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3 years ago

What is the net force on a bright red mercedes convertible traveling along a straight road at a steady speed of 100 hm/h?

1 answer:

6 0

I don't know what " hm/h " means as a unit of speed. But if the bright red object is traveling in a straight line at a constant speed, then its acceleration is zero, and that means that the net force acting on it is zero. The same would be true of a car of any other color traveling with the same velocity.

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How quickly would the 50kg box accelerate if the person applied a 570N force?

belka [17]

Acceleration = force / mass = 570 / 50 = .... m/s^2

6 0

3 years ago

Can someone please help me out with this question thanks

IrinaK [193]

Answer:

1.98V

Explanation:

V =IR

V =Voltage

I =Current

R = Resistance $V =IR\\V = 100 (0.0198)\\V = 1.98 volts\\Voltage 1.98V$

8 0

3 years ago

Can someone solve this problem and explain to me how you got it

likoan [24]

Answer:

Explanation:

Coulomb's law states that the force of attraction or repulsion between any two charges is proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between the charges

⇒ $F\alpha\frac{q1*q2}{r^{2}}$

∴ $F=k\frac{q1*q2}{r^{2}}$

where $F$ is the force of attraction or repulsion

$k$ is Coulumb's constant= $9*10^{9}Nm^{2}C^{-2}$

$q1$ and $q2$ are the magnitude of the charges

$r$ is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question 7:

Given: $F=1.9*10^{-29}N$ , $q1=q2=1.6*10^{-19}C$

By coulomb's law,

$1.9*10^{-29}=9*10^{9}*\frac{1.6*10^{-19}*1.6*10^{-19}}{r^{2}}$

⇒ $r=3.5m$

Question 6:

Given: $q1=q2=-1.5*10^{-6}C$ , $r=0.28m$

By coulomb's law,

$F=9*10^{9}*\frac{1.5*10^{-6}*1.5*10^{-6}}{0.28^{2}}$

⇒ $F=0.26N$

3 0

3 years ago

What is the magnification of an astronomical telescope whose objective lens has a focal length of 71 cm and whose eyepiece has a

Rus_ich [418]

To solve this problem it is necessary to apply the concepts related to optical magnification (is the process of enlarging the apparent size, not physical size, of something.). Specifically the angular magnification of an optical telescope is given by

$M = -\frac{f_o}{f_e}$