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lozanna [386]
3 years ago
12

A ball is dropped from a height of 20m. If its velocity increases uniformly at the rate 10m/s2 with what velocity and after what

time will it strike the ground?
Physics
2 answers:
belka [17]3 years ago
6 0

Answer: 2 seconds

Explanation:

adelina 88 [10]3 years ago
3 0

Since I've only memorized a couple of formulas for these things, I need to do the time first:

I'll use the formula  Distance = (1/2) (acceleration) (time)²

20 m = (1/2) (10 m/s²) (time)²

Time² = (20 m) / (5 m/s²)

Time² = 4 sec²

<em>Time = 2 seconds</em>

Now ...

-- When the ball hits the ground, it has been falling for 2 seconds.

-- Its acceleration has caused its speed to increase by 10 m/s every second.

-- So after 2 seconds, its speed has grown to (2 s) (10 m/s²) = <em>20 m/s</em> .

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An inductive argument guarantees its conclusion<br><br> True<br><br> False
lorasvet [3.4K]

Answer:

False

Explanation:

inductive arguments have some evidence but do not have full assurance of truth of the conclusion

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2 years ago
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You prepare tea with 0.250 kg of water at 85.0 ºC and let it cool down to room temperature (20.0 ºC) before drinking it. Essenti
vladimir2022 [97]

Answer:

232 J/K

Explanation:

The amount of heat gained by the air = the amount of heat lost by the tea.

q_air = -q_tea

q = -mCΔT

q = -(0.250 kg) (4184 J/kg/ºC) (20.0ºC − 85.0ºC)

q = 68,000 J

The change in entropy is:

dS = dQ/T

Since the room temperature is constant (isothermal):

ΔS = ΔQ/T

Plug in values (remember to use absolute temperature):

ΔS = (68,000 J) / (293 K)

ΔS = 232 J/K

5 0
3 years ago
A 975-kg elevator accelerates upward at 0.754 m/s2, pulled by a cable of negligible mass. Find the tension force in the cable.
Zina [86]

To solve this problem we will apply the concepts of equilibrium and Newton's second law.

According to the description given, it is under constant ascending acceleration, and the balance of the forces corresponding to the tension of the rope and the weight of the elevator must be equal to said acceleration. So

\sum F = ma

T-mg = ma

Here,

T = Tension

m = Mass

g = Gravitational Acceleration

a = Acceleration (upward)

Rearranging to find T,

T = m(g+a)

T = (975)(9.8+0.754)

T= 10290.15N

Therefore the tension force in the cable is 10290.15N

7 0
3 years ago
A mass that weighs 10 lb stretches a spring 2.4 in. It rests in a liquid which imparts a dampening force of 4 pounds when the ve
forsale [732]

Answer:

The solution is attached in the pictures below

Explanation:

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3 years ago
The cornea behaves as a thin lens of focal length approximately 1.80 {\rm cm}, although this varies a bit. The material of which
spayn [35]

Answer:

The height of the image will be "1.16 mm".

Explanation:

The given values are:

Object distance, u = 25 cm

Focal distance, f = 1.8 cm

On applying the lens formula, we get

⇒  \frac{1}{v} -\frac{1}{u} =\frac{1}{f}

On putting estimate values, we get

⇒  \frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}

⇒  \frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}

⇒  v=1.94 \ cm

As a result, the image would be established mostly on right side and would be true even though v is positive.

By magnification,

m=\frac{v}{u} and m=\frac{h_{1}}{h_{0}}

⇒  \frac{v}{u} =\frac{h_{1}}{h_{0}}

⇒  \frac{1.94}{25}=\frac{{h_{1}}}{15}

⇒  {h_{1}}=1.16 \ mm

8 0
3 years ago
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