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2 years ago

The force that accelerates a rocket in outer space is exerted on the rocket by the?

1 answer:

3 0

Gases at pressure are released by rockets as they travel towards space. According to Newton's third law, the combustion chamber's exhaust gases push the rocket with an accelerating force known as the thrust.

<h3>Explain exactly Newton's Third Law:</h3>

According to Newton's third law, if an object A pulls on an object B, then object B must exert an equal-sized and opposite-direction force on the first thing directed in the opposite direction. This law illustrates a symmetry in nature whereby forces always occur in pairs and whereby no body can exert a force without also being subjected to one.

<h3>What are Newton's 3rd law examples?</h3>

Action and response are always equal but always move in the opposite direction, according to Newton's third law of motion. A human walking on the ground, a hammer driving a nail, a magnet attracting a paper clip, and a horse pulling a cart are all examples of Newton's third rule of motion.

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A mass m = 0.6 kg is released from rest at the top edge of a hemispherical bowl with radius = 1.1 meters. The mass then slides w

Aneli [31]

Answer:

The angle is $\theta = 36.24 ^o$

Explanation:

From the question we are told that

The mass is $m = 0.6 \ kg$

The radius is $r = 1.1 \ m$

The speed is $v = 3.57 \ m /s$

According to the law of energy conservation

The potential energy of the mass at the top is equal to the kinetic energy at the bottom i.e

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

=> $h = \frac{1}{2 g } * v^2$

Here h is the vertical distance traveled by the mass which is also mathematically represented as

$h = r * sin (\theta )$

$\theta = sin ^{-1} [ \frac{1}{2* g* r } * v^2]$

substituting values

$\theta = sin ^{-1} [ \frac{1}{2* 9.8* 1.1 } * (3.57)^2]$

$\theta = 36.24 ^o$

3 0

4 years ago

You exert 375 J to move a 45-kg object 5 m. How much force is required?

Serjik [45]

The amount of force required is 69J

8 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:

$x = 360\cdot (4.420)$