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3 years ago

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A locomotive is accelerating at 3.08 m/s2. It passes through a 20.8-m-wide crossing in a time of 2.93 s. After the locomotive le

aves the crossing, how much time is required until its speed reaches 29.4 m/s?

1 answer:

svp [43]3 years ago

6 0

Answer:

Required time t=5.8s

Explanation:

Given data

Acceleration a=3.08 m/s²

Speed v=29.4 m/s

Required

How much time t is required

Solution

As the train passes through the crossing its motion is described by:

$v=v_{o}+at\\v-v_{o}=at\\and\\x=\frac{1}{2}(v+v_{o})t\\v+v_{o}=\frac{2x}{t}\\ So\\v=\frac{1}{2}(at+\frac{2x}{t} ) \\$

Substitute the given values

So

$v=\frac{1}{2}((3.08m/s^2)(2.93s)+\frac{2(20.8m)}{2.93s} )\\v=11.61m/s$

So time required can be calculated by:

$t=\frac{v-v_{o}}{a} \\t=\frac{29.4m/s-11.61m/s}{3.08m/s^2}\\ t=5.8s$

Required time t=5.8s

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Answer:

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Explanation:

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From the question we have

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3 years ago

A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It s

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Answer:

30.67m

Explanation:

Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;

s = ut ± $\frac{1}{2} at^2$ ------------(i)

Where;

s = horizontal/vertical displacement

u = initial horizontal/vertical component of the velocity

a = acceleration of the projectile

t = time taken for the projectile to reach a certain horizontal or vertical position.

Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;

h = vt ± $\frac{1}{2} at^2$ ------------(ii)

Where;

h = vertical displacement

v = initial vertical component of the velocity = usinθ

a = acceleration due to gravity (since vertical motion is considered)

t = time taken for the projectile to hit the target

<em>From the question;</em>

u = 47m/s, θ = 0.6rads

=> usinθ = 47 sin 0.6

=> usinθ = 47 x 0.5646 = 26.54m/s

t = 1.7s

Take a = -g = -10.0m/s (since motion is upwards against gravity)

Substitute these values into equation (ii) as follows;

h = vt - $\frac{1}{2} at^2$

h = 26.54(1.7) - $\frac{1}{2} (10)(1.7)^2$

h = 45.118 - 14.45

h = 30.67m

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3 years ago

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Answer: B

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You can use the conservation of momentum, under the assumption that no mass was lost when the collision occurred. The initial momentum of the system must equal the final momentum of the system. Our system is the region including, and only including, the satellite and the space debris. Classical momentum is defined as the product of mass and velocity:

$p_i=p_f$

$m_1v_1_i+m_2v_2_i=m_1v_1_f+m_2v_2_f$

Due to mass 1 equaling mass 2, we can factor these quantities out:

$m(v_1_i+v_2_i)=m(v_1_f+v_2_f)$

Cancel the mass term on both sides to get:

$v_1_i+v_2_i=v_1_f+v_2_f$

We have the initial and final velocities for everything besides the final velocity of the satellite. Plug everything in:

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2 years ago

A 8.10×10^3 ‑kg car is travelling at 25.8 m/s when the driver decides to exit the freeway by going up a ramp. After coasting 3.9

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Answer:

The magnitude of the average drag force is 2412.34 N.

Explanation:

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Height = 12.5 m

We need to calculate the magnitude of the average drag force

Using equation kinetic energy

$K.E_{i}=K.E_{f}+P.E+F_{d}$

$\dfrac{1}{2}mv_{i}^2=\dfrac{}{}mv_{f}^2+mgh+F\times d$

Where, $v_{i}$ = initial velocity

$v_{f}$ = final velocity

h = height

g = acceleration due to gravity

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m = mass of the car

d = distance

Put the value into the formula

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$F=\dfrac{\dfrac{1}{2}\times8.10\times10^{3}\times25.8-\dfrac{1}{2}\times8.10\times10^{3}\times13.1-8.10\times10^{3}\times9.8\times12.5}{3.90\times10^{2}}$

$F=-2412.34\ N$

$|F|=2412.34\ N$

Hence, The magnitude of the average drag force is 2412.34 N.

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3 years ago

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Law Incorporation [45]

Answer:

The mean distance of Earth from the Sun is 149.6 × 10^6 km and the mean distance of Mercury from the Sun is 57.9 × 10^6 km.

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3 0

3 years ago

Read 2 more answers