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2 years ago

A train travels 8.81 m/s in a -51.0° direction.

Physics

1 answer:

Bad White [126]2 years ago

5 0

The displacement of the train after 2.23 seconds is 25.4 m.

<h3>Resultant velocity of the train</h3>

The resultant velocity of the train is calculated as follows;

R² = vi² + vf² - 2vivf cos(θ)

where;

θ is the angle between the velocity = (90 - 51) + 37 = 76⁰

R² = 8.81² + 9.66² - 2(8.81 x 9.66) cos(76)

R² = 129.75

R = √129.75

R = 11.39 m/s

<h3>Displacement of the train</h3>

The displacement of the train is the change in position of the train after a given period of time.

The displacement is calculated as follows;

Δx = vt

Δx = 11.39 m/s x 2.23 s

Δx = 25.4 m

Thus, the displacement of the train after 2.23 seconds is 25.4 m.

Learn more about displacement here: brainly.com/question/2109763

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A ball is thrown vertically upward with a

Veronika [31]

Answer:

1) The maximum height reached is approximately 39.715 m

2) The time it takes the ball to reach the its highest point is approximately 2.85 s

Explanation:

1) The vertical velocity of the ball, u = 27.9 m/s

The acceleration due gravity, g = 9.8 m/s²

The kinematic equation that gives the height, h, reached by the ball is given as follows;

v² = u² - 2·g·h

Where;

v = The final velocity of the ball = 0 m/s at the maximum height

h = The height reached by the ball = $h_{max}$ at maximum height

Therefore, by substitution of the known values, we have;

v² = u² - 2·g·h

0² = 27.9² - 2 × 9.8 × $h_{max}$

2 × 9.8 × $h_{max}$ = 27.9²

$h_{max}$ = 27.9²/(2×9.8) ≈ 39.715

The maximum height reached = $h_{max}$ ≈ 39.715 m

2) From the kinematic equation of motion, v = u - g·t, we have the time, t, it takes the ball to reach the maximum height given as follows

At maximum height, the final velocity, v = 0 m/s, therefore, we have;

0 = 27.9 - 9.8 × t

9.8 × t = 27.9

t = 27.9/9.8 ≈ 2.85

The time it takes the ball to reach the maximum height = t ≈ 2.85 seconds.

6 0

3 years ago

A car is moving at constant velocity +48.8 m/s along a straight highway. Suddenly the driver spots a stalled vehicle 378.7 m ahe

Katen [24]

Answer:

a = - 3.14[m/s^2]

Explanation:

To solve this problem we must use the following kinematic equation:

$v_{f} ^{2}= v_{o} ^{2}+2*a*dx\\where:\\v_{f} = final velocity = 0 [m/s]\\v_{o}= initial velocity = 48.8 [m/s]\\ a = acceleration [m/s^2]\\dx = displacement [m]\\$

Now replacing:

$0 = 48.8^{2}+2*a*(378.7) \\-2381.44 = 757.4*a\\a = -3.14[m/s^2]$

The negative sign means the vehicle slows down.

4 0

4 years ago

The world's fastest production sportscar has a top speed of 415 kmh-1(a)Convert this speed to ms-1.[ 1](b)The distance from Lond

Alex73 [517]

Answer:

(a). The speed is 115.28 m/s.

(b). The time is 94 min.

(c). The International Space Station travels 43202.400 km.

(d). Speed is scalar quantity.

Velocity is vector quantity.

Explanation:

Given that,

Top speed = 415 km/h

1 miles = 1609 m

(a). We need to calculate the speed in m/s

Using conversion of km/h to m/s

$v=415\ km/h$

$v=415\times\dfrac{5}{18}$

$v=115.28\ m/s$

(b). We need to calculate the distance from London to Edinburgh in km

Using conversion for distance

$d=403\ miles$

Distance in meter,

$d=403\times1609$

$d=648427\ m$

$d=648.427\ km$

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{648.427}{415}$

$t=1.56\ h$

$t=1\ hours 34\ min$

$t= 94\ min$

(c). Speed v' =7.66 km/s

v'=7660 m/s

In the time it takes the car to travel from London to Edinburgh,

We need to calculate the distance

Using formula of distance

$d'=v'\times t$

Put the value into the formula

$d'=7660\times94\times60$

$d'=43202400\ m$

$d'=43202.400\ km$

(d). Speed :

Speed is equal to the distance divided by time.

It is scalar quantity.

Velocity :

Velocity is equal to the displacement divided by time.

It is vector quantity.

Hence, (a). The speed is 115.28 m/s.

(b). The time is 94 min.

(c). The International Space Station travels 43202.400 km.

(d). Speed is scalar quantity.

Velocity is vector quantity.

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