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
uranmaximum [27]
3 years ago
7

You are driving your car and the traffic light ahead turns red.You apply the breaks for 3s and the velocity of the car decreases

to 4.5 m/s . If the cars deceleration has a magnitude of 2.7m/s^2,what is the car's displacement during this time?
Physics
1 answer:
hammer [34]3 years ago
7 0

Answer:

The car's displacement during this time is 25.65 meters.

Explanation:

Given that,

Final velocity of the car, v = 4.5 m/s

Deceleration of the car, a=-2.7\ m/s^2

Let u is the initial speed of the car. It is given by :

v=u+at

u=v-at

u=4.5-(-2.7)\times 3

u = 12.6 m/s

Let d is the car's displacement during this time. It can be calculated using second equation of motion as :

d=ut+\dfrac{1}{2}at^2

d=12.6\times 3+\dfrac{1}{2}\times (-2.7)\times 3^2

d = 25.65 meters

So, the car's displacement during this time is 25.65 meters. Hence, this is the required solution.                        

You might be interested in
2. Three blocks, A,B and C of mass 2kg. 3kg. 5kg respectively kept side by side with one another are accelerated at 2m/s2 across
gulaghasi [49]

Answer:

Total mass of combination = 2+3+5 = 10kg.

Acceleration produced = 2m/s^2

hence force =( total mass × acceleration)= (2×10)= 20 N.

Net force on 3kg block = acceleration × mass = (2 × 2 )= 4 N

applied force on 2 kg block = 20N

Force between 2 kg and 3 kg block = (20-4) = 16N. ans

Net force on 3 kg block = 3 × 2 =6N.

Applied force on 3 kg block due to 2 kg block = 16N.

hence, force between 3 kg and 5 kg block = (16-6) = 10N .

answers:-

(a) 20 N

(b) 16N

(c) 10 N

4 0
1 year ago
The Young’s modulus of nickel is Y = 2 × 1011 N/m2 . Its molar mass is Mmolar = 0.059 kg and its density is rho = 8900 kg/m3 . G
Charra [1.4K]

Answer:

Atomic Size and Mass:

convert given density to kg/m^3 = 8900kg/m^3 2) convert to moles/m^3 (kg/m^3 * mol/kg) = 150847 mol/m^3 (not rounding in my actual calculations) 3) convert to atoms/m^3 (6.022^23 atoms/mol) = 9.084e28 atoms/m^3 4) take the cube root to get the number of atoms per meter, = 4495309334 atoms/m 5) take the reciprocal to get the diameter of an atom, = 2.2245e-10 m/atom 6) find the mass of one atom (kg/mol * mol/atoms) = 9.7974e-26 kg/atom Young's Modulus: Y=(F/A)/(dL/L) 1) F=mg = (45kg)(9.8N/kg) = 441 N 2) A = (0.0018m)^2 = 3.5344e-6 m^2 3) dL = 0.0016m 4) L = 2.44m 5) Y = 1.834e11 N/m^2 Interatomic Spring Stiffness: Ks,i = dY 1) From above, diameter of one atom = 2.2245e-10 m 2) From above, Y = 1.834e11 N/m^2 3) Ks,i = 40.799 N/m (not rounding in my actual calculations) Speed of Sound: v = ωd 1) ω = √(Ks,i / m,a) 2) From above, Ks,i = 40.799 N/m 3) From above, m,a = 9.7974e-26 kg 4) ω=2.0406e13 N/m*kg 5) From above, d=2.2245e-10 m 6) v=ωd = 4539 m/s (not rounding in actual calculations) Time Elapsed: 1) length sound traveled = L+dL = 2.44166 m 2) From above, speed of sound = 4539 m/s 3) T = (L+dL)/v = 0.000537505 s

7 0
3 years ago
A force of 45 newtons is applied on an object, moving it 12 meters away in the same direction as the force. What is the magnitud
NARA [144]
<h2>Answer: 540 J</h2>

Explanation:

The Work W done by a Force F refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.  

Now, when the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:  

W=(F)(d) (1)  

In this case both (the force and the distance in the path) are parallel (this means they are in the same direction), so the work W performed is the product of the force exerted to push the box F=45N by the distance traveled d=12m.

Hence:  

W=(45N)(12m)   (2)

W=540Nm=540J

6 0
3 years ago
What principle explains why lifting heavy objects is easier using ramps?
Anna11 [10]
By using ramps you can easily push or pull the object up the ramp.
5 0
3 years ago
Read 2 more answers
Dimension equation of work
kkurt [141]

Answer:

Explanation:

Work

Other units Foot-pound, Erg

In SI base units 1 kg⋅m2⋅s−2

Derivations from other quantities W = F ⋅ s W = τ θ

Dimension M L2 T−2

Idk if this is what u are looking for but i hope this help.:)

3 0
3 years ago
Other questions:
  • An experiment is conducted in which red light is diffracted through a single slit. part a listed below are alterations made, one
    8·2 answers
  • A man has a mass of 110kg. What is his weight?
    8·1 answer
  • The student makes a drawing of a carbon atom. Which of these should the student show in the nucleus of the atom?
    15·1 answer
  • A cannon is mounted on a cart which sits on the ground, supported by frictionless wheels. The mass of the cannon and cart is 4.6
    13·2 answers
  • A car travels 2 hours at 45 miles/ hour. How far did it go?
    13·1 answer
  • you are standing on top of a 344 m tall building. A drone operated by your friend is headed straight down, at a speed of 37.0 m/
    5·1 answer
  • Which of the following examples does not illustrate Newton's first law of motion?
    6·1 answer
  • How much work are you doing if you push on a 40 N rock that won't move?
    6·1 answer
  • What is the purpose of doing the whole body stretch from head to toe?
    8·1 answer
  • The number of employees for a certain company has been decreasing each year by 4%. If the company currently has 650 employees an
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!