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BartSMP [9]
3 years ago
15

What is the car's average velocity (in m/s) in the interval between t = 1.0 s to t = 1.5 s?

Physics
1 answer:
natali 33 [55]3 years ago
4 0

Answer:

1.4 m/s

Explanation:

From the question given above, we obtained the following data:

Initial Displacement (d1) = 0.9 m

Final Displacement (d2) = 1.6 m

Initial time (t1) = 1.5 secs

Final time (t2) = 2 secs

Velocity (v) =..?

The velocity of an object can be defined as the rate of change of the displacement of the object with time. Mathematically, it can be expressed as follow:

Velocity = change of displacement /time

v = Δd / Δt

Thus, with the above formula, we can obtain the velocity of the car as follow:

Initial Displacement (d1) = 0.9 m

Final Displacement (d2) = 1.6 m

Change in displacement (Δd) = d2 – d1 = 1.6 – 0.9

= 0.7 m

Initial time (t1) = 1.5 secs

Final time (t2) = 2 secs

Change in time (Δt) = t2 – t1

= 2 – 1.5

= 0.5 s

Velocity (v) =..?

v = Δd / Δt

v = 0.7/0.5

v = 1.4 m/s

Therefore, the velocity of the car is 1.4 m/s

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Natasha_Volkova [10]

Answer:

See explanation

Explanation:

Given that the density of the unknown substance is 0.79 g/cm3 and is soluble in water, the possible substances it could be are;

i) t-butanol

ii) ethanol

iii) 2-propanol

iv) acetone

However, the actual identity of the unknown substance can be obtained by carrying out a boiling point test. The four substances listed above have different boiling points. Hence the boiling point of the unknown substance ultimately discloses its identity.

5 0
3 years ago
A college dorm room measures 14 ft wide by 13 ft long by 6 ft high. What is the air in it under normal conditions?
kirza4 [7]

Complete question:

A college dormitory room measures 14 ft wide by 13 ft long by 6 ft high. Weight density of air is 0.07 lbs/ft3. What is the weight of air in it under normal conditions?

Answer:

the weight of the air is 76.44 lbs

Explanation:

Given;

dimension of the dormitory, = 14 ft by 13 ft by 6 ft

density of the air, = 0.07 lbs/ft³

The volume of the air in the dormitory room = 14 ft x 13 ft x 6 ft

                                                                          = 1092 ft³

The weight of the air = density  x  volume

                                   = 0.07 lbs/ft³  x  1092 ft³

                                   = 76.44 lbs

Therefore, the weight of the air is 76.44 lbs

6 0
2 years ago
Con lắc lò xo có độ cứng k = 100N/m được gắn vật có khối lượng m=0.1kg, kéo vật ra khỏi vị trí cân bằng 1 đoạn 5cm rồi buông tay
Delvig [45]

Answer:

The maximum velocity is 1.58 m/s.

Explanation:

A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.

Spring constant, K = 100 N/m

mass, m = 0.1 kg

Amplitude, A = 5 cm = 0.05 m

Let the angular frequency is w.

w = \sqrt{K}{m}\\\\w = \sqrt{100}{0.1}\\\\w = 31.6 rad/s

The maximum velocity is

v_{max} = w A\\\\v_{max} = 31.6\times 0.05 = 1.58 m/s

8 0
3 years ago
The speed of sound in air is around 330 m/s. If a bat emits a single high-pitched ‘click’ of sound in a cave that is 25m wide, c
Bingel [31]

Answer:

0.15 s

Explanation:

From the question given above, the following data were obtained:

Speed of sound (v) = 330 m/s

Distance (x) = 25 m

Time (t) =?

The time taken for the echo of the sound to the bat can be obtained as follow:

v = 2x / t

330 = 2 × 25 / t

330 = 50 / t

Cross multiply

330 × t = 50

Divide both side by 330

t = 50 / 330

t = 0.15 s

Thus, it will take 0.15 s for the echo of the sound to the bat

4 0
3 years ago
Place the following structures in the appropriate category based on their location in a cell
Studentka2010 [4]

Answer:

Cell surface.

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Nucleus

  • Cellular DNA
  • Nucleolus
  • Chromosome.

Cytoplasm

  • Mitochondria
  • Golgi apparatus
  • Cytoskeleton.

3 0
3 years ago
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