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Aneli [31]
3 years ago
11

A car is travelling at 27m/s and decelerates at a=5m/s2 for a distance of 10m. Calculate its final velocity. (Hint does decelera

tion imply that the acceleration is positive or negative?)
Physics
1 answer:
mylen [45]3 years ago
8 0

If deceleration = 5 m/s²

then acceleration = -5 m/s²

According to third law of motion,

= v² = 2as + u²

= v² = 2×-5×10 + 27²

= v² = -100 + 729

= v =√ 629

= 25.08 m/s

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2 years ago
A man ran a 5 mile race. The race looped around a city park and back
Tems11 [23]
10 miles common since
7 0
3 years ago
A long uniformly charged thread (linear charge density λ= 2.5 C/m) lies along the x axis in the figure.(Figure 1) A small charge
Kamila [148]

Answer 1) The electric field at distance r from the thread is radial and has magnitude  

E = λ / (2 π ε° r)  

The electric field from the point charge usually is observed to follow coulomb's law:  

E = Q / (4  π ε° r^{2})  

Now, adding the two field vectors:  

E_{thread}  =  {2.5 / (22 π ε° X 0.07 ) ; 0}  

Answer 2) E_{q}  = {2.3 / (4 2 π ε°) ( - 7/ (√(84); -12 / (√84))

Adding these two vectors will give the length which is magnitude of the combined field.  

The y-component / x-component gives the tangent of the angle with the positive x-axes.

Please refer the graph and the attachment for better understanding.

5 0
3 years ago
Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 5.0
e-lub [12.9K]

Answer:

f = 421.8 Hz

Explanation:

When she moved a distance of 1 m from mid point she observe first destructive interference due to two speakers

so we can say that path difference of sound due to two speakers will be equal to half of the wavelength

so path difference is given as

\Delta L = {3.5^2 + 12^2}^{0.5} - {1.5^2 + 12^2}^{0.5}

so it will be

\Delta L = 12.5 - 12.093

\Delta L = 0.4066

now we know that

\frac{\lambda}{2} = 0.4066

\lambda = 0.813

now frequency of sound is given as

f = \frac{v}{\lambda}

f = \frac{343}{0.813}

f = 421.8 Hz

4 0
3 years ago
a crate is being lifted into a truck. if it is moved with a 2470n force and 3650 j of work is done , then how far is the crate b
SVEN [57.7K]

Answer:

The crate was being lifted by a height of 1.48 meters.

Explanation:

In an attempt o move a crate;

Force applied = 2470 N

Work done by the force = 3650 J

We know that the work done is defined as the force used to move an object to a distance.

Given the Force used and the work done by that Force, we need to find out the distance the crate was lifted to.

Work done is defined as:

Work = Force*distance covered in the direction of the force

3650 = 2470*distance

distance = 3650/2470

distance = 1.48 meters

4 0
3 years ago
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