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N76 [4]
3 years ago
10

The brakes on your automobile are capable of creating a deceleration of 5.0 m/s2. If you are going 135 km/h and suddenly see a s

tate trooper, what is the minimum time in which you can get your car under the 75 km/h speed limit
Physics
1 answer:
poizon [28]3 years ago
8 0

Answer:

3.33 seconds

Explanation:

We can use the velocity formula [ v = u + at ] to solve.

Find the value "u".

135km/h -> 135km*1000m/3600s -> 37.5m/s

Find the value "v".

75km/h -> 75km*1000m/3600s -> 20.83m/s

Keep in mind we are dealing with "deceleration" so when we input 5.0m/s into the formula, it will be a negative value.

Now, find "t" which is the value we aren't given with the values we're given in the question.

20.83 = 37.5 - 5t

-16.67 = -5t

3.33 = t

Best of luck!

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Which property describes the distance between similar points on a wave?
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B

Explanation:

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2 years ago
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A clay vase on a potter's wheel experiences an angular acceleration of 5.69 rad/s2 due to the application of a 16.0-n m net torq
Digiron [165]
The equivalent of the Newton's second law for rotational motions is:
\tau = I \alpha
where
\tau is the net torque acting on the object
I is its moment of inertia
\alpha is the angular acceleration of the object.

Re-arranging the formula, we get
I= \frac{\tau}{\alpha}
and since we know the net torque acting on the (vase+potter's wheel) system, \tau=16.0 Nm, and its angular acceleration, \alpha = 5.69 rad/s^2, we can calculate the moment of inertia of the system:
I= \frac{\tau}{\alpha}= \frac{16.0 Nm}{5.69 rad/s^2} =2.81 kg m^2
8 0
3 years ago
heat engine operating between 100 °C and 700 °C has an efficiency equal to 40% of the maximum theoretical efficiency. How much e
ipn [44]

Answer:

Explanation:

Theoretical efficiency = T₁ - T₂ / T₁ where T₁ and T₂ is absolute temperature of hot and cold end of the heat engine.

= 600 / (273 + 700 )

= 600 / 973

= .6166

operating efficiency = 40% of .6166

= .4 x .6166

= .2466 = 24.66 %

efficiency = work output / heat input

= 5000 / heat input = .2466

heat input = 5000 / .2466

= 20275.75 J .

HEAT EXTRACED = 20275.75 J.

3 0
3 years ago
In a device like the one shown below, the cylinder is allowed to fall a distance of 300 m. As a result, the temperature of the w
Delvig [45]

Answer:

Increase in the temperature of water would be 0.9 degree C

Explanation:

As we know by energy conservation

Change in the gravitational potential energy of the cylinder = increase in the thermal energy of the water

Here we know that the gravitational potential energy of the cylinder is given as

U = mgh

here we have

h = 300 m

now we can say

Mc\Delta T = (m \times 9.8 \times 300)

now if the cylinder falls from height h = 100 m

then we have

Mc\Delta T' = (m \times 9.8 \times 100)

now from above two equations

\frac{\Delta T'}{\Delta T} = \frac{100}{300}

\Delta T' = 2.7 \times \frac{1}{3} = 0.9 Degree C

8 0
3 years ago
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