40 POINTS

A car travels initially at 24 m/s, until it enters the highway. If the car accelerates at 4 m/s^2 for a 96 meters, what is the c

marishachu [46]

initial velocity=u=24m/s
Acceleration=a=4m/s^2
Distance=s=96m
Final velocity=v

Using 3rd equation of kinematics

$\boxed{\Large{\sf v^2-u^2=2as}}$

$\\ \Large\sf\longmapsto v^2=u^2+2as$

$\\ \Large\sf\longmapsto v^2=24^2+2(4)(96)$

$\\ \Large\sf\longmapsto v^2=576+768$

$\\ \Large\sf\longmapsto v^2=1344$

$\\ \Large\sf\longmapsto v=\sqrt{1344}$

$\\ \Large\sf\longmapsto v=36.6m/s$

3 0

3 years ago

A ball is dropped from an upper floor, some unknown distance above your apartment. As you look out of your window, which is 1.50

laila [671]

Answer:

The ball is dropped at a height of 9.71 m above the top of the window.

Explanation:

<u>Given:</u>

Height of the window=1.5 m
Time taken by ball to cover the window height=0.15

Now using equation of motion in one dimension we have

$s=ut+\dfrac{at^2}{2}$

Let u be the velocity of the ball when it reaches the top of the window

then

$1.5=0.15u+\dfrac{9.8\times0.15^2}{2}\\u=3.96\ \rm m/s\\$

Now u is the final velocity of the ball with respect to the top of the building

so let t be the time taken for it to reach the top of the window with this velocity

$3.96=gt\\t=0.4\ \rm s\\$

Let h be the height above the top of the window

$h=\dfrac{gt^2}{2}\\\\h=\dfrac{9.8\times 0.4^2}{2}\\h=9.71\ \rm m$

3 0

3 years ago

Test Due! Which statement is true?

Lubov Fominskaja [6]

The correct answer is C. Mercury and Mars have the same gravitational force

Explanation:

This chart compares the different features of two planets in our solar system (Mercury and Mars). In this chart, the only numerical value or feature that is the same for both planets is gravity because for both planets gravity is 1.7 m/s2. This implies the gravitational force or the force that attracts objects towards the center of the planet is the same or that objects are pulled with the same force in both planets. Moreover, this factor depends on others such as mass, density, among others.

8 0

2 years ago

A particle moves along line segments from the origin to the points (1, 0, 0), (1, 5, 1), (0, 5, 1), and back to the origin under

Kaylis [27]

Answer:

0 J

Explanation:

Since work done W = ∫F.dr and F(x, y, z)= z²i + 4xyj + 5y²k and dr = dxi + dyj + dzk

F.dr = (z²i + 4xyj + 5y²k).(dxi + dyj + dzk) = z²dx + 4xydy + 5y²dz

W = ∫F.dr = ∫z²dx + 4xydy + 5y²dz = z²x + 2xy² + 5y²z

We now evaluate the work done for the different regions

W₁ = work done from (0,0,0) to (1,0,0)

W₁ = {z²x + 2xy² + 5y²z}₀₀₀¹⁰⁰ = 0²(1) + 2(1)(0)² + 5(0)²(0) - [(0)²(0) + 2(0)(0)² + 5(0)²(0)] = 0 - 0 = 0 J

W₂ = work done from (1,0,0) to (1,5,1)

W₂ = {z²x + 2xy² + 5y²z}₁₀₀¹⁵¹ = (1)²(1) + 2(1)(5)² + 5(5)²(1) - [0²(1) + 2(1)(0)² + 5(0)²(0)] = 1 + 50 + 125 - 0 = 176 J

W₃ = work done from (1,5,1) to (0,5,1)

W₃ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁵¹ = 1²(0) + 2(0)(5)² + 5(5)²(1) - [(1)²(1) + 2(1)(5)² + 5(5)²(1)] = 125 - (1 + 50 + 125) = 125 - 176 = -51 J

W₄ = work done from (0,5,1) to (0,0,0)

W₄ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁰⁰ = (0)²(0) + 2(0)(0)² + 5(0)²(0) - [1²(0) + 2(0)(5)² + 5(5)²(1)] = 0 - 125 = -125 J

The total work done W is thus

W = W₁ + W₂ + W₃ + W₄

W = 0 J + 176 J - 51 J - 125 J

W = 176 J - 176 J

W = 0 J

The total work done equals 0 J

4 0

3 years ago

A large in-falling fragment could be tracked using radar. Explain how distance, speed, and the direction of motion, of the fragm

ch4aika [34]

Answer:

speed, distance and direction of motion of the object can be determined by analyzing the radio wave.

Explanation:

We know that radar operates by transmitting radio waves to a destination and these waves are comes back to the receiver station. By Considering this transmission and receiver process, we can measure the distance, velocity and path of an object's movement.

Distance can be assessed by taking following consideration, the velocity of the waves is V. It can help to assess the time made for the waves to be emitted by the radar and felt by the receiver, let the time be t.

Therefore distance can be determine as D= v*t/2,

here 2 signifies that the distance travelled by the wave in either direction ( from transmitter to receiver and vice verse)

Using the source wave frequency, speed can be computed. In a specific frequency, the radar starts sending out the frequencies and the reflected wave will have a distinct frequency. The velocity can be determine by

$v= (\Delta f/f)(c/2),$

where $\Delta f$ is the change in frequency and

c is the speed of light (the wave).

Direction can be determine by applying above principle. change in frequency is used to determine the direction in the following way:

When the frequency transition is very low, the object moves away from the radar and vice verse.

3 0

3 years ago