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4 years ago

A car moves at 18m/s North. What is the x component of the velocity?

Physics

1 answer:

Marina86 [1]4 years ago

6 0

Velocity, v = 18 m/s, makes an angle 60 degrees with the horizontal (x-axis).

(horizontal) x-component of velocity, v(x) = v cos (60) = 18 m/s x cos(60) = 9 m/s

If you need to determine the y-component (vertical) of velocity,

(vertical) y-component of velocity, v(y) = v sin (60) = 18 m/s x sin(60) = 15.6 m/s

Hope it helps

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give an example of one living and one non-living thing that uses the force of buoyancy to function. explain how they work.

zaharov [31]

Answer:

the biotic one is a human in water and they flaot because of the air in our lungs and the abiotic one is a swimming buoys are filled with foam and foam floats

Explanation:

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2 years ago

Read 2 more answers

If a girl is running along a straight road with a uniform velocity 1.5m/s find her acceleration

Jlenok [28]

Answer:

Dear user,

Answer to your query is provided below

Acceleration is zero because of no change in velocity.

Explanation:

Remember that velocity is a vector quantity and a vector can change in 3 ways

•Magnitude only

•Direction only

•Both magnitude and direction.

Now the magnitude of velocity (speed) can stay constant while the direction is changing. This is the case in circular motion.

In the question above, it is mentioned that the girl is moving along a straight road. Therefore no change in direction of velocity.

7 0

3 years ago

A metal block suspended from a spring balance is submerged in water. You observe that the block displaces 55 cm3 of water and th

DiKsa [7]

Answer:

8977.7 kg/m^3

Explanation:

Volume of water displaced = 55 cm^3 = 55 x 10^-6 m^3

Reading of balance when block is immersed in water = 4.3 N

According to the Archimedes principle, when a body is immersed n a liquid partly or wholly, then there is a loss in the weight of body which is called upthrust or buoyant force. this buoyant force is equal to the weight of liquid displaced by the body.

Buoyant force = weight of the water displaced by the block

Buoyant force = Volume of water displaced x density of water x g

= 55 x 10^-6 x 1000 x .8 = 0.539 N

True weight of the body = Weight of body in water + buoyant force

m g = 4.3 + 0.539 = 4.839

m = 0.4937 kg

Density of block = mass of block / volume of block

= $\frac{0.4937}{55\times10^{-6}}$

Density of block = 8977.7 kg/m^3

4 0

3 years ago

In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a 40.0? angle from the horizontal.

HACTEHA [7]

A) Horizontal range: 16.34 m

B) Horizontal range: 16.38 m

C) Horizontal range: 16.34 m

D) Horizontal range: 16.07 m

E) The angle that gives the maximum range is $41.9^{\circ}$

Explanation:

The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:

$s=u_y t + \frac{1}{2}at^2$ (1)

where

s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)

$u_y = u sin \theta$ is the initial vertical velocity, where

u = 12.0 m/s is the initial speed

$\theta=40.0^{\circ}$ is the angle of projection

$u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s$

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (downward)

Substituting the numbers, we get

$-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0$

which has two solutions:

t = -0.21 s (negative, we ignore it)

t = 1.778 s (this is the time of flight)

The horizontal motion is instead uniform, so the horizontal range is given by

$d=u_x t$

where

$u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s$ is the horizontal velocity

t = 1.778 s is the time of flight

Solving, we find

$d=(9.19)(1.778)=16.34 m$

In this second case,

$\theta=42.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.1t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.851 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s$

So, the range of the shot is

$d=u_x t = (8.85)(1.851)=16.38 m$

In this third case,

$\theta=45^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.5t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.925 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s$

So, the range of the shot is

$d=u_x t = (8.49)(1.925)=16.34$ m

In this 4th case,

$\theta=47.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.8t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.981 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s$

So, the range of the shot is

$d=u_x t = (8.11)(1.981)=16.07 m$

From the previous parts, we see that the maximum range is obtained when the angle of releases is $\theta=42.5^{\circ}$ .

The actual angle of release which corresponds to the maximum range can be obtained as follows:

The equation for the vertical motion can be rewritten as

$s-u sin \theta t + \frac{1}{2}gt^2=0$

The solutions of this quadratic equation are

$t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}$

This is the time of flight: so, the horizontal range is

$d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta$

It can be found that the maximum of this function is obtained when the angle is

$\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})$

Therefore in this problem, the angle which leads to the maximum range is

$\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

8 0

4 years ago

A model train engine was moving at a constant speed on a straight horizontal track. As the engine moved along, a marble was fir

bagirrra123 [75]

Answer:

The marble was moving in a projectile and the speed of the engine was 2.716 m/s

Explanation:

The vertical component of the marble's flight path relative to the train

is given by the equation y(t) = v*t - (4.9)*t^2,

where, v is the initial upward velocity of the marble relative to the train.

So with y(1) = v - 4.9 = 0 we have

v = 4.9 m/s.

The marble will reach maximum height after 0.5 seconds, at which the

height will be y(0.5) = (4.9)*(0.5) - (4.9)*(0.5)^2 = (4.9)*(0.25) = 1.225 m.

Now, the marble has a vertical velocity component of 4.9 m/s and a horizontal velocity component

of V m/s such that tan(61) = 4.9 / V

V = 4.9 / tan(61) = 2.716 m/s

This horizontal velocity component of the marble is the same as the

speed of the train i.e. 2.716 m/s.

3 0

4 years ago