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2 years ago

What is the eaquation for acceleration ?

2 answers:

8 0

Acceleration is the rate of change of velocity of an object. where a is acceleration, v is the final velocity of the object, u is the initial velocity of the object and t is the time that has elapsed. This equation can be rearranged to give: v = u + at.

Jlenok [28]2 years ago

3 0

A = Δv / Δt = (vf - vi)/(tf - ti).

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Two trains A and B of length 400 m each are moving on two parallel tracks with

Vinvika [58]

Answer:

The initial distance between the trains is 1450 m. 

Explanation:

In the question two trains are of equal length 400 m and moves at a uniform speed of 72 km/h. train A is moving ahead of train B. If the train B has to overtake train A it should accelerate.

Train B’s acceleration is $1m/s^2$ and it accelerated for 50 seconds.

 $a=1 m/s^2$ 

t=50 s 

initial speed u=72km/h 

we have to convert this speed into m/s 

 $u=72 \times 5/18=20 m/s$ 

Distance covered in accelerating phase $S=ut+1/2 at^2$ 

 $=20 \times 50+1/2 \times 1 \times 50^2$ 

 $=1000+1250=2250 m$ 

If a train is just behind another, the distance covered by the train located behind during overtaking phase will be equal to the sum of the lengths of the trains.

Here length of train A+length of train $B=400+400=800 m$ 

Hence the initial distance between the trains = $2250-800=1450 m$ 

6 0

2 years ago

When is your weight is equal to mg?

Vinil7 [7]

Answer:

The weight of an object is defined as the force of gravity on the object and may be calculated as the mass times the acceleration of gravity, w = mg.

6 0

3 years ago

Read 2 more answers

The speed of a tennis ball that is served is 73.14 m/s. During a serve, the ball typically starts from rest and is in contact wi

Serga [27]

Answer: Third option

F = 250w

Explanation:

The impulse can be written as the product of force for the time interval in which it is applied.

$I = F (t_2-t_1)$

You can also write impulse I as the change of the linear momentum of the ball

$I = mv_2 -mv_1$

So:

$F (t_2-t_1) = mv_2 -mv_1$

We want to find the force applied to the ball. We know that

$(t_2-t_1) = 30$ milliseconds = 0.03 seconds

The initial velocity $v_1$ is zero.

The final speed $v_2 = 73.14\ m / s$

$F * 0.03 = 73.14m$

$F * 0.03 = 73.14m\\\\F=\frac{73.14m}{0.03}\\\\F=2438m$

We must express the result of the force in terms of the weight of the ball.

We divide the expression between the acceleration of gravity

$g = 9.8\ m / s ^ 2$

$F=\frac{2438m*g}{g},\ \ m*g=w\\\\g=9.8\ m/s^2\\\\F=\frac{2438w}{9.8}\\\\F=249w$

The answer is the third option

3 0

3 years ago

Calculate the magnitude of electric field strength at a point 3cm from an infinite line of charge of linear density 18 μC/cm sit

valina [46]

Answer:

The magnitude of the electric field strength = 7.2 x 10⁸ N/C

Explanation:

The linear density:

$\begin{gathered} \lambda=18\mu C\text{ /cm} \\ \\ \lambda=\frac{18*10^{-6}C}{0.01m} \\ \\ \lambda=0.0018\text{ C/m} \end{gathered}$

Point r = 3 cm = 3/100 m

r = 0.03 m

The electric field strength is calculated below

$\begin{gathered} E=\frac{\lambda}{2\pi\epsilon_o\epsilon_rr} \\ \\ E=\frac{0.0018}{2\times3.14\times8.85\times10^{-12}\times1.5\times0.03} \\ \\ E=719709237.468\text{ N/C} \\ \\ E=7.2\times10^8\text{ N/C} \\ \end{gathered}$

The magnitude of the electric field strength = 7.2 x 10⁸ N/C

8 0

11 months ago

A 2 kg block is on a horizontal surface. A horizontal force is applied by a person to the block to pull it on the surface with a

Agata [3.3K]

Answer:

work done = 500 J

option e is correct

Explanation:

given data

mass = 2 kg

acceleration = 2 m/s

coefficient of kinetic friction = 0.3

block D = 50 m

to find out

work

solution

we know that work done is

work done = f × d ...........1

here f is force and d is block i.e 50 m

so here

force , f - k mg =ma

f = 2 ( 2 + 0.3 ×10 )

f = 10 N

so from equation 1

work done = f × d

work done = 10 × 50

work done = 500 J

option e is correct

3 0

3 years ago