1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
yawa3891 [41]
3 years ago
8

Two objects are thrown vertically upward, first one, and then, a bit later, the other. Is it (a) possible or (b) impossible that

both objects reach the same maximum height at the same instant of time?
Physics
1 answer:
Studentka2010 [4]3 years ago
5 0

Answer:

No, it is impossible

Explanation:

Kinematics equation:

Vf^{2} =Vo^{2} -2gy

if height is maximum:

y=H and Vf=0

so:

<h3>Vo^{2} =2gH</h3><h3>H=Vo^{2} /2g</h3>

Analysis: From the last equation we see that the maximum height depends ONLY on the initial speed. This means that if both objects reach the same maximum height, then they necessarily need to have the SAME initial velocity. If they have the same initial velocity and in order to reach the maximum height at the SAME time the only way is that they are released at the SAME TIME.

You might be interested in
A traveling wave on a string can be described by the equation : y = (5.26 ~\text{m}) \cdot \sin \big( (1.65 ~\frac{\text{rad}}{\
zloy xaker [14]

Answer:

  t = 1.77 s

Explanation:

The equation of a traveling wave is

       y = A sin [2π (x /λ -t /T)]

where A is the oscillation amplitude, λ the wavelength and T the period

the speed of the wave is constant and is given by

      v = λ f

Where the frequency and period are related

     f = 1 / T

we substitute

      v = λ / T

let's develop the initial equation

    y = A sin [(2π / λ) x - (2π / T) t +Ф]

where Ф is a phase constant given by the initial conditions

the equation given in the problem is

    y = 5.26 sin (1.65 x - 4.64 t + 1.33)

if we compare the terms of the two equations

 

         2π /λ = 1.65

          λ = 2π / 1.65

          λ = 3.81 m

         2π / T = 4.64

          T = 2π / 4.64

          T = 1.35 s

we seek the speed of the wave

           v = 3.81 / 1.35

           v = 2.82 m / s

           

Since this speed is constant, we use the uniformly moving ratios

          v = d / t

           t = d / v

           t = 5 / 2.82

           t = 1.77 s

3 0
3 years ago
An electron of mass 9.11x10^-31 kg has an initial speed of 2.40x10^5 m/s. It travels in a straight line, and its speed increases
egoroff_w [7]

Answer: A) Force = 3.841*10^-18 N.

B) force (f) is 4.30* 10^12 times greater than the weight (Fg).

Explanation: mass of electronic charge = 9.11*10^-31kg

v = final velocity = 6.80*10^5 m/s

u = initial velocity = 2.40 * 10^5 m/s

S= distance covered = 4.8cm = 0.048m

a = acceleration

Since the acceleration of the electron is assumed to be constant, newton's laws of motion are valid.

Thus, recall that

v² = u² + 2aS

(6.80*10^5)² = ( 2.40*10^5)² + 2*a( 0.048)

46.24 * 10^10 = 5.76 * 10^10 + 0.096a

46.24 *10^10 - 5.76* 10^10 = 0.096a

40.48* 10^10 = 0.096a

a = 40.48 * 10^10/0.096

a = 4.2167*10^12m/s².

Force = mass * acceleration

Force = 9.11*10^-31 * 4.2167*10^12

Force = 3.841*10^-18 N.

Weight =Fg= mg where g = acceleration due gravity = 9.8m/s²

Fg= 9.11*10^-31 * 9.8

Fg = 8.9278* 10^-30 N

By comparing the force and the weight, we have that

F/Fg = 3.841 * 10^-18/8.9278 * 10^-30 = 4.30* 10^12.

This implies that the force (f) is 4.30* 10^12 times greater than the weight (Fg).

7 0
3 years ago
A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.
kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

\frac{1}{p}+\frac{1}{q}=\frac{1}{f}

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm

And the positive sign means the image is upright.

6 0
3 years ago
If the primary coil of a transformer has 200 turns and is supplied with 120 v ac power, how many turns must the secondary coil h
Oduvanchick [21]
The ratio of the turns to the voltage should be equal
i.e: 200/120 = t/12
so the secondary coil should have 20 turns
5 0
3 years ago
Small rockets are used to make small adjustments in the speed of satellites. One such rocket has a thrust of 42 N. If it is fire
Over [174]

To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.

If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.

F = ma \rightarrow a = \frac{F}{m}

Replacing,

a =\frac{42N}{83000kg}

a =5.06*10^{-4}m/s^2

The total speed change

\Delta v = v_f -v_0 \rightarrow v_f =\text{Final velocity and } v_0 = \text{Initial velocity } we have that the value is 0.71m/s

If we know that acceleration is the change of speed in a fraction of time,

a= \frac{\Delta v}{t} \rightarrow t = \frac{\Delta v}{a}

We have that,

t= \frac{0.71m/s}{5.06*10^{-4}m/s^2 }

t = 1403.16s

Therefore the Rocket should be fired around to 1403.16s

7 0
3 years ago
Other questions:
  • Two angles are supplementary. The first angle measures 60° what is the measurement of the second angle?
    13·1 answer
  • A conductor is formed into a loop that encloses an area of 1.0 m2. The loop is orientedat a 30.0° angle with the xy-plane. A var
    7·1 answer
  • Who speaks the line "Lord, what fools these mortals be"?
    8·1 answer
  • Select all the facts about a motor
    6·1 answer
  • What is one common product use microwaves<br><br><br><br> dont have much timeee
    10·1 answer
  • What two types of softball games can you play?<br><br> I really need to know
    5·2 answers
  • A 4 kg object moving to the left collides with and sticks to a 3 kg object moving to the right. Which of the following is true o
    11·1 answer
  • A 64.0 cm long cord is vibrating in such a manner that it forms a standing wave with two antinodes. (The cord is fixed at both e
    10·1 answer
  • Long ding dong what it do?
    5·2 answers
  • A flight into space by a spacecraft where the spacecraft returns to Earth without achieving orbit is called a
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!