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3 years ago

The picture is there. Please help

Physics

1 answer:

SashulF [63]3 years ago

5 0

Answer:

c. only magnetic

Explanation:

<h3>becuase they are both magnet is pulling the paper clip which is metal and if using magnetic force.</h3>

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A 13-cm-diameter cd has a mass of 25 g . part a what is the cd's moment of inertia for rotation about a perpendicular axis throu

faltersainse [42]

<span>Answer: For a disc, the moment of inertia about the perpendicular axis through the center is given by 0.5MR^2. where M is the mass of the disc and R is the radius of the disc. For the axis through the edge, use parallel axis theorem. I = I(axis through center of mass) + M(distance between the axes)^2 = 0.5MR^2 + MR^2 (since the axis through center of mass is the axis through the center) = 1.5 MR^2</span>

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2 years ago

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The difference in electronegativity values between H and O is 1.4. What type of bond will these elements form?

Gnoma [55]

Who what where when why how ?

6 0

3 years ago

A 6 N and a 10 N force act on an object. The moment arm of the 6 N force is 0.2 m. If the 10 N force produces five times the tor

Levart [38]

Answer:

The moment arm is 0.6 m

Explanation:

Given that,

First force $F_{1}=6\ N$

Second force $F_{2}=10\ N$

Distance r = 0.2 m

We need to calculate the moment arm

Using formula of torque

$\tau=Force\times lever\ arm$

So, Here,

$\tau_{2}=5 \tau_{1}$

We know that,

The torque is the product of the force and distance.

Put the value of torque in the equation

$F_{2}\times d_{2}=5\times F_{1}\times r_{1}$

$r_{2}=\dfrac{5\times F_{1}\times r_{1}}{F_{2}}$

Where, $F_{1}$ =First force

$F_{1}$ =First force

$F_{2}$ =Second force

$r_{1}$ = distance

Put the value into the formula

$r_{2}=\dfrac{5\times6\times0.2}{10}$

$r_{2}=0.6\ m$

Hence, The moment arm is 0.6 m

6 0

3 years ago

If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the in

wel

The question is incomplete! The complete question along with answer and explanation is provided below.

Question:

A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.

What is the change in the potential energy (in Joules) of the mass as it goes up the incline?

If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?

Given Information:

Mass = m = 0.5 kg

Horizontal distance = d = 40 cm = 0.4 m

Vertical distance = h = 7 cm = 0.07 m

Normal force = Fn = 1 N

Required Information:

Potential energy = PE = ?

Work done = W = ?

Answer:

Potential energy = 0.343 Joules

Work done = 0.39 N.m

Explanation:

The potential energy is given by

PE = mgh

where m is the mass of the object, h is the vertical distance and g is the gravitational acceleration.

PE = 0.5*9.8*0.07

PE = 0.343 Joules

As you can see in the attached image

sinθ = opposite/hypotenuse

sinθ = 0.07/0.4

θ = sin⁻¹(0.07/0.4)

θ = 10.078°

The horizontal component of the normal force is given by

Fx = Fncos(θ)

Fx = 1*cos(10.078)

Fx = 0.984 N

Work done is given by

W = Fxd

where d is the horizontal distance

W = 0.984*0.4

W = 0.39 N.m

3 0

2 years ago

The Nucleus of the Atom is in the center of the Atom, not in the outer rings & orbitals.

chubhunter [2.5K]

Answer:

true

Explanation:

this the nucleus is located at the centre and contains protons and neutrons

3 0

3 years ago