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Sergio [31]
3 years ago
15

A car with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. if a car is init

ially traveling at 55 mi/h part a how much time does it take the car to stop?
Physics
1 answer:
Sladkaya [172]3 years ago
8 0

The time required by the car to stop is 4.916 sec.

Since the car is moving with the constant deceleration we can apply the first equation of motion to calculate the time required by the car to stop.

The first equation of motion is given as

V=u+at

Here, V=final speed of the car=0 mi/h as the car stops

u =initial speed of the car=55 mi/hr=24.58 m/s

a= acceleartion =-5 m/s^2 (here negative sign indicates for deceleration)

Now applying the values in the first equation

V=u+at

0=24.58-5*t

t=4.916 sec

Therefore the car will stops in 4.916 sec.

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The answer is

<em>All the three light bulbs will glow</em>

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The driver of a train moving at 23m/s applies the breaks when it pases an amber signal. The next signal is 1km down the track an
pochemuha

Answer:

3.2 m/s

Explanation:

Given:

Δx = 1000 m

v₀ = 23 m/s

a = -0.26 m/s²

t = 76 s

Find: v

This problem is over-defined.  We only need 3 pieces of information, and we're given 4.  There are several equations we can use.  For example:

v = at + v₀

v = (-0.26 m/s²) (76 s) + (23 m/s)

v = 3.2 m/s

Or:

Δx = ½ (v + v₀) t

(1000 m) = ½ (v + 23 m/s) (76 s)

v = 3.3 m/s

Or:

v² = v₀² + 2aΔx

v² = (23 m/s)² + 2(-0.26 m/s²)(1000 m)

v = 3.0 m/s

Or:

Δx = vt − ½ at²

(1000 m) = v (76 s) − ½ (-0.26 m/s²) (76 s)²

v = 3.3 m/s

As you can see, you get slightly different answers depending on which variables you use.  Since 1000 m has 1 significant figure, compared to the other variables which have 2 significant figures, I recommend using the first equation.

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4 years ago
How many millimeters are there in one centimeter
Anika [276]
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3 years ago
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What kind of lenses are in a light microscope
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4 0
4 years ago
a 2.6 kg block is attached to a horizontal spring that has a spring constant of 126 N/m. At the instant when the displacement of
NeTakaya

Answer:

The acceleration of the object is 5.57\ m/s^2.

Explanation:

Given that,

Mass of the block, m = 2.6 kg

Spring constant of the spring, k = 126 N/m

At the instant when the displacement of the spring from its unstained length is 0.115 m. We need to find the acceleration of the object.

When the block is displaced, the force acting on the spring is equal to the force due to its motion. Such as :

kx=ma

a is acceleration of the object

a=\dfrac{kx}{m}\\\\a=\dfrac{126\times 0.115}{2.6}\\\\a=5.57\ m/s^2

So, the acceleration of the object is 5.57\ m/s^2.

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