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Alik [6]
3 years ago
13

How much force is needed to accelerate a 1,800kg car at rate of 1.5 m/s2?

Physics
1 answer:
topjm [15]3 years ago
7 0

<u>Given data:</u>

acceleration (a) = 1.5 m/s² ,

           mass (m) = 1800 Kg ,

        Determine F = ?

         <em>From Newtons II law</em>

                          F = m.a   N

                              = 1800× 1.5

                             = 2700 N

<em>2700 N force needed to accelerate the car</em>

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9. A sports car travels on a straight road at 22.0 km/h and increases its speed to 57.0 km/h in
Varvara68 [4.7K]

Answer:

4.4 M/s

Explanation:

6 0
3 years ago
A driver notices an upcoming speed limit change from 45 mi/h (20 m/s) to 25 mi/h (11 m/s). If she estimates
zloy xaker [14]

Answer:

-2.79 m/s²

Explanation:

Given:

v₀ = 20 m/s

v = 11 m/s

Δx = 50 m

Find: a

v² = v₀² + 2aΔx

(11 m/s)² = (20 m/s)² + 2a (50 m)

a = -2.79 m/s²

Round as needed.

8 0
3 years ago
A secret agent is locked in a room. He pushes against the door but cannot open it. Finally, he falls to the floor exhausted. Has
Vitek1552 [10]

Answer: There is not work done at the door because the door did not move.

Explanation:  Work is defined as the movement done by a force.

So if you move to apply a force F in an object and you move it a distance D, the work applied on the object is  

W = F*D

In this case, the secret agent pushes against the door, so there is a force, but the agent does not move the door, so D = 0, so there is no motion of the door, which implies that there is no work done at the door.

8 0
3 years ago
The air in a room has a pressure of 1 atm, a dry-bulb temperature of 24°C, and a wet-bulb temperature of 17°C. Using the psychro
Licemer1 [7]

Answer:

Given that

Dry-bulb temperature(T) =24°C

Wet-bulb temperature(Tw) = 17°C

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From the diagram

ω= specific humidity

Lets take these two lines   Dry-bulb temperature(T) line and Wet-bulb temperature(Tw) cut at point P

From chart at point P

a)

Specific humidity,ω = 0.00922 kg/kg

b)

The enthalpy ( h)

h=47.59 KJ/kg

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d)

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5 0
3 years ago
A 40-cmcm-long tube has a 40-cmcm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. A
Licemer1 [7]

Answer:

1070 Hz

Explanation:

First, I should point out there might be a typo in the question or the question has inconsistent values. If the tube is 40 cm long, standing waves cannot be produced at 42.5 cm and 58.5 cm lengths. I assume the length is more than the value in the question then. Under this assumption, we proceed as below:

The insert in the tube creates a closed pipe with one end open and the other closed. For a closed pipe, the difference between successive resonances is a half wavelength \frac{\lambda}{2}.

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f=\dfrac{v}{\lambda}

f=\dfrac{343}{0.32}=1070 \text{ Hz}

7 0
3 years ago
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