Acceleration units are in m/s^2 , so you take 25.5 m/s divided by 5.75s. Acceleration is the rate of change of velocity :)

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7 0

3 years ago

During a braking test, a car is brought to rest beginning from an initial speed of 60 mi/hr in a distance of 120 ft. With the sa

maw [93]

Answer:

Explanation:

Given

Initial speed $u=60\ mi/hr\approx 88\ ft/s$

distance traveled before coming to rest $d_1=120\ ft$

using equation of motion

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$0-(88)^2=2\times a\times 120---1$

for $u_2=80\ mi/hr\approx 117.33\ ft/s$

using same relation we get

$0-(117.33)^2=2\times a\times (d_2)----2$

divide 1 and 2 we get

$(\frac{88}{117.33})^2=\frac{120}{d_2}$

$d_2=213.32\ ft$

So a distance if 213.32 ft is required to stop the vehicle with 80 mph speed

8 0

3 years ago

There are 3.3V passing through an orange power supply cable, and there are 0.025 ohms of resistance in the orange wire. How much

PIT_PIT [208]

P = V^2 / R.

So, 3.3^2 / 0.025 = 435.6W.

Note, you can get the power equation from:
P = V*I. Also, I = V/R.
Substituting V/R in for I in the 1st equation, you get P = V^2 / R.

5 0

3 years ago

Which best describes a circuit in series?

Tpy6a [65]

A circuit which only has one path for current to follow

6 0

3 years ago