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lorasvet [3.4K]
3 years ago
10

The cable holding a 2125 kg elevator has a maximum strength of 21,750 N. What is the maximum upward acceleration the cable can g

ive the elevator without breaking
Physics
1 answer:
vivado [14]3 years ago
4 0

Answer:

10.23m/s^2

Explanation:

GIven data

mass of elevator = 2125 kg

Force= 21,750 N

Required

The maximum acceleration upward

F= ma

a= F/m

a=21,750/2125

a= 10.23m/s^2

Hence the acceleration is 10.23m/s^2

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On a hot day, the deck of a small ship reaches a temperature of 48
AlekseyPX

The final temperature of the seawater-deck system is 990°C.

<h3>What is heat?</h3>

The increment in temperature adds up the thermal energy into the object. This energy is Heat energy.

The deck of a small ship reaches a temperature Ti= 48.17°C seawater on the deck to cool it down. During the cooling, heat Q =3,710,000 J are transferred to the seawater from the deck. Specific heat of seawater= 3,930 J/kg°C.

Suppose for 1 kg of sea water, the heat transferred from the system is given by

3,710,000 = 1 x 3,930 x (T - 48.17)

T = 990°C  to the nearest tenth.

The final temperature of the seawater-deck system is 990°C.

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6 0
1 year ago
Plz help<br>I need it fast<br>​
Rufina [12.5K]

Answer:

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Explanation:

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7 0
3 years ago
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Consider two parallel plate capacitors. The plates on Capacitor B have half the area as the plates on Capacitor A, and the plate
vichka [17]

Answer:

CB = 4.45 x 10⁻⁹ F = 4.45 nF

Explanation:

The capacitance of a parallel plate capacitor is given by the following formula:

C = ε₀A/d

where,

C = Capacitance

ε₀ = Permeability of free space

A = Area of plates

d = Distance between plates

FOR CAPACITOR A:

C = CA = 17.8 nF = 17.8 x 10⁻⁹ F

A = A₁

d = d₁

Therefore,

CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F   ----------------- equation 1

FOR CAPACITOR B:

C = CB = ?

A = A₁/2

d = 2 d₁

Therefore,

CB = ε₀(A₁/2)/2d₁

CB = (1/4)(ε₀A₁/d₁)

using equation 1:

CB = (1/4)(17.8 X 10⁻⁹ F)

<u>CB = 4.45 x 10⁻⁹ F = 4.45 nF</u>

5 0
2 years ago
A ball is projected horizontally from the top of a bertical building 25.0m above the ground level with an initial velocity of 8.
kirill115 [55]

Answer:

Solution given:

height [H]=25m

initial velocity [u]=8.25m/s

g=9.8m/s

now;

a. How long is the ball in flight before striking the ground?

Time of flight =?

Now

Time of flight=\sqrt{\frac{2H}{g}}

substituting value

  • =\sqrt{\frac{2*25}{9.8}}
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<h3><u>the ball is in flight before striking the ground for 2.26seconds</u>.</h3>

b. How far from the building does the ball strike the ground?

<u>H</u><u>o</u><u>r</u><u>i</u><u>z</u><u>o</u><u>n</u><u>t</u><u>a</u><u>l</u><u> </u>range=?

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<h3><u>The ball strikes 18.63m far from building</u>. </h3>
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Because the tip of the moon's shadow ... the area of "totality" ... is never more than a couple hundred miles across, It never covers a single place for more than 7 minutes, and can never stay on the Earth's surface for more than a few hours altogether during one eclipse.

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3 years ago
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