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Marizza181 [45]
3 years ago
14

Henry can lift a 200 N load 20 m up a ladder in 40 s. Ricardo can lift twice the load up one-half the distance in the same amoun

t of time. Which comparison between Ricardo and Henry is correct?
Physics
2 answers:
prohojiy [21]3 years ago
4 0
Henry will lift 200 N load 20 m up a ladder in 40 s.  While the Ricardo will take 400 N load in 80 seconds. So, For Henry to take 400 N load it will take him 80 seconds in two attempts. And,also, he will have to cover 40 m of distance. 
madreJ [45]3 years ago
4 0

Answer;

-Ricardo exerted a greater force, did the same amount of work, and had the same power output as Henry.

Explanation;

For Ricardo to lift twice the load lifted by Henry for the one-half the distance in the same amount of time, he will need more force as the load is heavier.

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A player kicks a soccer ball from ground level and sons at flying at an angle of 30° at a speed of 26MS how far did the ball tra
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Explanation:

60 meters is he answer for this question

7 0
4 years ago
A train traveling at 27.5 m/s accelerates to 42.4 m/s over 75.0 s. What is the displacement of the train in this time period
Sergio [31]

Answer:

2621.25 meters

Explanation:

First, write down what we are given.

Initial velocity = 27.5 m/s

Final velocity = 42.4 m/s

Time = 75 seconds

We need to look at the kinematic equations and determine which one will be best.  In this case, we need an equation with distance.  I am going to use v_{f}^{2} = v_{i}^{2} +2ad, but you can also use the other equation, x = v_{o}t+\frac{1}{2}at^{2}

We need to find acceleration.  To find it, we need to use the formula for acceleration: a = \frac{v_{f}-v_{i}}{t}.  Plugging in values, a = \frac{42.4-27.5}{75} = .199\ m/s^{2}

Next, plug in what we know into the kinematics equation and solve for distance.  42.4^{2} = 27.5^{2} + 2(.199)(d)\\d = 2621.25\ meters

7 0
3 years ago
A photographer uses his camera, whose lens has a 50 mm focal length, to focus on an object 2.5 m away. He then wants to take a p
Temka [501]

Answer:

0.004 m away from the film

Explanation:

u = Object distance

v = Image distance

f = Focal length = 50 mm

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{2.5}\\\Rightarrow \frac{1}{v}=\frac{98}{5} \\\Rightarrow v=\frac{5}{98}=0.051\ m

The image distance is 0.051 m

When u = 50 cm

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{0.5}\\\Rightarrow \frac{1}{v}=18\\\Rightarrow v=\frac{1}{18}=0.055\ m

The image distance is 0.055 m

The lens has moved 0.055-0.051 = 0.004 m away from the film

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3 years ago
A little girl holds an ice cream cone with an upward force of 2 N as she walks 5 m across a room. How
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Answer:

I dont know

Explanation:

8 0
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