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3 years ago

If an object accelerates from rest, with a constant of 8 m/s2, what will its velocity be after 35s?

Physics

1 answer:

AlladinOne [14]3 years ago

5 0

Answer:

Its speed will be 280 m/s

Explanation:

Constant Acceleration Motion

It's a type of motion in which the speed of an object changes by an equal amount in every equal period of time.

If a is the constant acceleration, vo the initial speed, vf the final speed, and t the time, vf can be calculated as:

$v_f=v_o+at$

The object accelerates from rest (vo=0) at a constant acceleration of $a=8\ m/s^2$ . The final speed at t=35 seconds is:

$v_f=0+8*35$

$v_f=280\ m/s$

Its speed will be 280 m/s

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3 years ago

a boy Held a book of weight 50 newton’s in his fully outstretched hand, at a distance of 50 cm away calculate the moment of forc

miss Akunina [59]

Answer:

the moment of force is 25 Nm.

Explanation:

Given;

weight of the book, F = 50 N

distance through which his hand was stretched, d = 50 cm = 0.5 m

The moment of force is calculated as;

moment of force = F x r

moment of force = 50 N x 0.5 m

moment of force = 25 Nm

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3 years ago

A fireperson is 50 m from a burning building and directs a stream of water from a fire hose at an angle of 300 above the horizon

notsponge [240]

Answer:

We can think the water stream as a solid object that is fired.

The distance between the fireperson and the building is 50m. (i consider that the position of the fireperson is our position = 0)

The angle is 30 above the horizontal. (yo wrote 300, but this has no sense because 300° implies that he is pointing to the ground).

The initial speed of the stream is 40m/s.

First, using the fact that:

x = R*cos(θ)

y = R*sin(θ)

in this case R = 40m/s and θ = 30°

We can use the above relation to find the components of the velocity:

Vx = 40m/s*cos(30°) = 34.64m/s

Vy = 20m/s.

First step:

We want to find the time needed to the stream to hit the buildin.

The horizontal speed is 34.64m/s and the distance to the wall is 50m

So we want that:

34.64m/s*t = 50m

t = 50m/(34.64m/s) = 1.44 seconds.

Now we need to calculate the height of the stream at t = 1.44s

Second step:

The only force acting on the water is the gravitational one, so the acceleration of the stream is:

a(t) = -g.

g = -9.8m/s^2

For the speed, we integrate over time and we get:

v(t) = -g*t + v0

where v0 is the initial speed: v0 = 20m/s.

The velocity equation is:

v(t) = -g*t + 20m/s.

For the position, we integrate again over time:

p(t) = -(1/2)*g*t^2 + 20m/s*t + p0

p0 is the initial height of the stream, this data is not known.

Now, the height at the time t = 1.44s is

p(1.44s) = -5.9m/s^2*(1.44s)^2 + 20m/s*1.44s + po

= 16.57m + p0

So the height at wich the stream hits the building is 16.57 meters above the initial height of the fire hose.

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4 years ago

1) Subatomic particles called muons can be created in the upper atmosphere by collisions of cosmic rays (energetic particles com

Vsevolod [243]

Answer and explanation:

Muon travelled straight down towards the earth. Therefore the tree moves up in the rest frame of muon (option a)

In muon rest frame it travels Zero meters

Distance, d = Velocity, v * Time, s

$where, v = 0.9c = 0.9 \times 8 \times 10^8 , s = 2.2 \mu s$

$d = 0.9 \times 3 \times 10^8 \times 2.2 \times 10^{-6}\\\\d = 594m$

Distance from the top of the mountain to the tree is the same as the distance travelled by the tree in the muons rest frame

that is same as in part C which is 594m

Using lorentz contraction

In the rest frame of someone standing on the mountain

the distance is given by

$d' = \frac{d}{\gamma} = d\sqrt{1 - \frac{v^2}{c^2}}, where, \frac{1}{\gamma}= \sqrt{1 - \frac{v^2}{c^2}}$

$d' = 594\sqrt{1 - \frac{(0.9c)^2}{c^2}}$

$d' = 594\sqrt{1 - 0.81}$

$d' = 594 \times 0.4359$

d' = 258.92m

in the rest frame of someone standing on the mountain,

muon moves straight down

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4 years ago