Please help on this one?

Car A is traveling north on a straight highway and car B is traveling west on a different straight highway. Each car is approach

Inessa05 [86]

Answer:

102 km/h

Explanation:

x = distance of car A at any time from the intersection

x₀ = distance of car A at some time = 0.4 km

$v_{A}$ = Speed of car A = 75 km/h

y = distance of car B at any time from the intersection

y₀ = distance of car B at some time = 0.3 km

$v_{B}$ = Speed of car B = 70 km/h

d = distance between the two cars at any time

d₀ = distance between the two cars at some time

v = rate of change of distance between the cars

Using Pythagorean theorem

d²₀ = x₀² + y₀²

d²₀ = 0.4² + 0.3²

d₀ = 0.5 m

Distance between the two cars at any time is given using Pythagorean theorem as

d² = x² + y²

Taking derivative both side relative to "t"

$2d \left ( \frac{dd}{dt} \right ) = 2x ( \frac{dx}{dt} \right ) + 2y ( \frac{dy}{dt} \right )$

$d_{o} v = x_{o} ( \frac{dx}{dt} \right ) + y_{o} ( \frac{dy}{dt} \right )$

(0.5) v = (0.4) $v_{A}$ + (0.3) $v_{B}$

(0.5) v = (0.4) (75) + (0.3) (70)

v = 102 km/h

7 0

3 years ago

With what minimum speed must you toss a 130 g ball straight up to just touch the 14-m-high roof of the gymnasium if you release

Dmitry [639]

THIS IS THE COMPLETE QUESTION BELOW

With what minimum speed must you toss a 130 g ball straight up to just touch the 14-m-high roof of the gymnasium if you release the ball 1.1 m above the ground

And what speed does the ball hit the ground? Solve this problem using energy.

Answer

a)minimum speed must you toss a 130 g is 15.9090m/s

b)speed the ball hit the ground is 16.57m/s

Explanation:

a)We know that For any closed/isolated system, the total energy is CONSERVED.

K.E. lost by the ball=The change in P.E of the ball at 1.1m above the ground as well as the P.E. of the ball at 14 m-high roof

This statement can be expressed as the expression below from K.E and P.E energy formula

P.E. = mgh

K.E. = (1/2)mv^2

Therefore,

(mgh1 - mgh2)=(1/2)mv^2

Where h2=the ball height above the ground=1.1m

h1=ball height at roof of the gymnasium= 14m

Then if we substitute we have

[(10) x (0.14) x (9.81)] - [(1.1) x (0.14) x (9.81)] = (1/2)(0.14)(v^2)

16.45137=0.065V^2

V=15.9090m/s

minimum speed must you toss a 130 g is 15.9090m/s

b)To calculate the speed the ball hit the ground?

This is the highest point (14m-high roof),and the type of the energy the ball possesses is Po.tential energy only.

At the lowest point (ground), the energy the ball possesses is K.E. only.

P.E at 10m-high roof = K.E. at ground.

(14) x (0.13) x (9.81) = (1/2) x (0.13) x v^2

17.8542= 0.065V^2

V= 16.57

Therefore,And speed the ball hit the ground is 16.57m/s

6 0

3 years ago

When the moon is waxing, it's lighted part appears

kvv77 [185]

To increase it's size

Waxing is the opposite of waning, which is to decrease.

7 0

3 years ago

Read 2 more answers

Which best explains why the receiver of a signal must understand the code or language being used

Oksi-84 [34.3K]

The receiver of a signal must understand the code or language being used to avoid confusion and losses.

<h3>What is a Signal?</h3>

This is usually in the form of a sound or body movement and is involved in conveying messages to people.

The receiver must understand the code or language in order to prevent confusion or loss of lives and properties.

Read more about Signal here brainly.com/question/15304191

#SPJ1

4 0

2 years ago

Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center

kiruha [24]

Answer:

The angular velocity is $w_f = 1.531 \ rad/ s$

Explanation:

From the question we are told that

The mass of each astronauts is $m = 50 \ kg$

The initial distance between the two astronauts $d_i = 7 \ m$

Generally the radius is mathematically represented as $r_i = \frac{d_i}{2} = \frac{7}{2} = 3.5 \ m$

The initial angular velocity is $w_1 = 0.5 \ rad /s$

The distance between the two astronauts after the rope is pulled is $d_f = 4 \ m$

Generally the radius is mathematically represented as $r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m$

Generally from the law of angular momentum conservation we have that

$I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}$

Here $I_{k_1 }$ is the initial moment of inertia of the first astronauts which is equal to $I_{p_1}$ the initial moment of inertia of the second astronauts So

$I_{k_1} = I_{p_1 } = m * r_i^2$

Also $w_{k_1 }$ is the initial angular velocity of the first astronauts which is equal to $w_{p_1}$ the initial angular velocity of the second astronauts So

$w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts So

$I_{k_2} = I_{p_2} = m * r_f^2$

Also $w_{k_2 }$ is the final angular velocity of the first astronauts which is equal to $w_{p_2}$ the final angular velocity of the second astronauts So