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2 years ago

How long does it take a car traveling at 45km/h to travel 100.0 m?

Physics

2 answers:

givi [52]2 years ago

7 0

Answer:

8m/s

Explanation:

speed = 45km/h

Distance = 100m

we have to find time = ?

Formula for speed is = Distance/ Time

Here Distance is given in 'm' so we need to convert speed value in 'm'

So to convert km/h in m formula is divide 45km/h by 3.6 then the km/h value gets converted in m so now the value is 12.5 m/s

So now,

speed = Distance/Time

we have to find Time

Then,

Time = Distance/ Speed

= 100/12.5

= 8m/s

djyliett [7]2 years ago

6 0

Answer:

8 seconds

Explanation:

Since the carspeed is in km/h, we need equal units, so we will make 100.0m 0.1000km.

Then we need to find how long it takes the car to travel 0.1km

We can use the formula distance=speed * time and get

0.1=45 * time

Therefore we get .002222... hours

Multiplying this by 3600 (to get seconds, 60x60), we get 8 seconds

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Y_Kistochka [10]

Answer:

Force exerted = 25.41 kN

Explanation:

We have equation of motion

v² = u²+2as

u = 345 m/s, s = 8.9 cm = 0.089 m, v = 0 m/s

0² = 345²+2 x a x 0.089

a = -668679.78 m/s²

Force exerted = Mass x Acceleration

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3 years ago

A positive charge of 8.0 × 10-4 C is in an electric field that exerts a force of 3.5 × 10-4 N on it. What is the strength of the

Gennadij [26K]

Answer:

E = 0.437 N/C

Explanation:

Given that,

Charge, $q=8\times 10^{-4}\ C$

Electric force, $F=3.5\times 10^{-4}\ N$

Let the strength of the electric field is E. We know that, the electric force is given by :

F = qE

Where

E is the electric field strength

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So, the strength of the electric field is equal to 0.437 N/C.

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3 years ago

A skateboarder, starting from rest, rolls down a 12.8-m ramp. When she arrives at the bottom of the ramp her speed is 8.89 m/s.

melamori03 [73]

Answer:

a) a = 3.09 m/s²

b) aₓ = 2.60 m/s²

Explanation:

a) The magnitude of her acceleration can be calculated using the following equation:

$V_{f}^{2} = V_{0}^{2} + 2ad$

<u>Where</u>:

$V_{f}$ : is the final speed = 8.89 m/s

$V_{0}$ : is the initial speed = 0 (since she starts from rest)

a: is the acceleration

d: is the distance = 12.8 m

$a = \frac{V_{f}^{2}}{2d} = \frac{(8.89 m/s)^{2}}{2*12.8 m} = 3.09 m/s^{2}$

Therefore, the magnitude of her acceleration is 3.09 m/s².

b) The component of her acceleration that is parallel to the ground is given by:

$a_{x} = a*cos(\theta)$

<u>Where</u>:

θ: is the angle respect to the ground = 32.6 °

$a_{x} = 3.09 m/s^{2}*cos(32.6) = 2.60 m/s^{2}$

Hence, the component of her acceleration that is parallel to the ground is 2.60 m/s².

I hope it helps you!

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