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2 years ago

5

A hiker climbs a mountain. Starting at the base of the mountain, he first moved up 520m at a 32.0 degree angle. What is the fina

l displacement of the hiker?

1 answer:

balu736 [363]2 years ago

5 0

Answer:

$\displaystyle \vec{d}=$

Explanation:

<u>Displacement Vector</u>

Suppose an object is located at a position

$\displaystyle P_1(x_1,y_1)$

and then moves at another position at

$\displaystyle P_2(x_2,y_2)$

The displacement vector is directed from the first to the second position and can be found as

$\displaystyle \vec{d}=$

If the position is given as magnitude-angle data ( z , α), we can compute its rectangular components as

$\displaystyle x=z\ cos\alpha$

$\displaystyle y=z\ sin\alpha$

The question describes the situation where the initial point is the base of the mountain, where both components are zero

$\displaystyle P_1(0,0)$

The final point is given as a 520 m distance and a 32-degree angle, so

$\displaystyle x_2=520\ cos32^o= 440.99\ m$

$\displaystyle y_2=520\ sin32^o=275.6\ m$

The displacement is

$\displaystyle \vec{d}=$

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Describe a technology used in space exploration.

Alex777 [14]

Answer:

High speed optical communication technology

To be able to communicate from the space to the earth and from earth to space is one of the most essential features required during space exploration.

Explanation:

Space exploration involves going into the space, beyond the earth's atmosphere. Landing on other planets and studying their details, going into deeper space beyond the planets to discover new cosmic events or structures is all a part of space exploration.

The key to analyse the studies and observations is being able to communicate the data collected, photos taken etc to the launch centers or space centers on earth. The space centers on earth should also be able to communicate with the persons or the satellites in space.

This is made possible using the optical communication technology which involves the use of optical fibers, lasers etc, since high speeds are more efficient during communication

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3 years ago

A bowling ball of 35.2 kg, generates 218 kg* m/s units of momentum. What is the velocity of the ball?

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6.19 m/s

Explanation:

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2 years ago

A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci

ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity ( $v$ ), measured in meters per second, by using this kinematic equation:

$v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s}$ (1)

Where:

$a$ - Acceleration, measured in meters per square second.

$\Delta s$ - Travelled distance, measured in meters.

$v_{o}$ - Initial velocity, measured in meters per second.

If we know that $v_{o} = 0\,\frac{m}{s}$ , $a = 2.35\,\frac{m}{s^{2}}$ and $\Delta s = 595\,m$ , the final velocity of the rocket is:

$v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}$

$v\approx 52.882\,\frac{m}{s}$

The time associated with this launch ( $t$ ), measured in seconds, is:

$t = \frac{v-v_{o}}{a}$

$t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }$

$t = 22.503\,s$

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

$v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o})$ (2)

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

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

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

If we know that $v_{o} = 52.882\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s^{2}}$ and $s_{o} = 595\,m$ , then the maximum height reached by the rocket is:

$v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})$

$s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}$

$s = 737.577\,m$

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

$s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$ (2)

Where:

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

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

$t$ - Time, measured in seconds.

If we know that $s_{o} = 595\,m$ , $v_{o} = 52.882\,\frac{m}{s}$ , $s = 0\,m$ and $a = -9.807\,\frac{m}{s^{2}}$ , then the time needed by the rocket is:

$0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}$

$-4.904\cdot t^{2}+52.882\cdot t +595 = 0$

Then, we solve this polynomial by Quadratic Formula:

$t_{1}\approx 17.655\,s$ , $t_{2} \approx -6.872\,s$

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0

2 years ago

An object is dropped from the top of a building and is observed to take 7. 2s to hit the ground. How tall is the building?.

maw [93]

the height of the building is H=36 m.

<h3>What is The Law of Gravity?</h3>

According to Newton's law of gravity, every particle of matter in the universe is attracted to every other particle with a force that varies directly as the product of their masses and inversely as their distance from one another.

Properties of Gravity -

It is a universal attractive force. It is directly proportional to the product of the masses of the two bodies.
It obey inverse square law.
It is the weakest force known in nature.

Examples of Gravity -

The force that holds the gases in the sun.
The force that causes a ball you throw in the air to come down again.
The force that causes a car to coast downhill even when you aren't stepping on the gas.

v₀=0 m/s

H₀=0 m

g=10 m/s²

t=7,2 s

H - ?

$H= H_{0} + v_{0}t+\frac{gt^{2} }{2} \\$

H = 0 +0 × 7.2 + 10(7.2)²/2

H = 36m

to learn more about Gravity go to - brainly.com/question/12528243

#SPJ4

8 0

1 year ago

While playing baseball with your friends your hands begin to sting after you ctach several fast balls.

muminat

Answer:

Explanation:

To stop a ball with high momentum in a small-time imparts a high amount of impact on hands. This is the reason for the stinging of hands.

The momentum of the ball is due to the mass and velocity. To prevent stinging in the hand one needs to lower his hands to increase the time of contact. In this way, the momentum transfer to the hands will be lesser.

7 0

3 years ago