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3 years ago

An object is dropped near the earth's surface. At the end of 4.0 s, how fast is it falling in m/s? (Ignore air resistance.)

2 answers:

7 0

For free falling bodies, the final velocity may be calculated through the equation,
Vf = gt
Where g is the acceleration due to gravity (9.8 m/s²) and t is the time elapsed. Substituting the known values,
Vf = (9.8 m/s²) x (4 s) = 39.2 m/s
Therefore, the object's velocity is approximately 39.2 m/s.

DochEvi [55]3 years ago

4 0

The answer would be 39.2 m/s

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A 970-kg sports car collides into the rear end of a 2300-kg SUV stopped at a red light. The bumpers lock, the brakes are locked,

vlabodo [156]

Answer:

22.73 m/s or 81.72 kph

Explanation

We can find the combined mass of both cars as

970 kg + 2300 kg = 3270 kg.

Then the normal force of the cars can be calculated as

F(n)= mg

Where g is acceleration due to gravity 9.8m/s^2

3270 kg ×9.8 = 32046 kg*m/s^2.

coefficient of kinetic friction between tires and road to be 0.80 × F(n)

Then the frictional force can be calculated as

= (32046kg*m/s^2 × 0.80 )

= 25636.8 kg*m/s^2

We can now calculate the work done that was used stopping the cars as

Frictional force × distance

(25636.8 kg*m/s^2 ) × 2.9m= 74346.72kg*m^2/s^2

From kinetic energy formula, the combined velocity of the car can be determined

E=0.5 M V²

√(2E/M) = V

√(2*74346.72kg*m^2/s^2 / 3270 kg) = V

V= √ (45.472)

V=6.743293m/s

the momentum of both cars can be determined as

6.743293m/s * 3270 kg

= 22050.57kg*m/s

Now the final momentum of both cars must be equal to the the momentum of

the sports car just prior to the collision. Therefore, the speed of the sports car at impact.

=(22050.57 kg*m/s) / 970 kg = 22.73 m/s

We can convert that to km/h.

22.73 m/s * 3600 s/h / 1000 m/km = 81.72 kph

7 0

3 years ago

At what speed would a 3.00 x 10^4 kg airplane have to fly and with a momentum of 1.60 x 10^9 kg.m/s

Ymorist [56]

Answer:

5.3×10⁴ m/s

Explanation:

From the question,

Momentum = mass× velocity

M = mV................ Equation 1

Where M = momentum of the airplane, m = mass of the airplane, V = Velocity of the airplane

make V the subject of the equation

V = M/m.................. Equation 2

Given: M = 1.6×10⁹ Kg.m/s, m = 3.0×10⁴ kg

Substitute these values into equation 2

V = 1.6×10⁹/3.0×10⁴

V = 5.3×10⁴ m/s

3 0

3 years ago

When flying in an airplane, you are most likely in which layer of the atmosphere? mesosphere thermosphere stratosphere trosphere

Sergeeva-Olga [200]

Lower stratosphere, this is to avoid turbulence

6 0

3 years ago

Read 2 more answers

Mass of trolley (m):

Nookie1986 [14]

Answer:

Explanation:

4 0

3 years ago

Agri-Chem's contract with Enerco specified a maximum of 90,000 cu.ft. X 103 per day for its complexes. However, curtailments are

zvonat [6]

Answer:

Caustic soda would be least affected by a gas curtailment

Explanation:

Let,

X1= ammonia

X2= ammonium phosphate

X3= ammonium nitrate

X4 = Urea

X5= hydro-fluoric acid

X6= Chlorine

X7= Caustic soda

X8= Vinyl chloride monomer.

Agri chem’s current natural gas usage

= (1200 × 8 + 540 × 10 + 490 × 12 + …)

= 85,680,000 cu. ft. per day

When, the curtailment is 20%, availability is

= 0.8 X 85,680

= 68,554,000 cu. ft. per day

Therefore, the gas constant

= 8X1 + 10X2 + 12X3 + 12X4 + 7X5 + 18X6 + 20X7 + 14X8 ≤68,544

When, the curtailment is 40% availability is

= 0.6 X 85,680

= 51,408,000 cu. ft. per day

Constant = 8X1 + 10X2 + 12X3 + 12X4 + 7X5 + 18X6 + 20X7 + 14X8≤ 51,408

7 0

3 years ago