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2 years ago

If two scooters are moving at the same speed but in opposite directions is their velocity the same? Explain

Physics

2 answers:

boyakko [2]2 years ago

8 0

Answer:

yes

Explanation:

because there going the same speed but in different directions.

kirill [66]2 years ago

4 0

Yes they are going the same speed in different directions

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In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.42 m. The mug

telo118 [61]

Answer:

a) $V_{x}=3.72m/s$ , b) ∠=-54.83°

Explanation:

In order to solve this problem, we must start with a drawing of the situation, this will help us visualize the problem better. (See picture attached).

Now, the idea is that the beer mug has a horizontal speed and no vertical speed at initial conditions. So knowing this, we can start finding the initial velocity of the mug.

In order to do so, we need to find the time it takes for the mug to reach the ground. We can find it by using the following equation:

$y=y_{0}+V_{y0}t+\frac{1}{2}a_{y}t^{2}$

We can see from the drawing that y and the initial velocity in y are zero, so we can simplify our formula:

$0=y_{0}+\frac{1}{2}a_{y}t^{2}$

so we can solve for t, so we get:

$t=\sqrt{\frac{-(2)y_{0}}{a}}$

so now we can substitute the known values, so we get:

$t=\sqrt{\frac{-(2)(1.42)}{-9.8}}$

which yields:

t=0.538s

So we can use this value to find the velocity in x:

$V_{x}=\frac{x}{t}$

When substituting we get:

$V_{x}=\frac{2m}{0.538s}$

which yields:

$V_{x}=3.72m/s$

In order to solve part b, we need to find the y-component of the velocity, for which we can use the following formula:

$\Delta y=\frac{V_{f}^{2}-V_{0}^{2}}{2a}$

We know that $V_{0}$ is zero, so we can simplify the expression:

$\Delta y=\frac{V_{yf}^{2}}{2a}$

So we can solve the equation for $V_{yf}^{2}$ so we get:

$V_{yf}=\sqrt{2\Delta y a}$

and when substituting the known values we get:

$V_{yf}=\sqrt{2(-1.42m)(-9.8m/s^{2})}$

which yields:

$V_{yf}=-5.28m/s$

Once we got the final velocity in y, we can use it together with the velocity in x to find the angle.

So we can use the following formula:

$tan \theta =\frac{V_{y}}{V_{x}}$

when solving for theta we get:

$\theta = tan^{-1}(\frac{V_{y}}{V_{x}})$

We can substitute so we get:

$\theta = tan^{-1}(\frac{-5.28m/s}{3.72m/s})$

which yields:

$\theta = -54.83^{o}$

7 0

3 years ago

When sitting in the tree the cat has a total of 1375J in its Gravitational potential store. What is the maximum amount of energy

Murrr4er [49]

Answer:

1375J

Explanation:

The gravitational potential/potential energy of the at the top of the tree which is the energy by virtue of its position.

P.E = mgh

mass = m

Acceleration due to gravity = g

height = h

At the top of the tree, the value of h (height) is high resulting in the gravitational potential. When the cat lands on the ground, the value of h is zero, the the gravitational potential would be zero and all the potential energy have been converted to other forms of energy.

Therefore, the total gravitational potential store is equal to the maximum amount of energy that can be transferred which is equal to 1375J.

4 0

3 years ago

A 1232 kg car moving north at 25.6 m/s collides with a 2028 kg car moving north at 17.5 m/s . They stick together. In what direc

Citrus2011 [14]

Answer:

I. Angle = 41.7° Northeast.

II. Vr = 7.08m/s

Explanation:

Let the two cars be denoted by A and B

Given the following data;

Mass of car A = 1232 Kg

Velocity of car A = 25.6 m/s

Mass of car B = 2028 Kg

Velocity of car B = 17.5m/s

First of all, we would solve for momentum;

Momentum = mass × velocity

Momentum, M1 = 1232 × 25.6

Momentum, M1 = 31539.2 Kgm/s

Momentum, M2 = 2028 × 17.5

Momentum, M2 = 35490 Kgm/s

Now, let's find the resultant momentum using the Pythagoras theorem;

R² = M1² + M2²

R² = 31539.2² + 35490²

R² = 994721136.6 + 1259540100

R² = 2254261237

Taking the square root of both sides, we have

Resultant momentum, R = 47479.06 Kgm/s

To find the direction;

Angle = tan¯¹(M1/M2)

Angle = tan¯¹(31539.2/35490)

Angle = tan¯¹(0.89)

Angle = 41.7° Northeast.

To find the speed;

R = (M1 + M2)Vr

47479.06 = (31539.2 + 35490)Vr

47479.06 = 67029.2Vr

Vr = 47479.06/67029.2

Vr = 7.08m/s

6 0

2 years ago

An object is dropped from a 45 m high building. At the same time, another object is thrown

Dafna11 [192]

Answer:

-92.33 (meaning the objects will not meet above the ground).

Explanation:

We can use the kinematic equation displacement = initial velocity*time + 1/2*acceleration*time^2.

We can plug in the known values of the 2 objects into the equation, where t is the time and x is the displacement:

x = 0*t + 1/2*(-9.8)*t^2+45

x = 8.5*t + 1/2*(-9.8)*t^2

We need to first solve for t to solve for x. Since both equations are equal to x, we can set them equal to each other and solve for t:

0*t + 1/2*(-9.8)*t^2+45 = 8.5*t + 1/2*(-9.8)*t^2

-4.9*t^2 +45 = 8.5*t + -4.9*t^2

45 = 8.5*t

t = 45/8.5 ≈5.294

Now, we can plug t as 5.294 into any of the equations above to solve for x:

x = 0*5.294 + 1/2*-9.8*(5.294)^2+45 ≈ -92.33

That means, the objects will not meet above the ground.

4 0

3 years ago

Which three temperature readings all mean the same thing?

Drupady [299]

Answer : The correct option is, (D) 273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit

Explanation :

Conversion of degree Celsius to Kelvin :

$K=^oC+273$

Conversion of degree Celsius to degrees Fahrenheit :

$^oF=(\frac{9}{5}\times ^oC)+32$

By using these two conversion factors, we get the three temperature readings all mean the same thing.

For option A :

$K=^oC+273=100+273=373K$

$^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 100)+32=212^oF$

For option B :

$K=^oC+273=100+273=373K$

$^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 100)+32=212^oF$

For option C :

$K=^oC+273=0+273=273K$

$^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 0)+32=32^oF$

For option D :

$K=^oC+273=0+273=273K$

$^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 0)+32=32^oF$

From the given options, only option (D) is correct.

Hence, the correct option is, (D) 273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit

7 0

3 years ago

Read 2 more answers