1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
never [62]
3 years ago
8

A person doing chin-up weighs 700.0 N, disregarding the weight of the arms. During the first 25.0 cm of the lift, each arm exert

s an upward force of 355 N on the torso. If the upward movement starts from rest, what is the person's speed at this point?

Physics
2 answers:
defon3 years ago
7 0
Please tell me if you have any trouble reading my writing. I tried to color code things to make it easier to understand.Let me know if you have any questions!

Answer: 0.26 m/s

dlinn [17]3 years ago
4 0
First, we will get the resultant force:
The direction of the force due to the person's weight is vertically down.
weight of person = 700 newton

Assume that the force exerted by the arms has a vertically upwards direction.
Force exerted by arms = 2*355 = 710 newtons

Therefore, the resultant force = 710 - 700 = 10 newtons (in the vertically upwards direction) 

Now, we will get the mass of the person.
weight = 700 newtons
weight = mass * acceleration due to gravity
700 = 9.8*mass
mass = 71.428 kg

Then we will calculate the acceleration of the resultant force:
Force = mass*acceleration
10 = 71.428*acceleration
acceleration = 0.14 m/sec^2

Finally, we will use the equation of motion to get the final speed of the person.
V^2 = U^2 + 2aS where:
V is the final velocity that we need to calculate
U is the initial velocity = 0 m/sec (person starts at rest)
a is the person's acceleration = 0.14 m/sec^2
S is the distance covered = 25 cm = 0.25 meters

Substitute with the givens in the above equation to get the final speed as follows:
V^2 = U^2 + 2aS
V^2 = (0)^2 + 2(0.14)(0.25)
V^2 = 0.07
V = 0.2645 m/sec

Based on the above calculations:
The person's speed at the given point is 0.2645 m/sec

You might be interested in
Tawny notices that Jim has been forgetting to check two forms of
juin [17]
I think the answer is c but I’m not sure
6 0
2 years ago
Convert the following in to SI units without changing its values<br> a. 4000 g <br>b. 2.4 km​
lana66690 [7]

hi <3

i believe i explained this answer properly in my last answer but it would be 4kg and 2400m as these are the SI units for these values.

hope this helps :)

3 0
3 years ago
Read 2 more answers
What is solar wind i ask that you help me
Novay_Z [31]
My answer -

the corona, the sun's outer layer, reaches temperatures of up to 2 million degrees Fahrenheit (1.1 million Celsius). At this level, the sun's gravity can't hold on to the rapidly moving particles, and it streams away from the star.

The sun's activity shifts over the course of its 11-year cycle, with sun spot numbers, radiation levels, and ejected material changing over time. These alterations affect the properties of the solar wind, including its magnetic field properties, velocity, temperature and density. The wind also differs based on where on the sun it comes from and how quickly that portion is rotating.

The velocity of the solar wind is higher over coronal holes, reaching speeds of up to 500 miles (800 kilometers) per second. The temperature and density over coronal holes are low, and the magnetic field is weak, so the field lines are open to space. These holes occur at the poles and low latitudes, and reach their largest when activity on the sun is at its minimum. Temperatures in the fast wind can reach up to 1 million degrees F (800,000 C).

At the coronal streamer belt around the equator, the solar wind travels more slowly, at around 200 miles (300 km) per second. Temperatures in the slow wind reach up to 2.9 million F (1.6 million C).

p.s

Glad to help you and if you need anything else on brainly let me know so I can elp you again have an AWESOME!!! :^)
4 0
3 years ago
The gravitational acceleration is 9.81 m/s2 here on Earth at sea level. What is the gravitational acceleration at a height of 35
azamat

To solve this problem it is necessary to apply the definition of severity of Newtonian laws in which it is specified that gravity is defined by

g= \frac{GM}{R^2}

Where

G= Gravitational Constant

M = Mass of Earth

R= Radius from center of the planet

According to the information we need to find the gravity 350km more than the radius of Earth, then

g_{ss} = \frac{GM}{R+h^2}

g_{ss} = \frac{6.67*10^{-11}*5.972*10^{24}}{(6371*10^3+350*10^3)^2}

g_{ss} = 8.82m/s^2

Therefore the gravitational acceleration at 350km is 8.82m/s^2

5 0
3 years ago
Give an example of free fall​
Elena-2011 [213]

Answer:

A stone that is dropped down into an empty well

3 0
3 years ago
Other questions:
  • Can magnets move other objects from a distance​
    7·1 answer
  • A fisherman's scale stretches 3.7 cm when a 3.5 kg fish hangs from it. (a) what is the spring constant? n/m (b) what will be the
    8·1 answer
  • What is the acceleration of an object with mass of 42.6 kg when an unbalanced force of 112 N is applied to it
    5·2 answers
  • How are force and mass related​
    5·1 answer
  • A swimmer swims 1000 m in the pool in 8.6 minutes. What was the average speed of the swimmer in m/s?
    11·2 answers
  • The length and width of a rectangle are 1.82 cm and 1.5 cm respectively. Calculate area of the rectangle and write in correct si
    10·1 answer
  • Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A pipe closed at both ends can have standing wav
    11·1 answer
  • Please help me......
    8·1 answer
  • Wave speed?_____<br> Frequency?_____<br> Wavelength?_____
    5·1 answer
  • The average lifespan of an incandescent lightbulb (at 60 W) is 1,200 hours. How much energy does the incandescent lightbulb use
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!