4. The earth exerts a gravitational force of 3.5 N on an object. What is the mass of

At time t = 1, a particle is located at position (x, y) = (5, 2). If it moves in the velocity field F(x, y) = xy − 1, y2 − 11 fi

Amanda [17]

Answer:

Its approx location is (5.18,1.9)

Explanation:

Using F( 5,2) = ( xy-1, y²-11)

= ( 5*2-¹, 2²-11)

= (9,-5)

= so at point t=1.02

(5,2)+(1.02-1)*(9,-5)

(5,2)+( 0.02)*(9,-5)

(5+0.18, 2-0.1)

= ( 5.18, 1.9)

3 0

3 years ago

A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground,

n200080 [17]

Answer:

a. $t = \frac{v_{0} +/- \sqrt{v_{0} ^{2} - gD} }{g}$ b. D = v₀²/2g

Explanation:

Here is the complete question

A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance

Find the time at which the two balls collide.

Express your answer in terms of the variables D ,v₀ , and appropriate constants..

t = ?!

Part B

Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Express your answer in terms of the variables v₀ and g .

D =?!

Solution

The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.

The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.

The two balls collide when their vertical distances are equal. That is H₁ = H₂

So, D - v₀t + gt²/2 = v₀t - gt²/2

Collecting like terms

D + gt²/2 + gt²/2 = v₀t + v₀t

D +gt² = 2v₀t

gt² - 2v₀t + D = 0.

Using the quadratic formula,

$t = \frac{-(-2v_{0} ) +/- \sqrt{(-2v_{0} )^{2} - 4 X g XD} }{2g} = \frac{2v_{0} +/- \sqrt{4v_{0} ^{2} - 4gD} }{2g} = \frac{v_{0} +/- \sqrt{v_{0} ^{2} - gD} }{g}$

B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.

0 = v₀² - 2gs

s = v₀²/2g.

Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,

0 = v₀ - gt₀

t₀ = v₀/g

Also H = s₁ + s where s₁ = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.

So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g

6 0

3 years ago

A football field lies so one endzone is to the east and one is to the west. A running back gets the ball and starts running 20o

Jobisdone [24]

Answer:

( 80.87 i - 12.96 j ) m

Explanation:

Running back movements :

20⁰ north of east for 10 meters,

Runs straight east for 60 meters

runs 35⁰ east of south for 20 meters

A) show vector pointing from the starting point to the end ( where he scored )

The final vector displacement : ( 80.87 i - 12.96 j ) m

which is : 81.90 m, 9.10⁰ south of east

attached below is the required diagram

8 0

2 years ago

2. A 2.5 meter-long wave travels at 1 m/s towards a fixed boundary. After 2 seconds, how does the wave appear?

Elan Coil [88]

Answer:

Since incident wave and its reflected part in opposite phase superimpose on each other

So correct answer will be

Option B

Explanation:

Here we know that the wave reflection is done by rigid boundary

So when wave is reflected by the boundary then its phase is reversed by 180 degree

so the reflected wave is in reverse phase from the boundary

so we can superimpose the reflected part with incident wave to dine the resultant wave

So the phenomenon is given as follow

6 0

3 years ago

An electron is ejected from the cathode by a photon with an energy slightly greater than the work function of the cathode. How w

Ksivusya [100]

It will be approximately equal.

<h3>How will the final kinetic energy change?</h3>

We can infer that all of the energy in the electron is Potential energy (PE) because the energy provided by the photon is hardly enough to outweigh the work function.

It will gain kinetic energy (KE) as it advances in the direction of the anode because it is moving through an electric field. All of the PE will have been transformed to KE by the time it reaches the anode.

According to the question

K = hf - W

W = Work function

The energy of photons is comparable. After conversion, there was only a little amount of KE remaining.

Therefore, PE (W) essentially equals KE (K).

It will about be equal.

Learn more about work function here:

brainly.com/question/19595244

#SPJ4

3 0

2 years ago