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3 years ago

Explain the similarities between a car at rest and a car traveling at a constant velocity.

1 answer:

8 0

The similarity between a car at rest and a car moving at constant velocity is that the both cars have zero acceleration. the car at rest is not moving so its acceleration should also be zero. for the car at constant velocity, since the velocity of the car is not changing, and acceleration is change in velocity per time, that it why it is also zero

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Which statement about speed and/or velocity is true?

maw [93]

Answer: The right Answer is Velocity has both speed and direction.

Explanation:

i took the test

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3 years ago

Read 2 more answers

A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and a

Stels [109]

The power dissipated across a component can be calculated through the formula P=I^2xR

Substituting the values in we get P=(0.5)^2x10=2.5W

4 0

3 years ago

Does anyone know this one? Thanks

Inessa [10]

Answer:

$3.844\,*\,10^5$

So a=3.844 and b=5

Explanation:

Scientific notation requests to write a number using powers of ten as a factor accompanying a real number (a) between 1 and smaller than 10 that contains the digits to exactly represent the original number. So in this case, the number 384,400 can be written as:

$384,400=3.844 \,*\,100,000= 3.844 \,*\,10^5$

with a=3.844, and "5" as the exponent of ten (so b=5)

6 0

4 years ago

You paddle a conoe with a force of 325 N. You and the canoe have a combined mass of 250 kg. What is the acceleration of the cano

Brums [2.3K]

$f = ma$

Rearranging it, we get;

$a = \frac{f}{m}$
Where a is the acceleration, f is the force, and m is the mass

$a = \frac{325}{250}$
$a = 1.3 \frac{m}{ {s}^{2} }$

7 0

3 years ago

At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction

bogdanovich [222]

Answer:

magnitude of force on charge 2Q = $\frac{KQ^{2} }{I^{2} }$

Direction of force on charge = 61 ⁰

Explanation:

The magnitude on the force on the charge can be evaluated by finding the net force acting on the charge 2Q i.e x-component of the net force and the y-component of the net force

║F║ = $\sqrt{f_{x}^{2} + f_{y}^{2} }$ = after considering the forces coming from Q, 3Q and 4Q AND APPLYING COULOMBS LAW

magnitude of force acting on 2Q = $\frac{KQ^{2} }{I^{2} }$

The direction of the force on charge 2Q is calculated as

tan ∅ = $\frac{f_{y} }{f_{x} }$ = 1.8284

therefore ∅ = $tan^{-1}$ 1.8284

= 61⁰

3 0

3 years ago