$t$ - Time, measured in seconds.

$g$ - Gravitational acceleration, measured in meters per square second.

$x$ , $y$ - Current horizontal and vertical position of the projectile, measured in meters.

Given that $x_{o} = 0\,m$ , $y_{o} = 80\,m$ , $v_{o,x} = 360\,\frac{m}{s}$ , $v_{o,y} = 0\,\frac{m}{s}$ and $g = -9.807\,\frac{m}{s^{2}}$ , the kinematic equations are, respectively:

$x = 360\cdot t$

$y = 80-4.094\cdot t^{2}$

(a) If $y = 0\,m$ , the time taken for the projectile to reach the water is:

$80 - 4.094\cdot t^{2} = 0$

$t = \sqrt{\frac{80}{4.094} }\,s$

$t \approx 4.420\,s$

The projectile takes approximately 4.420 seconds to reach the water.

(b) The horizontal scope is the horizontal distance done by the projectile before reaching the water. If $t \approx 4.420\,s$ , the horizontal scope of the projectile is:

$x = 360\cdot (4.420)$

$x = 1591.2\,m$

The horizontal scope of the projectile is 1591.2 meters.

(c) If $t = 2\,s$ , the height that remains to descend is:

$y = 80-4.094\cdot (2)^{2}$

$y = 63.624\,m$

The remaining height to descend after 2 seconds of being launched is 63.624 meters.

6 0

3 years ago

It takes fifteen minutes to complete a drilling operation using a lathe. the duration of this operation is known as _____ time.

dalvyx [7]

I think the correct answer would be processing time. The time of fifteen minutes describes how long the whole operation was done or how long is the whole processing of drilling. Hope this answers the question. Have a nice day.

8 0

4 years ago