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3 years ago

Which class of lever do spoon and scissors belong to?

Physics

2 answers:

balandron [24]3 years ago

8 0

In a Class One Lever, the Fulcrum is located between the Load and the Force.

Goshia [24]3 years ago

7 0

Answer:

spoon and scissors belong to first class lever.

hope it is helpful to you

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3 years ago

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3 years ago

Two stationary positive point charges, charge 1 of magnitude 3.90 nC and charge 2 of magnitude 1.80 nC, are separated by a dista

soldi70 [24.7K]

Answer:

$v = 7793150 m/s$

Explanation:

First, we are going to calculate the electrical potential in the point middle between the two charges

Remember that the electrical potential can be calculated as:

$v = \frac{kQ}{r}$

Where $k = 8.9874 x 10^{9} \frac{Nm^{2} }{C^{2} }$

and it is satisfy the superposition principle, thus

$v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.23} + \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.23}$

$v = 222.73v$

The electrical potential at 10 cm from charge 1 is:

$v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.1} + \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.36}$

$v = 395.44 v$

Since the work - energy theorem, we have:

$q\Delta v = \frac{mv^{2} }{2}$

where q is the electron's charge and m is the electron's mass

Therefore:

$v = \sqrt{\frac{2q\Delta v}{m} }$

$v = 7793150 m/s$

6 0

3 years ago

A man pushes on a piano with mass 190 kgkg ; it slides at constant velocity down a ramp that is inclined at 18.0 ∘∘ above the ho

creativ13 [48]

Answer:

The magnitude of applied force,parallel to the incline is 575.38 N and parallel to the floor is 605 N.

Explanation:

Given:

Mass of the piano $(m)$ = 190 kg

Inclined angle $(\theta)$ = 18 degree

Considering gravity, $g$ = 9.8 $ms^-^2$

And

Using, $sin(18) =0.30$ and $cos(18)=0.95$

FBD diagram is attached with all the force acting on the floor and and the inclined.

We have to find the magnitude of forces,when the man pushes it parallel to the incline and to the floor.

When the man pushes it parallel to the incline.

Balancing the forces as $\sum F=0$ .

⇒ $F+mgsin(\theta) =0$

⇒ $F=-mgsin(\theta)$

⇒ Here it is negative as the force is acting downward.

⇒ Plugging the values of mass $(m)$ and angle $(\theta)$ .

⇒ $F=190\times 9.8\times sin(18)$

⇒ $F=575.38$ N

When the force is parallel to the floor.

⇒ $Fcos(\theta)=mgsin(\theta)$

⇒ $F=\frac{mgsin(\theta)}{cos(\theta)}$

⇒ Plugging the values.

⇒ $F=\frac{190\times 9.8\times sin(18)}{cos(18)}$

⇒ $F=605$ N

So,

The magnitude of applied force in inclined direction is 575.38 Newton and parallel to the floor is 605 N.

6 0

3 years ago

A model rocket accelerates at 15.3 m/s2 with a force of 44 N.

melamori03 [73]

To calculate the mass of the body moving, we use Newton's second law of motion which is F = ma where F is the force, m is the mass of the object and a is its acceleration.

F = ma
44 = m(15.3)
m = 2.9 kg

The mass of the rocket would be 2.9 kg.

8 0

3 years ago

Read 2 more answers

Which class of lever do spoon and scissors belong to?​

Which class of lever do spoon and scissors belong to?