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3 years ago

Tim has a weight of 500N, which his half of his father's

Physics

1 answer:

Assoli18 [71]3 years ago

3 0

Answer:

3.75m on the left end

Explanation:

Given data

Tim's weight = 500N= 0.5kN

Tim's fathers weight= 500*2= 1000N= 1kN

Tim and the fathers weight = 500+1000= 1500N= 1.5kN

Tim's uncle weight= 1.5kN

Length of seesaw= 5m

let the distance Tim's uncle sits be x

We know that summation of clockwise moment equals the summation of anticlockwise moment

Taking moment about the center of the seesaw

1.5*2.5= 1*x

3.75= 1x

Divide boths sides by 1

x= 3.75/1

x= 3.75m

Hence Tim's uncle will sit 3.75m on the left end

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A small fan has blades that have a radius of 0.0600 m. When the fan is turned on, the tips of the blades have a tangential accel

zhuklara [117]

Answer:

α = 395 rad/s²

Explanation:

Main features of uniformly accelerated circular motion

A body performs a uniformly accelerated circular motion when its trajectory is a circle and its angular acceleration is constant (α = cte). In it the velocity vector is tangent at each point to the trajectory and, in addition, its magnitude varies uniformly.

There is tangential acceleration (at) and is constant.

at = α*R Formula (1)

where

α is the angular acceleration

R is the radius of the circular path

There is normal or centripetal acceleration that determines the change in direction of the velocity vector.

Data

R = 0.0600 m :blade radius

at = 23.7 m/s² : tangential acceleration of the blades

Angular acceleration of the blades (α)

We replace data in the formula (1)

at = α*R

23.7 = α*(0.06)

α = (23.7) / (0.06)

α = 395 rad/s²

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3 years ago

Two friends are having a conversation. Anna says a satellite in orbit is in freefall because the satellite keeps falling toward

Leno4ka [110]

Answer:

Anna is correct.

Explanation:

Anna is right as a satellite is in free fall because it keeps falling toward Earth, meaning that gravity is the only force acting on it. Tom is incorrect because an object does not have to accelerate at a certain speed to be considered to be in free fall. As long as gravity is the only force acting upon the object, it is considered to be in free fall.

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3 years ago

A space shuttle with a mass of approximately 7.08 E5 kg is sitting on the launch pad. What would be the weight of the space shut

Afina-wow [57]

The weight of the shuttle is $6.94\cdot 10^6 N$

Explanation:

The weight of an object on Earth is the gravitational force exerted by the Earth on the object.

The magnitude of the weight of an object is given by:

$W=mg$

where

m is the mass of the object

$g=9.8 m/s^2$ is the acceleration of gravity on Earth's surface

And the direction is downward (towards the Earth's centre).

For the shuttle in this problem, its mass is

$m=7.08\cdot 10^5 kg$

So, its weight is

$W=(7.08\cdot 10^5)(9.8)=6.94\cdot 10^6 N$

Note that while the mass of an object (m) does not change, its weight (W) changes according to the location, since the value of g can be different at different location (for example, on the Moon, the value of g is about 1/6 the value of g on the Earth).

Learn more about weight and forces:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

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6 0

3 years ago

Please someone help, I’m very confused and it’s due soon, thanks

Anit [1.1K]

Answer:

1 s
19.6 m
2 s
0.8 m/s^2
28 m/s
79 m/s
0.37 s
26 m/s
242 m/s
19,930 m

Explanation:

In physics, many of the relationships between speed, distance, and acceleration are tied up in the equations for potential and kinetic energy. For an object of mass M* at height h in a gravity field with acceleration g, the potential energy is

PE = Mgh

At velocity v, the kinetic energy of the object is ...

KE = 1/2Mv^2

When an object is dropped or launched from rest, the height and velocity are related by the fact that kinetic energy gets translated to potential energy, or vice versa. This gives rise to ...

PE = KE

Mgh = (1/2)Mv^2

The mass (M) can be factored out of this, so we have ...

2gh = v^2

This can be solved for height:

h = v^2/(2g) . . . . [eq1]

or for velocity:

v = √(2gh) . . . . [eq2]

When acceleration is constant, as assumed here, the velocity changes linearly (to/from 0). So, over the time of travel, the average velocity is half the final velocity. That is,

t = 2h/v

Depending on whether you start with h or with v, this resolves to two more equations:

t = 2(v^2/(2g))/v = v/g . . . . [eq3]

t = 2h/(√(2gh)) = √(4h^2/(2gh)) = √(2h/g) . . . . [eq4]

The last of these can be rearranged to give distance as a function of time:

h = gt^2/2 . . . . [eq5]

or acceleration as a function of time and distance:

g = 2h/t^2 . . . . [eq6]

These 6 equations can be used to solve the problems posed. Just "plug and chug." For problems in Earth's gravity, we use g=9.8 m/s^2. (You may want to keep these equations handy. Be aware of the assumptions they make.)

_____

* M is used for mass in these equations so as not to get confused with m, which is used for meters.

_____

1) Use [eq4]: t = √(2·6 m/(9.8 m/s^2)) ≈ 1.107 s ≈ 1 s

2) Use [eq5]: h = (9.8 m/s^2)(2 s)^2/2 = 19.6 m

3) Use [eq4]: t = √(25 m/(4.9 m/s^2)) ≈ 2.259 s ≈ 2 s

4) Use [eq6]: g = 2(10 m)/(5 s)^2 = 0.8 m/s^2

5) Use [eq2]: v = √(2·9.8 m/s^2·40 m) = 28 m/s

6) Use [eq2]: v = √(2·9.8 m/s^2·321 m) ≈ 79.32 m/s ≈ 79 m/s

7) Using equation [eq3], we will find the time until Tina reaches her maximum height. Her actual off-the-ground total time is double this value. Using [eq3]: t = v/g = (1.8 m/s)/(9.8 m/s^2) = 9/49 s. Tina is in the air for double this time:

2(9/49 s) ≈ 0.37 s

8) Use [eq2]: v = √(2·9.8 m/s^2·33.5 m) ≈ 25.624 m/s ≈ 26 m/s

9) Use [eq2]: v = √(2·9.8·3000) m/s ≈ 242.49 m/s ≈ 242 m/s

(Note: the terminal velocity in air is a lot lower than this for an object like a house.)

10) Use [eq1]: h = (625 m/s)^2/(2·9.8 m/s^2) ≈ 19,930 m

_____

<em>Additional comment</em>

Since all these questions make use of the same equation development, I have elected to answer them. Your questions are more likely to be answered if you restrict your posts to 3 or fewer questions each.

5 0

3 years ago

A cat is shot horizontally out of a cannon that is on top of a 125 meter tall cliff. If the cat lands 286 m away from the base o

Elden [556K]

The initial velocity of the cat is 56.6 m/s

Explanation:

The motion of the cat is a projectile motion, consisting of:

- a uniform horizontal motion with constant velocity

- a vertical accelerated motion with constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time it takes for the cat to reach the ground. We use the following suvat equation:

$s=ut+\frac{1}{2}at^2$