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3 years ago

PLEASE HELP

2 answers:

4 0

Don't over think it!

For example, if you do 5 laps in 5 minutes then you do one lap per minute. If you use that example and you apply it to your problem you'll notice that the answer is simply C. 1km/min.

Best wishes!

Grace [21]3 years ago

4 0

C. 1km/min because the car covers a distance of 5km over a period of 5 minutes, so the car travels 1 kilometer each minute.

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The two vectors and in fig. 3-28 have equal magnitudes of 10.0 m and the angles are 30° and 105°. find the (a) x and (b) y compo

mylen [45]

You can just use basic trigonometry to solve for the x & y components.

vector a = 10cos(30) i + 10sin(30) j = <5sqrt(3), 5>

vector b is only slightly harder because the angle is relative to vector a, and not the positive x-axis. Anyway, this just makes vector b with an angle of 135deg to the positive x-axis.

vector b = 10cos(135) i + 10sin(135) j = <-5sqrt(2), 5sqrt(2)>

So now we can do the questions:

r = a + b

r = <5sqrt(3)-5sqrt(2), 5+5sqrt(2)>

(a) 5sqrt(3)-5sqrt(2)

(b) 5+5sqrt(2)

(c)

|r| = sqrt( (5sqrt(3)-5sqrt(2))2 + (5+5sqrt(2))2 )

= 12.175

(d)

θ = tan-1 ( (5+5sqrt(2)) / (5sqrt(3)-5sqrt(2)) )

θ = 82.5deg

6 0

2 years ago

Ice of mass 5 g at 0 °C melts to water at 0 °C.

amid [387]

Answer:

Q=1670J

Explanation:

Mass of ice: m=5g=0.005kg

Latent heat: lambda=3.34×10⁵J/kg

Heat received by ice: Q=m×lambda

Q=0.005×3.34×10⁵=5×334=1670J

5 0

2 years ago

What is the change in potential energy of an energetic 72 kg hiker who makes it from the floor of death valley to the top of mt.

storchak [24]

The lowest point in Death Valley is 85 m below sea level. The summit of nearby Mt. Whitney has an elevation of 4420 m.

3 0

3 years ago

Barack is playing basketball in his back yard. He takes a shot 7.0 m from the basket (measured along the ground), shooting at an

VMariaS [17]

Answer:

The time of flight of the ball is 1.06 seconds.

Explanation:

Given $\Delta x=7\ m$

$\theta=45 \°$

Also, $\Delta y=(3.5-2)=1.5\ m$

$a_x=0\ and\ a_y=-9.81\ m/s^2$

Let us say the velocity in the x-direction is $v_x$ and in the y-direction is $v_y$ . And acceleration in the x-direction is $a_x$ and in the y-direction is $a_y$ .

Also, $\Delta x\ and\ \Delta y$ is distance covered in x and y direction respectively. And $t$ is the time taken by the ball to hit the backboard.

We can write $v_x=v_0cos(45)\ and\ v_y=v_0sin(45)$ . Where $v_0$ is velocity of ball.

Now,

$\Delta x=v_x\times t+\frac{1}{2}\times a_x\times t^2\\ \Delta x=v_x\times t+\frac{1}{2}\times 0\times t^2\\\Delta x=v_xt$

$\Delta x=v_0cos(45)\times t\\7=v_0cos(45)\times t\\\\t=\frac{7}{v_0cos(45)}$

Also,

$\Delta y=v_y\times t+\frac{1}{2}\times a_y\times t^2\\ 1.5=v_0sin(45)\times \frac{7}{v_0cos(45)}+\frac{1}{2}\times (-9.81)\times(\frac{7}{v_0cos(45)} )^2\\\\1.5=7-\frac{481}{(v_0)^2}\\ \\\frac{481}{(v_0)^2}=5.5\\\\(v_0)^2=\frac{481}{5.5}\\ \\(v_0)^2=87.45\\\\v_0=\sqrt{87.45}=9.35\ m/s$ .

Plugging this value in

$t=\frac{7}{v_0cos(45)}\\ \\t=\frac{7}{9.35\times 0.707}\\ \\t=\frac{7}{6.611}$

$t=1.06\ seconds$

So, the time of flight of the ball is 1.06 seconds.

6 0

3 years ago

Doss [256]

Answer:

The resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.

Explanation:

Considering west direction along negative x-axis and north direction along positive y-axis

Given:

The car travels at a speed of 120 km/h in the west direction.

The car then travels at the same speed in the north direction.

Now, considering the given directions, the velocities are given as:

Velocity in west direction is, $\overrightarrow{v_1}=-120\ \vec{i}$

Velocity in north direction is, $\overrightarrow{v_2}=120\ \vec{j}$

Now, since $v_1\ and\ v_2$ are perpendicular to each other, their resultant magnitude is given as:

$|\overrightarrow{v_{res}}|=\sqrt{|\overrightarrow{v_1}|^2+|\overrightarrow{v_2}|^2}$

Plug in the given values and solve for the magnitude of the resultant.This gives,

$|\overrightarrow{v_{res}}|=\sqrt{(120)^2+(120)^2}\\\\|\overrightarrow{v_{res}}|=120\sqrt{2} = 169.71\ km/h$

Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.

So, the direction is given as:

$x=\tan^{-1}(\frac{|v_2|}{|v_1|})\\\\x=\tan^{-1}(\frac{120}{-120})=\tan^{-1}(-1)=-45\ deg(clockwise\ angle\ with\ the\ x-axis)$

Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.

4 0

2 years ago