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Aneli [31]
3 years ago
11

You (in the x' frame) are moving with v = +5 m/s with respect to the x frame. If your initial position in each frame is Om, what

are the positions in each frame after 3 s have passed? X:-15 m x': 15 m Answer 1: - 15 m Answer 2: 15 m
Physics
1 answer:
WARRIOR [948]3 years ago
3 0

Answer:

Distance is 15 meters.

Explanation:

It is given that,

Velocity with respect to the x frame, v = 5 m/s

We need to find the position in each frame after 3 seconds have passed.

Distance covered is calculated by the product of velocity and time taken. It is given by :

d=v\times t

d=5\ m/s\times 3\ s

d = 15 meters

So, after 3 seconds the position in each frame is 15 meters from origin. Hence, this is the required solution.

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A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimumcoefficient of static friction
zhenek [66]

Answer:

c. 0.816

Explanation:

Let the mass of car be 'm' and coefficient of static friction be 'μ'.

Given:

Speed of the car (v) = 40.0 m/s

Radius of the curve (R) = 200 m

As the car is making a circular turn, the force acting on it is centripetal force which is given as:

Centripetal force is, F_c=\frac{mv^2}{R}

The frictional force is given as:

Friction = Normal force × Coefficient of static friction

f=\mu N

As there is no vertical motion, therefore, N=mg. So,

f=\mu mg

Now, the centripetal force is provided by the frictional force. Therefore,

Frictional force = Centripetal force

f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu=\frac{v^2}{Rg}

Plug in the given values and solve for 'μ'. This gives,

\mu = \frac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}\\\\\mu=\frac{1600\ m^2/s^2}{1960\ m^2/s^2}\\\\\mu=0.816

Therefore, option (c) is correct.

7 0
3 years ago
I need so much help 50 points!!!!!!<br><br>complete the graph
motikmotik

<u>First Symbol </u>: Cobalt (Co)

Its Group Number - 9

Its Period Number - 4

Its Family Name - Transition Metal

<u>Second Symbol</u> : Silicon (Si)

Its Group Number - 14

Its Period Number - 2

Its Family Name - Semiconductor

<u>Third Symbol</u> : Astatine (At)

Its Group Number - 17

Its Period Number - 6

Its Family Name - Halogen

<u>Fourth Symbol </u>: Magnesium (Mg)

Its Group Number - 2

Its Period Number - 3

Its Family Name - Alkaline Earth Metal

<u>Fifth Symbol</u> : Xenon (Xe)

Its Group Number - 18

Its Period Number - 5

Its Family Name - Noble Gas

6 0
3 years ago
Water initially at 200 kPa and 300°C is contained in a piston–cylinder device fitted with stops. The water is allowed to cool at
netineya [11]

Answer:

Δu=1300kJ/kg  

Explanation:

Energy at the initial state

p_{1}=200kPa\\t_{1}=300^{o}\\u_{1}=2808.8kJ/kg(tableA-5)

Is saturated vapor at initial pressure we have

p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)

Process 2-3 is a constant volume process

p_{3}=100kPa\\v_{3}=v_{2}=0.8858m^{3}/kg\\u_{3}=1508.6kJ/kg(tableA-5)

The overall in internal energy

Δu=u₁-u₃

We replace the values in equation

Δu=u₁-u₃

=2808.8kJ/kg-1508.6kJ/kg\\=1300kJ/kg

Δu=1300kJ/kg  

3 0
2 years ago
A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the
lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

l^2 = x^2 + y^2-----(1)

Differentiating with respect to t ( time ),

0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}  ( l = 26 feet = constant )

\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}

\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}

We have,

y = 10, \frac{dx}{dt}= -3\text{ feet per min}

\frac{dy}{dt}=\frac{3x}{10}-----(X)

(i) From equation (1),

26^2 = x^2 + 10^2

676=x^2 + 100

576 = x^2

\implies x = 24\text{ feet}

From equation (X),

\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}

(ii) From equation (X),

\frac{dy}{dt}\propto x

Thus, for different value of x the value of \frac{dy}{dt} would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

\frac{dy}{dt}=0\text{ feet per min}

3 0
3 years ago
The diagram below shows a 5.00-kilogram block
bixtya [17]

The name and strength of the force holding the block up is 50 N upward - Normal force.

The given parameters:

  • <em>Mass of the block, m = 5 kg</em>

The weight of the block acting downwards due to gravity is calculated as follows;

W = mg

where;

  • <em>g is acceleration due to gravity = 10 m/s²</em>

W = 5 x 10

W = 50 N <em>(</em><em>downwards</em><em>)</em>

Since the block is at rest, an a force equal to the weight of the block must be acting upwards. This force is known as normal reaction.

Fₙ = 50 N <em>(</em><em>upwards</em><em>)</em>

Thus, the name and strength of the force holding the block up is 50 N upward - Normal force.

Learn more about Normal force here: brainly.com/question/14486416

4 0
2 years ago
Read 2 more answers
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