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3 years ago

How do contact forces such as friction differ from gravitational and magnetic forces?

1 answer:

7 0

"Contact" means "touching". A contact force can't act on an object unless
they're touching. Friction is a contact force.

Gravitational forces pull masses together even if they're not touching.
Magnetic forces pull a piece of iron toward a magnet even if they're not touching.

Gravitational and magnetic forces are "non-contact" forces.

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Which of the following statements apply to electric charges?

Gre4nikov [31]

Answer:

The statement "If a positively charged rod is brought close to a positively charged object, the two objects will repel " applies to electric charges.

Explanation:

There are only two types of electric charges. Both having own magnitude but different charge.

1. Positive charge

2. Negative charge

Like charges repel each other and opposite charges always attract each other.

When a positively charged rod is brought close to a positively charged object, the rod and the object will repel.

6 0

3 years ago

What is the angular momentum of a 0.25 kg mass rotating on the end of a piece of

finlep [7]

L = r x p = rmv = mr²ω

L = 0.25 x 0.75² x 12.5 = 1.758

4 0

2 years ago

Which fundamental force has a small range and is always an attractive force?

erica [24]

Answer:

gravitational force is a fundamental force and also , it does have a small range and is always an attractive force.

Explanation:

5 0

3 years ago

An object is placed in front of a convex mirror with a radius of curvature of magnitude 10 cm. The mirror produces an image that

jek_recluse [69]

Answer:

u = - 20 cm

m = $\frac{1}{5}$

Given:

Radius of curvature, R = 10 cm

image distance, v = 4 cm

Solution:

Focal length of the convex mirror, f:

f = $\frac{R}{2} = \frac{10}{2} = 5 cm$

Using Lens' maker formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

Substitute the given values in the above formula:

$\frac{1}{5} = \frac{1}{u} + \frac{1}{4}$

$\frac{1}{u} = \frac{1}{5} - \frac{1}{4}$

u = - 20 cm

where

u = object distance

Now, magnification is the ratio of image distance to the object distance:

magnification, m = $\frac{|v|}{|u|}$

magnification, m = $\frac{|4|}{|-20|}$

m = $\frac{4}{20}$

m = $\frac{1}{5}$

4 0

3 years ago

The Ha line of the Balmer series is emitted in the transition from n-3 to n 2. Compute the wavelength of this line for H and 2H.

Citrus2011 [14]

Explanation:

According to Rydberg's formula, the wavelength of the balmer series is given by:

$\frac{1}{\lambda}=R(\frac{1}{2^2}-\frac{1}{3^2})$

R is Rydberg constant for an especific hydrogen-like atom, we may calculate R for hydrogen and deuterium atoms from:

$R=\frac{R_{\infty}}{(1+\frac{m_e}{M})}$

Here, $R_{\infty}$ is the "general" Rydberg constant, $m_e$ is electron's mass and M is the mass of the atom nucleus

For hydrogen, we have, $M=1.67*10^{-27}kg$ :

$R_H=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{1.67*10^{-27}kg})}\\R_H=1.09677*10^7m^{-1}$

Now, we calculate the wavelength for hydrogen:

$\frac{1}{\lambda}=R_H(\frac{1}{2^2}-\frac{1}{3^2})\\\lambda=[R_H(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.0967*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5646*10^{-7}m=656.46nm$

For deuterium, we have $M=2(1.67*10^{-27}kg)$ :

$R_D=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{2*1.67*10^{-27}kg})}\\R_D=1.09707*10^7m^{-1}\\\\\lambda=[R_D(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.09707*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5629*10^{-7}=656.29nm$

5 0

3 years ago