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2 years ago

A tight knot can be easily opened by using a long spanner. Give reason.

Physics

2 answers:

olga nikolaevna [1]2 years ago

5 0

Answer:

Torque

Explanation:

loosening loosening a boat requires a certain amount of torque which is the product of the force acting on the wrench and the moment arm, or distance from the center of rotation at which the force is applied. the longer the handle, the last force it takes to generate the same amount of torque

erma4kov [3.2K]2 years ago

4 0

Answer:

It is because the effort distance is greater than the load distance

Explanation:

As we know, Effort×effort distance = load × load distance

So when effort distance is increases,

The effort decreases

So when the spanner's handle is long

A tight knot can easily be opened by less effrot

I hope it helped

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An object is dropped from rest into a pit, and accelerates due to gravity at roughly 10 m/s2. It hits the ground in 5 seconds. A

vitfil [10]

Answer:

Second pit is 375 m deeper compared to first pit.

Explanation:

We have equation of motion s = ut + 0.5at²

First object hits the ground after 5 seconds,

Initial velocity, u = 0 m/s

Acceleration, a = 10 m/s²

Time, t = 5 s

Substituting,

s = ut + 0.5 at²

s = 0 x 5 + 0.5 x 10 x 5²

s = 125 m

Depth of pit 1 = 125 m

Second object hits the ground after 10 seconds,

Initial velocity, u = 0 m/s

Acceleration, a = 10 m/s²

Time, t = 10 s

Substituting,

s = ut + 0.5 at²

s = 0 x 10 + 0.5 x 10 x 10²

s = 500 m

Depth of pit 2 = 500 m

Difference in depths = 500 - 125 = 375 m

Second pit is 375 m deeper compared to first pit.

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1 year ago

Quinn is testing the motion of two projectiles x and y by shooting them from a sling shot. What can we say best describes the mo

Studentka2010 [4]

Explanation:

A projectile motion may be defined as that form of a motion that is experienced by an object or a particle which is projected near the surface of the Earth and the particle moves along the curved path subjected to gravity force only.

Thus a projectile motion is always acted upon by a constant acceleration due to gravity in the down ward direction.

In the context, Quinn shoots two particle x and y from his sling shot and he observes that both his projectiles travels in a parabola curve in the air. Both the object x and y touches the ground a distance apart from him which is known as the range and it depends upon the velocity of the projectile. Both the projectile reaches a maximum height and then drop on the ground in a parabola shape.

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1 year ago

Read 2 more answers

A plane initially traveling at 200 m/s due west experiences a 10 m/s head wind coming from the opposite direction. A). What will

Hitman42 [59]

Ok so it would be late and the relative velocity would be 190 m/s because 200 m/s - 10 m/s is 190 m/s. Hope this helps.

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2 years ago

You want to move a heavy box with mass 30.0 kg across a carpeted floor. You pull hard on one of the edges of the box at an angle

charle [14.2K]

Answer:

a= $5.54m/s^{2}$

Explanation:

The net force, $F_{net}$ of the box is expressed as a product of acceleration and mass hence

$F_{net}=ma$ where m is mass and a is acceleration

Making a the subject, a= $\frac {F_{net}}{m}$

From the attached sketch,

∑ $F_{net}=Fcos\theta-F_{f}$ where $F_{f}$ is frictional force and $\theta$ is horizontal angle

Substituting ∑ $F_{net}$ as $F_{net}$ in the equation where we made a the subject

a= $\frac {Fcos\theta-F_{f}}{m}$

Since we’re given the value of F as 240N, $F_{f}$ as 41.5N, $\theta$ as $30^{o}$ and mass m as 30kg

a= $\frac {240cos30-41.5}{30.0}=\frac {166.346}{30.0}=5.54m/s^{2}$

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2 years ago

If the loss of 3500 kcal is equal to a loss of 1.0 lb, how many days will it take charles to lose 5.0 lb

rusak2 [61]

We are missing an important piece of information needed to answer this question: the number of kcal Charles losses per day. However, we can come up with a general equation in which kcal/day is the only independent variable.

We know that it takes 3500 kcal to lose one pound. To lose 5 pounds, Charles needs to lose 5 x 3500 kcal = 17,500 kcal.

To find how many days it takes Charles to lose 17,500 kcal (5 pounds), we must divide that amount by the number of kcal Charles loses per day.
Here is the equation to calculate that number

Number of days= 17500 / (kcal per day)

If given calories, remember that 1000 calories = 1 kcal, and .001 kcal = 1 cal

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2 years ago