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Nookie1986 [14]
2 years ago
5

a train is moving with an initial velocity of 30 m/s, the brakes are applied so as to produce a uniform acceleration of -1.5 m/s

². Calculate the time in which it will come to rest.​
Physics
1 answer:
Pepsi [2]2 years ago
4 0

Answer:

\boxed{\sf Time \ in \ which \ train \ will \ come \ to \ rest = 20 \ sec}

Given:

Initial velocity (u) = 30 m/s

Final speed (v) = 0 m/s

Acceleration (a) = - 1.5 m/,s²

To Find:

Time in which train will come to rest (t).

Explanation:

\sf From \ equation \ of \ motion: \\ \sf \implies \bold{v = u + at} \\ \\ \sf Substituting \ value \ of \ v, \ u \ and \ a:  \\  \sf \implies 0 = 30 + ( - 1.5)(t) \\   \sf  \implies 0 = 30 - 1.5(t) \\  \sf \implies 30 - 1.5(t) = 0 \\  \\  \sf Subtract  \: 30  \: from  \: both  \: sides: \\  \sf \implies (30 -  \boxed{ \sf 30}) - 1.5(t) =  \boxed{ \sf  - 30} \\  \\  \sf 30 - 30 = 0 :  \\  \sf \implies  - 1.5(t) =  - 30 \\  \\  \sf Divide  \: both  \: sides \:  of \:  - 1.5(t) =  - 30 \: by \:  - 1.5 :  \\  \sf \implies  \frac{  - 1.5(t)}{ \boxed{ \sf - 1.5}}  =  \frac{ - 30}{ \boxed{ \sf -1.5 }}  \\  \\  \sf \frac{ \cancel{ \sf 1.5}}{\cancel{ \sf 1.5}}  = 1 :  \\  \sf \implies t =  \frac{ - 30}{ - 1.5}  \\  \\   \sf  \frac{ - 30}{ - 1.5}  =  \frac{\cancel{ \sf 1.5} \times 20}{\cancel{ \sf 1.5}}  = 20 :  \\  \sf  \implies t = 20 \: sec

So,

Time in which train will come to rest = 20 seconds

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In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strike
zysi [14]

a) t=\sqrt{\frac{2h}{g}}

b) v=\frac{d}{\sqrt{\frac{2h}{g}}}

c) v=\sqrt{d^2(\frac{g}{2h})+(2gh)}

d) \theta=tan^{-1}(\frac{2h}{d}) (radians)

Explanation:

a)

The motion of the cup sliding off the counter is the motion of a projectile, consisting of two independent motions:

- A uniform motion along the horizontal direction

- A uniformly accelerated motion (free fall) along the vertical direction

The time of flight of the cup is entirely determined by the vertical motion, therefore we can use the suvat equation:

s=ut+\frac{1}{2}at^2

where here:

s=h (the vertical displacement is the height of the counter)

u=0 (the initial vertical velocity of the cup is zero)

a=g (the vertical acceleration is the acceleration of gravity)

Solving for t, we find the time of flight of the cup:

h=0+\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}

b)

To solve this part, we just analyze the horizontal motion of the cup.

Here we know that the horizontal motion of the cup is uniform: this means that is horizontal speed is constant during the whole motion, and it is actually equal to the speed at which the mug leaves the counter.

For a uniform motion, the speed is given by

v=\frac{d}{t}

where

d is the distance covered

t is the time taken

Here, the distance covered is d, the distance from the base of the counter, while the time taken is the time of flight:

t=\sqrt{\frac{2h}{g}}

Substituting into the previous equation, we find the speed of the mug as it leaves the counter:

v=\frac{d}{\sqrt{\frac{2h}{g}}}

c)

Here we want to find the speed of the cup immediately before it hits the floor.

Here we have to consider that while the mug falls, its vertical speed increases, while the horizontal speed remains constant.

Therefore, the horizontal speed of the cup before it hits the ground is:

v_x=\frac{d}{\sqrt{\frac{2h}{g}}}=d\sqrt{\frac{g}{2h}}

The vertical speed instead is given by the suvat equation:

v_y=u_y + at

where:

u_y=0 is the initial vertical velocity

a=g is the acceleration

t=\sqrt{\frac{2h}{g}} is the time of flight

Substituting,

v_y = 0 +g(\sqrt{\frac{2h}{g}})=\sqrt{2gh}

The actual speed of the cup just before it hits the floor is the resultant of the horizontal and vertical speeds, so it is:

v=\sqrt{v_x^2+v_y^2}=\sqrt{d^2(\frac{g}{2h})+(2gh)}

d)

Just before hitting the floor, the velocity of the cup has two components:

v_x=d\sqrt{\frac{g}{2h}} is the horizontal component (in the forward direction)

v_y=\sqrt{2gh} is the vertical component (in the downward direction)

Since the two components are perpendicular to each other, the angle of the direction is given by the equation

tan \theta = \frac{v_y}{v_x}

where here \theta is measured as below the horizontal direction.

Substituting the expressions for v_x,v_y, we find:

tan \theta = \frac{\sqrt{2gh}}{d\sqrt{\frac{g}{2h}}}=\frac{2h}{d}

So

\theta=tan^{-1}(\frac{2h}{d}) (radians)

4 0
3 years ago
Two instruments produce a beat frequency of 5 Hz. If one has a frequency of 264 Hz, what could be the frequency of the other ins
Lerok [7]

Answer:

259 Hz or 269 Hz

Explanation:

Beat: This is the phenomenon obtained when two notes of nearly equal frequency are sounded together. The S.I unit of beat is Hertz (Hz).

From the question,

Beat = f₂-f₁................ Equation 1

Note: The frequency of the other instrument is either f₁ or f₂.

If the unknown instrument's frequency is f₁,

Then,

f₁ = f₂-beat............ equation 2

Given: f₂ = 264 Hz, Beat = 5 Hz

Substitute into equation 2

f₁ = 264-5

f₁ = 259 Hz.

But if the unknown frequency is f₂,

Then,

f₂ = f₁+Beat................. Equation 3

f₂ = 264+5

f₂ = 269 Hz.

Hence the beat could be 259 Hz or 269 Hz

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Question 8 of 10
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here is the site : textbook.com

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already did the work, i just need someone to see if i did the tangent line right? the line has to touch the point 0.6! thank you
Dmitrij [34]

The tangent looks good.

The curve is a bit crooked, at the 0.9 and 1.

But overall, cool graph.

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2 years ago
Type O negative blood is considered the universal red blood cell donor because it _____.
marissa [1.9K]

Answer:

Honestly i think the answer is B

Explanation:

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