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aleksandrvk [35]
3 years ago
14

An empty elevator has a mass of 722 kg. It moves between floors at a maximum speed of 6.00 m/s. The elevator is stopped on the 2

0th floor of the building when someone pushes the call button in the lobby (the first floor). Assuming that it takes 15.25 meters for the elevator to reach its maximum speed, and assuming constant acceleration, calculate the tension in the cable as the elevator car begins to descend (during the period of constant acceleration). Take the acceleration due to gravity to be 9.81 m/s2. Show your work.
Physics
1 answer:
FinnZ [79.3K]3 years ago
7 0

Answer:

Explanation:

The lift is going down with acceleration

Initial speed u = 0

Final speed v = 6 m/s

distance s = 15.25 m

acceleration a = ?

v² = u² + 2 a s

6² = 0 + 2 x a x 15.25

a = 1.18 m /s²

Elevator is going down with acceleration  .

mg - T = ma where T is tension in the cable .

722 x 9.8 - T = 722 x 1.18

7075.6 - T = 851.96

T = 6223.64 N .

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A 1.2 x10 3 kilogram automobile in motion strikes a 1.0 x 10 -4 kilogram insect as a result the insect is accelerated at a rate
Mariana [72]
According to Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
F=ma
where F is the magnitude of the force, m is the mass of the object and a its acceleration.

In this problem, the object is the insect, with mass m=1.0 \cdot 10^{-4} kg. The acceleration of the insect is a=1.0 \cdot 10^2 m/s^2, therefore we can calculate the force exerted by the car on the insect:
F=ma=(1.0 \cdot 10^{-4} kg)(1.0 \cdot 10^2 m/s^2)=0.01 N

How do we find the force exerted by the insect on the car?
According to Newton's third law (known as action-reaction law), when an object A exerts a force on an object B, object B also exerts a force equal and opposite on object A. Therefore, the force exerted by the insect on the car is equal to the force exerted by the car on the object, so it is 0.01 N.
6 0
3 years ago
You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 7 m at your feet, then a
Zolol [24]

Answer:

1. v = 6.67 m/s

2. d = 9.54 m

Explanation:

1. To find the horizontal velocity of the rock we need to use the following equation:

d = v*t \rightarrow v = \frac{d}{t}    

<u>Where</u>:

d: is the distance traveled by the rock

t: is the time

The time can be calculated as follows:

t = \sqrt{\frac{2d}{g}}

<u>Where:</u>

g: is gravity = 9.8 m/s²

t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*7 m}{9.8 m/s^{2}}} = 1.20 s

Now, the horizontal velocity of the rock is:

v = \frac{d}{t} = \frac{8 m}{1.20 s} = 6.67 m/s      

Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.          

2. To calculate the distance at which the projectile will land, first, we need to find the time:

t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*(7 m + 3 m)}{9.8 m/s^{2}}} = 1.43 s

So, the distance is:

d = v*t = 6.67 m/s*1.43 s = 9.54 m    

Therefore, the projectile will land at 9.54 m of the second cliff.

I hope it helps you!        

7 0
3 years ago
A professional golfer hits a golf ball of mass 46 g with her 5-iron, and the ball first strikes the ground 155 m away. The ball
RideAnS [48]

Answer:

C=2.32\times 10^{-4}\ Ns^2/m^2

Explanation:

It is given that,

Mass of the golf ball, m = 46 g = 0.046 kg

Terminal speed of the ball, v = 44 m/s

The drag force, F_r=Cv^2

Where, C is the drag coefficient. At terminal speed, the weight of the ball is balanced by the drag force.

Cv^2=mg

C=\dfrac{mg}{v^2}

C=\dfrac{0.046\times 9.8}{(44)^2}

C=2.32\times 10^{-4}\ Ns^2/m^2

Hence, this is the required solution.

4 0
3 years ago
What is the relationship between distance of the objects and the gravitational force between them? (Another way of asking: What
Nadya [2.5K]

Explanation:

increase in the force of gravity

6 0
3 years ago
A battery with an emf of 12.0 V shows a terminal voltage of 11.7 V when operating in a circuit with two lightbulbs, each rated a
wariber [46]
<h2>Answer:</h2>

0.46Ω

<h2>Explanation:</h2>

The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;

E = V + Ir                      --------------------(a)

Where;

I = current flowing through the circuit

But;

V = I x Rₓ                    ---------------------(b)

Where;

Rₓ = effective or total resistance in the circuit.

<em>First, let's calculate the effective resistance in the circuit:</em>

The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.

Let;

R₁ = resistance in the first bulb

R₂ = resistance in the second bulb

Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;

P = \frac{V^{2} }{R}

=> R = \frac{V^{2} }{P}             -------------------(ii)

Where;

P = Power of the bulb

V = voltage across the bulb

R = resistance of the bulb

To get R₁, equation (ii) can be written as;

R₁ = \frac{V^{2} }{P}    --------------------------------(iii)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ = \frac{12.0^{2} }{4}

R₁ = \frac{144}{4}

R₁ = 36Ω

Following the same approach, to get R₂, equation (ii) can be written as;

R₂ = \frac{V^{2} }{P}    --------------------------------(iv)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ = \frac{12.0^{2} }{4}

R₂ = \frac{144}{4}

R₂ = 36Ω

Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;

\frac{1}{R_{X} } = \frac{1}{R_1} + \frac{1}{R_2}       -----------------(v)

Substitute the values of R₁ and R₂ into equation (v) as follows;

\frac{1}{R_X} = \frac{1}{36} + \frac{1}{36}

\frac{1}{R_X} = \frac{2}{36}

Rₓ = \frac{36}{2}

Rₓ = 18Ω

The effective resistance (Rₓ) is therefore, 18Ω

<em>Now calculate the current I, flowing in the circuit:</em>

Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;

11.7 = I x 18

I = \frac{11.7}{18}

I = 0.65A

<em>Now calculate the battery's internal resistance:</em>

Substitute the values of E = 12.0, V = 11.7V and I = 0.65A  into equation (a) as follows;

12.0 = 11.7 + 0.65r

0.65r = 12.0 - 11.7

0.65r = 0.3

r = \frac{0.3}{0.65}

r = 0.46Ω

Therefore, the internal resistance of the battery is 0.46Ω

5 0
3 years ago
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