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3 years ago

A runner completes the 200-meter dash with a time of 19.80 seconds. What was the runner's average speed in miles per hour?

Physics

1 answer:

andriy [413]3 years ago

4 0

Answer:

v = 22.54 mph.

Explanation:

Given that,

Distance moved, d = 200 m

Time, t = 19.8 s

We need to find the runner's average speed.

We know that,

1 mile = 1609.34 m

200 m = 0.124 miles

19.8 seconds = 0.0055 h

So,

Speed = distance/time

$v=\dfrac{0.124}{0.0055}\\\\v=22.54\ mph$

So, the runner's average speed is 22.54 mph.

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Three displacements through a hedge maze. (b) The displacement vectors. (c) The first displacement vector and its components. (d

Vlad1618 [11]

Answer:

18.3 m , 25.4°

Explanation:

d1 = 6 m, θ1 = 40°

d2 = 8 m, θ2 = 30°

d3 = 5 m, θ3 = 0°

Write the displacements in the vector form

$d_{1}=6\left ( Cos40\widehat{i}+Sin40\widehat{j} \right )=4.6\widehat{i}+3.86\widehat{j}$

$d_{2}=8\left ( Cos30\widehat{i}+Sin30\widehat{j} \right )=6.93\widehat{i}+4\widehat{j}$

$d_{3}=5\widehat{i}$

The total displacement is given by

$\overrightarrow{d}=\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}$

$d=\left ( 4.6+6.93+5 \right )\widehat{i}+\left ( 3.86+4 \right )\widehat{j}$

$d=16.53\widehat{i}+7.86\widehat{j}$

magnitude of resultant displacement is given by

$d ={\sqrt{16.53^{2}+7.86^{2}}}=18.3 m$

d = 18.3 m

Let θ be the angle of resultant displacement with + x axis

$tan\theta =\frac{7.86}{16.53}=0.4755$

θ = 25.4°

8 0

3 years ago

Three forces act on a moving object. One force has a magnitude of 83.7 N and is directed due north. Another has a magnitude of 5

LekaFEV [45]

Answer:

$|\vec{F}_3| = 102.92 \ N$

$\theta = 57 \° 24 ' 48''$

Explanation:

For an object to move with constant velocity, the acceleration of the object must be zero:

$\vec{a} = \vec{0}$ .

As the net force equals acceleration multiplied by mass , this must mean:

$\vec{F}_{net} = m \vec{a} = m * \vec{0} = \vec{0}$ .

So, the sum of the three forces must be zero:

$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0}$ ,

this implies:

$\vec{F}_3 = - \vec{F}_1 - \vec{F}_2$ .

To obtain this sum, its easier to work in Cartesian representation.

First we need to define an Frame of reference. Lets put the x axis unit vector $\hat{i}$ pointing east, with the y axis unit vector $\hat{j}$ pointing south, so the positive angle is south of east. For this, we got for the first force:

$\vec{F}_1 = 83.7 \ N \ (-\hat{j})$ ,

as is pointing north, and for the second force:

$\vec{F}_2 = 59.9 \ N \ (-\hat{i})$ ,

as is pointing west.

Now, our third force will be:

$\vec{F}_3 = - 83.7 \ N \ (-\hat{j}) - 59.9 \ N \ (-\hat{i})$

$\vec{F}_3 = 83.7 \ N \ \hat{j} + 59.9 \ N \ \hat{i}$

$\vec{F}_3 = (59.9 \ N , 83.7 \ N )$

But, we need the magnitude and the direction.

To find the magnitude, we can use the Pythagorean theorem.

$|\vec{R}| = \sqrt{R_x^2 + R_y^2}$

$|\vec{F}_3| = \sqrt{(59.9 \ N)^2 + (83.7 \ N)^2}$

$|\vec{F}_3| = 102.92 \ N$

this is the magnitude.

To find the direction, we can use:

$\theta = arctan(\frac{F_{3_y}}{F_{3_x}})$

$\theta = arctan(\frac{83.7 \ N }{ 59.9 \ N })$

$\theta = 57 \° 24 ' 48''$

and this is the angle south of east.

7 0

3 years ago

In the image of the underwater submarine (i can't upload it), if the only 2 forces acting on the submarine are the downward forc

shutvik [7]

K so I'm not completely sure about this...

but I believe the answer would be C. It will float upward

because the buoyancy, in the image, is stronger than the Gravitational pull.

Let me know if that is right or wrong plz

Hope it's right and helps tho!

4 0

3 years ago

A 12,500 N alien UFO is hovering about the surface of Earth. At time , its position can be given as () = ((0.24 m/s^3)^3 + 25 m)

stepladder [879]

a) $F=(3675i-4543k)N$

b) 5843 N

Explanation:

The position of the UFO at time t is given by the vector:

$r(t)=(0.24t^3+25)i+(4.2t)j+(-0.43t^3+0.8t^2)k$

Therefore it has 3 components:

$r_x=0.24t^3+25\\r_y=4.2t\\r_z=-0.43t^3+0.8t^2$

We start by finding the velocity of the UFO, which is given by the derivative of the position:

$v_x=r'_x=\frac{d}{dt}(0.24t^3+25)=3\cdot 0.24t^2=0.72t^2\\v_y=r'_y=\frac{d}{dt}(4.2t)=4.2\\v_x=r'_z=\frac{d}{dt}(-0.43t^3+0.8t^2)=-1.29t^2+1.6t$

And then, by differentiating again, we find the acceleration:

$a_x=v'_x=\frac{d}{dt}(0.72t^2)=1.44t\\a_y=v'_y=\frac{d}{dt}(4.2)=0\\a_z=v'_z=\frac{d}{dt}(-1.29t^2+1.6t)=-2.58t+1.6$

The weight of the UFO is W = 12,500 N, so its mass is:

$m=\frac{W}{g}=\frac{12500}{9.8}=1276 kg$

Therefore, the components of the force on the UFO are given by Newton's second law:

$F=ma$

So, Substituting t = 2 s, we find:

$F_x=ma_x=(1276)(1.44t)=(1276)(1.44)(2)=3675 N\\F_y=ma_y=0\\F_z=ma_z=(1276)(-2.58t+1.6)=(1276)(-2.58(2)+1.6)=-4543 N$

So the net force on the UFO at t = 2 s is

$F=(3675i-4543k)N$

The magnitude of a 3-dimensional vector is given by

$|v|=\sqrt{v_x^2+v_y^2+v_z^2}$

where

$v_x,v_y,v_z$ are the three components of the vector

In this problem, the three components of the net force are:

$F_x=3675 N\\F_y=0\\F_z=-4543 N$

Therefore, substituting into the equation, we find the magnitude of the net force:

$|F|=\sqrt{3675^2+0^2+(-4543)^2}=5843 N$

7 0

3 years ago

The outside diameter of the playing area of an optical Blu-ray disc is 11.75 cm, and the inside diameter is 4.50 cm. When viewin

erica [24]

Answer:

The minimum angular speed 57.0 rad/s

Explanation:

In this exercise we must use the relationships between the angular and linear magnitudes, for the speeds we have

v = w r

As they give us that linear velocity is constant, we will calculate the angular velocity for the two radii, external and internal

r1 = 11.75 cm (1 m / 100cm) = 0.1175 m

w1 = v / r1

w1 = 6.70 / 0.1175

w1 = 57.0 rd / s

R2 = 4.50 cm = 0.0450 m

w2 = 6.70 / 0.0450

w2 = 148 rad / s

The minimum angular speed 57.0 rad / s

4 0

3 years ago