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jekas [21]
2 years ago
11

A car starts from rest and travels for t1 seconds with a uniform acceleration a1. The driver then applies the brakes, causing a

uniform acceleration a2. If the brakes are applied for t2 seconds, determine the following. Answers are in terms of the variables a1, a2, t1, and t2. (a) How fast is the car going just before the beginning of the braking period?
Physics
1 answer:
Triss [41]2 years ago
5 0

Explanation:

It is given that,

Initially the car is at rest and travels for t₁ seconds with a uniform acceleration a₁. The driver then applies the brakes, causing a uniform acceleration a₂, If the brakes are applied for t₂ seconds.

We need to find the speed of the car just before the beginning of the braking period.

Using the formula of acceleration. It is given by :

a_1=\dfrac{v-u}{t_1}

u = 0

v=a_1\times t_1

So, just before the beginning of the braking period the speed of the car is a_1\times t_1. Hence, this is the required solution.

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A student pushes against a wall with a force of 30N. The wall does not move. What amount of force does the wall exert on the stu
True [87]

Answer:

C

Explanation:

they both have to be the same for both to not move

8 0
2 years ago
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How can you say that joule and Newton are derived units​
maxonik [38]

Answer:

F = M a    where M is acceleration and a is acceleration

a = x / s^2 = distance / time squared

The Newton is derived because mass, distance, and time are all fundamental units One would have to look at the fundamental requirements for these definitions, but they can all be repeated in a laboratory.

So the Newton is determined from these fundamental units and since the Joule equals Newton * Distance it is also derived from the fundamental units.

If one has the three fundamental units then one can derive the Joule and Newton.

3 0
2 years ago
Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.400mm wide. The diffraction pattern is observed
kogti [31]

Answer:

a)y_{first}=5.3mm

b)y_{second}=10.6-5.3 =5.3 mm  

Explanation:

a)

The width of the central bright in this diffraction pattern is given by:

y=\frac{m\lambda D}{a} when m is a natural number.

here:

  • m is 1 (to find the central bright fringe)                
  • D is the distance from the slit to the screen
  • a is the slit wide
  • λ is the wavelength

So we have:

y_{first}=\frac{633*10^{9}*3.35}{0.0004}

y_{first}=5.3mm

b)

Now, if we do m=2 we can find the distance to the second minima.

y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}

y_{2}=10.6 mm

Now we need to subtract these distance, to get the width of the first bright fringe :

y_{second}=10.6-5.3 =5.3 mm    

I hope it heps you!

     

4 0
2 years ago
Use the periodic table from the lesson to answer the following question. The atomic mass of F is ___. 9 10 19 55
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18.9 but round it and you get 19

7 0
3 years ago
Read 2 more answers
On a horizontal frictionless floor, a worker of weight 0.900 kN pushes horizontally with a force of 0.200 kN on a box weighing 1
Ad libitum [116K]

Answer:

D) The worker will accelerate at 2.17  m/s²  and the box will accelerate at 1.08  m/s² , but in opposite directions.

Explanation:

Newton's third law

Newton's third law or principle of action and reaction states that when two interaction bodies appear equal forces and opposite directions. in each of them.

F₁₂= -F₂₁

F₁₂: Force of the box on the worker

F₂₁: Force of the worker on the box

Newton's second law

∑F = m*a

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Formula to calculate the mass (m)

m =  W/g

Where:

W : Weight (N)

g : acceleration due to gravity  (m/s²)

Data

W₁ =1.8 kN   : box weight

W₂ = 0.900 kN : worker weight

F₂₁ = 0.200 kN

F₁₂ = - 0.200 kN

g = 9.8 m/s²

Newton's second law for the box

∑F = m*a

F₂₁ = m₁*a₁    m₁=W₁/g

0.2 kN = (1.8kN)/(9.8 m/s² ) *a₁

a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{1.8 kN}

a₁= 1.08 m/s² : acceleration of the box

Newton's second law for the worker

∑F = m*a

F₁₂ = m₂*a₂ , m₂=W₂/g

- 0.2 kN =( (0.9 kN) /(9.8 m/s² ) )*a₂

a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{0.9 kN}

a₂=  -2.17 m/s² : acceleration of the worker

5 0
3 years ago
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