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2 years ago

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2 part question

1 answer:

schepotkina [342]2 years ago

8 0

the answer is yes number5 hope this helps did before

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The total distance from a house to a school to is 9.5 km. A student travels all the way from his house to the school and back to

marysya [2.9K]

Answer:

0

Explanation:

Displacement is a vector from initial to final point. Because initial and final point are the same, so displacement is 0.

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3 years ago

The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. If HST has a tangential speed of 7,750 m/s, how long is HS

deff fn [24]

Answer: 5,640 s (94 minutes)

Explanation:

the tangential speed of the HST is given by

$v=\frac{2\pi r}{T}$ (1)

where

$2\pi r$ is the length of the orbit

r is the radius of the orbit

T is the orbital period

In our problem, we know the tangential speed: $v=7,750 m/s$ . The radius of the orbit is the sum of the Earth's radius and the distance of the HST above Earth's surface:

$r=6.38\cdot 10^6 m+569,000 m=6.95\cdot 10^6 m$

So, we can re-arrange equation (1) to find the orbital period:

$T=\frac{2 \pi r}{v}=\frac{2 \pi (6.95\cdot 10^6 m/s)}{7,750 m/s}=5,640 s$

Dividing by 60, we get that this time corresponds to 94 minutes.

6 0

3 years ago

Read 2 more answers

A person weighs 60 kg. The area under the foot of the person is 150 cm2. Find the pressure exerted on the ground by the person.

9966 [12]

Answer:

40000 N/m²

Explanation:

Applying,

P = F/A................... Equation 1

Where P = Pressure, F = Force, A = Area.

From the question,

The force(F) exerted by the person's foot is thesame as it's weight.

F = W = mg............ Equation 2

Where m = mass of the person, g = acceleration due to gravity.

Substitute equation 2 into equation 1

P = mg/A................ Equation 3

Given: m = 60 kg, g = 10 m/s², A = 150 cm² = (150/10000) m² = 0.015 m²

Substitute these values into equation 3

P = (60×10)/0.015

P = 600/0.015

P = 40000 N/m²

3 0

2 years ago

A 55 meter rope is lying on the floor and has has a weight of 4040 N in total. How much work is required to lift up one end of t

Kazeer [188]

Answer:

Explanation:

Given

Length of rope $L=55\ m$

Weight of rope $W=4040\ N$

weight density $\lambda =\frac{4040}{55}=73.45\ N/m$

Work done to lift rope 33 m

$W=\int_{0}^{33}\lambda hdh$

$W=\int_{0}^{33}73.45hdh$

$W=73.45\left [ \left ( \frac{h^2}{2}\right )\right ]^{33}_0$

$W=39.993\ kJ$

6 0

3 years ago

An airplane is moving at 350 km/hr. If a bomb is

Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity</h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

2 years ago