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
Ivahew [28]
2 years ago
14

If I have an object that starts from rest and reaches a velocity of 25m/s over a distance of 11m what is the acceleration of the

object
28.4m/s
28m/s
28.4m/s^2
28m/s^2
Physics
1 answer:
Elodia [21]2 years ago
3 0

Answer:

Here is something that may help you!!

Explanation:

I found it in a cite (not that I'm plagiarizing, or anything).

You might be interested in
A ramp is used to load furniture onto a moving truck. The person does 1240 J of work pushing
Len [333]

Answer:

The efficiency of the ramp is, Eff = 6.63 %

Explanation:

Given,

The work done by the person pushing the furniture up the ramp is, W₁ = 1240 J

The work done by the ramp is, W₀ = 822 J

The efficiency of the ramp is given by the formula,

                                 <em> Eff = ( W₀ / W₁ ) x 100%</em>

                                        = ( 822 / 12400 ) x 100%

                                        = 6.63 %

Hence, the efficiency of the ramp is, Eff = 6.63 %

6 0
3 years ago
Children are sled riding on a hill One little girl pulls her sled back up the hill and does 379.5 joules of work while pulling i
Serggg [28]

Answer:

2.2N

Explanation:

Given parameters:

Work done  = 379.5J

Height  = 173m

Unknown:

Amount of force exerted on the sled  = ?

Solution:

The amount of force she exerted on the sled is the same as her weight.

Work done is the force applied to move a body through a distance.

      Work done  = mgh

m is the mass

g is the acceleration due to gravity

h is the height

     mg  = weight;

        Work done  = weight x h

           379.5 = weight  x  173

           weight  = \frac{379.5}{173}   = 2.2N

4 0
2 years ago
Jet aircraft maintenance crews are required to wear protective earplugs. Members of a particular crew wear earplugs that reduce
Taya2010 [7]

Answer:

So the sound intensity level they would experience without the earplugs is 110.32dB.

Explanation:

Given data

Sound intensity by factor =215

Sound intensity level =87 dB

To find

Sound intensity level they would experience without the earplugs

Solution

First we need to find the new sound intensity level

So

I_{n}=215(10^{\frac{87}{10} } )\\I_{n}=1.08*10^{11})

The dB can be calculated as:

dB=10log(I_{n})\\

Substitute the given values

dB=10log(1.08*10^{11})\\dB=110.32dB

So the sound intensity level they would experience without the earplugs is 110.32dB.

7 0
3 years ago
A small but measurable current of 3.8 × 10-10 A exists in a copper wire whose diameter is 2.5 mm. The number of charge carriers
Karolina [17]

Answer:

a) 4.9*10^-6

b) 5.71*10^-15

Explanation:

Given

current, I = 3.8*10^-10A

Diameter, D = 2.5mm

n = 8.49*10^28

The equation for current density and speed drift is

J = I/A = (ne) Vd

A = πD²/4

A = π*0.0025²/4

A = π*6.25*10^-6/4

A = 4.9*10^-6

Now,

J = I/A

J = 3.8*10^-10/4.9*10^-6

J = 7.76*10^-5

Electron drift speed is

J = (ne) Vd

Vd = J/(ne)

Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)

Vd = 7.76*10^-5/1.3584*10^10

Vd = 5.71*10^-15

Therefore, the current density and speed drift are 4.9*10^-6

And 5.71*10^-15 respectively

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
Other questions:
  • Which characteristic accounts for the fact that red lights are used in dark rooms and DO NOT expose negatives during developing?
    9·2 answers
  • 19. The term, The Mad as a Hatter, began in 19th century Europe because hatmakers used mercury.
    14·1 answer
  • What is the most likely reason that horses and mountain goats have hooves?
    13·2 answers
  • Find the force on an object which has a mass of 20 kg and an acceleration of 10 m/s2.
    6·1 answer
  • What happens to an object at rest if there is no net force<br> on the object?
    13·1 answer
  • A student is holding a marble in his hand. What are the equal and opposite forces that are acting on it?
    10·1 answer
  • Which runner finished the 100 m race in the least amount of time?
    12·1 answer
  • a car decelerates at a rate of 2.0 m/s^2 and comes to a complete stop after traveling 25 m. What was the speed of the car right
    6·1 answer
  • when heated the temperature of a water sample increased from 15°C to 39°C. Is absorbed 41840 joules of heat. what is the mass of
    14·1 answer
  • Describe what happened. When was there more potential energy in the system?
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!