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avanturin [10]
4 years ago
13

A cheetah can run at a maximum speed

Physics
1 answer:
vivado [14]4 years ago
3 0

Answer:

25 seconds (rounded up to nearest one second)

Explanation:

Cheetah's speed = 91.8 km/h

Gazelle's speed = 77.9 km/h

Both animals are running at full speed with the gazelle 97.3 m ahead.

Converting their respective speeds to m/s ;

Cheetah's speed = \frac{91.8 * 1000}{3600} = 25.5 m/s

Gazelle's speed = \frac{77.9 * 1000}{3600}  = 21.6 m/s (correct to one decimal place)

Assuming the gazelle stops, cheetah will run at 25.5 m/s - 21.6 m/s = 3.9 m/s to catch its prey (gazelle).

Time = distance ÷ speed

The time cheetah will take to catch its prey = 97.3 m ÷ 3.9 m/s = 24.94871795 s = 25 seconds (rounded up to nearest one second)

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It is always a good idea to get some sense of the "size" of units. For example, the mass of an apple is about 100 g , whereas a
marta [7]

Answer and Explanation:

The estimation of the mass of the objects is as follows

Given that

The mass of an apple is about 100 g

Just like that, we do some estimation of different objects

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An adult male is approx 100 kg

A college physics textbook is around 1 kg

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4 0
4 years ago
If a 4 engine jet accelerates down a runway at 8.7 m/s^2, Suppose now that all 4 engins are operational on the jet from the prev
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5.22m/s^2

Explanation:

One of the first propulsion characteristics given in the example is that all engines are equal.

In this way if we have 4 engines running at the same time, it means that its capacity is 100%.

Under this premise, if 100% is found, the Jet is capable of reaching a speed of 8.7m / s ^ 2.

However, the question is, what would happen if 2.4 "Engines" now work.

To do this then we make a simple equivalence,

If 4 engines is the equivalent of 100%, when would it be 2.4 engines?

X = \frac{2.4 * 100\%} { 4 }= 60\%

In this way it would mean that the body could be driven to 60% of its total.

So

Speed_{Decreased} = 8.7 * 60\% = 5.22 \frac{m}{s^2}

3 0
3 years ago
Consider three capacitors C1, C2, and C3 and a battery. If
VLD [36.1K]

Answer:

Charge on C₁ = charge on all the three capacitors in series with it = 7.5 μC

Explanation:

Since the same voltage in the battery is used for the entire rundown,

From this information "only C₁ is connected to the battery, the charge on C₁ is 30.0 μC",

Q = C₁V = 30 μC

V = (30/C₁)

the series combination of C₂ and C₁ is connected across the battery, the charge on C₁ is 15.0 μC

The charge on both capacitors are the same and equal to 15 μC (because they are in series)

Q = (Ceq) V = 15 μC

(Ceq) = (15/V) μF

The voltage is still the same as in the first connection process

V = (30/C₁)

(Ceq) = (15/V) μF

(Ceq) = 15 ÷ (30/C₁)

(Ceq) = 15 × (C₁/30) = 0.5 C₁

(1/Ceq) = (2/C₁)

For series connection

(1/Ceq) = (1/C₁) + (1/C₂)

(2/C₁) = (1/C₁) + (1/C₂)

(2/C₁) - (1/C₁) = (1/C₂)

(1/C₁) = (1/C₂)

C₁ = C₂

C₂ = C₁

C₃, C₁, and the battery are connected in series, resulting in a charge on C₁ of 10.0 μC.

The charge on both capacitors are the same and equal to 10 μC (because they are in series)

Q = (Ceq) V = 10 μC

(Ceq) = (10/V) μF

The voltage is still the same as in the first connection process

V = (30/C₁)

(Ceq) = (10/V) μF

(Ceq) = 10 ÷ (30/C₁)

(Ceq) = 10 × (C₁/30) = 0.333 C₁

(1/Ceq) = (3/C₁)

For series connection

(1/Ceq) = (1/C₁) + (1/C₃)

(3/C₁) = (1/C₁) + (1/C₃)

(3/C₁) - (1/C₁) = (1/C₃)

(2/C₁) = (1/C₃)

C₁ = 2C₃

C₃ = (C₁/2)

C₁, C₂, and C₃ are connected in series with one another and

with the battery, what is the charge on C₁

The charge on C₁ is the same as the charge on all the capacitors and equal to Q,

Q = (Ceq) V

(1/Ceq) = (1/C₁) + (1/C₂) + (1/C₃)

Substituting for C₂ and C₃

C₂ = C₁ and C₃ = (C₁/2)

(1/C₂) = (1/C₁) and (1/C₃) = (2/C₁)

(1/Ceq) = (1/C₁) + (1/C₁) + (2/C₁)

(1/Ceq) = (4/C₁)

Ceq = (C₁/4)

Q = (Ceq) V = (C₁/4) V

But recall that V = (30/C₁) from the first connection

Q = (C₁/4) (30/C₁)

Q = (30/4) = 7.5 μC

Hope this helps!

6 0
3 years ago
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