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3 years ago

INK]: A car is moving at 200 km/hr east. What is its velocity?

Physics

1 answer:

Greeley [361]3 years ago

5 0

Answer:

200 km\h

or 0.621 mp\h its the same speed just different mesuarements

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It takes you 8.3 min to walk with an average velocity of 1.6 m/s to the north from the bus stop to the museum entrance. How far

Anarel [89]

solution:

1.6 m/s = 96 m/min (in other words, 1.6 m/s x 60 s/min)

96 m/min x 8.3 min = 796.8 m

$s=ut +\frac{1}{2}at^2\\there is no accleration mentioned so,\\s= uv\\8.3\times60=498(s)\\510\times1.6=816(m)$

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3 years ago

How do I convert 14.8 cm into MEters?

stellarik [79]

Multiply it by a fraction equal to ' 1 ', like this:

(14.8 cm) x (1 meter/100 cm) = 14.8/100 = 0.148 meter

6 0

3 years ago

Read 2 more answers

7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee

wlad13 [49]

Answer:

$N_{top}=9.8N\\N_{bottom}=33.4N$

b) $v_{min}=4.4m/s$

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is $a_{cp}=\frac{v^2}{R}$ . At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

$-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg$

Which means:

$N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N$

The limit for falling off would be $N_{top}=0$ , so the minimum speed would be:

$0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s$

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3 years ago

Help pls it’s urgent giving brainiest!!

Lorico [155]

Answer:

2nd no. positive.... 3rd no. negitive......

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2 years ago

What affect would using a 12V car battery have on the operation of your circuit? (Do not try this.) What would happen to the cur

k0ka [10]

Answer:

Incomplete question

This is the completed question

If the resistor in the circuit had a larger resistance then the current would be then have to be proportionally smaller. Because the batteries each give off 1.5 volts then the current would have to be the variable that would change. What affect would using a 12V car battery have on the operation of your circuit? (Do not try this.) What would happen to the current? What would happen to the resistor?

Explanation:

Using ohms law as our basis

Ohms law state that, the voltage in an ohmic conductor is directly proportional to the current

V∝I

Resistance is the constant of proportionality

Then

V=iR

Since we want a relationship between current and resistance.

then, I=V/R

So, current is inversely proportional to Resistance

as the current increase the resistance reduce and as the current reduces the resistance increases.

a. So, increasing the voltage from 1.5V to 12V increases the current In the circuit because voltage Is directly proportional to I.

From ohms law

V=iR

When v =1.5V

I=1.5/R

When V increase to 12V

I=12/R

I.e, it increases by a factor of 8. Eight times it's initial value

b. Now, the resistance in the circuit is the constant of proportionality and it doesn't change in a given circuit expect when using a variable resistoa r like rheostat.

6 0

3 years ago