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3 years ago

What is the SI unit for speed and velocity?

Physics

2 answers:

rosijanka [135]3 years ago

5 0

Answer:

Meter per second

Explanation:

Velocity and speed use the same units when being measured. The SI unit of distance and displacement is the meter. The SI unit of time is the second. The SI unit of speed and velocity is the ratio — the meter per second .

Anarel [89]3 years ago

4 0

Answer:

the answer is meter per second

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Answer:

15.3 N

Explanation:

Force = mass x acceleration

= 2.55 kg x 6.00 m/s^2

= 15.3 kgm/s^2 = 15.3 N

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Sunlight has its maximum intensity at a wavelength of 4.83 x 10'm; wha energy does this correspond to in e?

Lera25 [3.4K]

Answer:

E = 2,575 eV

Explanation:

For this exercise we will use the Planck equation and the relationship of the speed of light with the frequency and wavelength

E = h f

c = λ f

Where the Planck constant has a value of 6.63 10⁻³⁴ J s

Let's replace

E = h c / λ

Let's calculate for wavelengths

λ = 4.83 10-7 m (blue)

E = 6.63 10⁻³⁴ 3 10⁸ / 4.83 10⁻⁷

E = 4.12 10-19 J

The transformation from J to eV is 1 eV = 1.6 10⁻¹⁹ J

E = 4.12 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

E = 2,575 eV

5 0

4 years ago

Imagine a particle that has three times the mass of the electron. All other quantities given above remain the same. What is the

melamori03 [73]

Answer:

The only parameter that changes is mass m

It is only necessary to calculate the ratio Eh/Ee

$m_{h}=3m_{e}\\E_{h}=\frac{3m_{e}v^{2}}{2}\\E_{e}=\frac{m_{e}v^{2}}{2}\\\frac{E{h}}{E{e}}=3$

The kinetic energy of the heavy paricle is three times the kinetic energy of an electron

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4 years ago

A car that weighs 1.0 x 10^4 N is initially moving at a speed of 38 km/h when the brakes are applied and the car is brought to a

hram777 [196]

Answer:

Part a) Force on car = 2833.84 Newtons

Part b) Time to stop the car = 3.8 seconds

Part c) Factor for stopping distance is 4.

Part d) Factor for stopping time is 1.

Explanation:

The deceleration produced when the car is brought to rest in 20 meters can be found by third equation of kinematics as

$v^2=u^2+2as$

where

v = final speed of the car ( = 0 in our case since the car stops)

u = initial speed of the car = 38 km/hr = $\frac{38\times 1000}{3600}=10.56m/s$

a = deceleration produces

s = distance in which the car stops

Applying the given values we get

$0^2=10.56^2+2\times a\times 20\\\\a=\frac{0-10.56^2}{2\times 20}\\\\\therefore a=-2.78m/s^2$

Now the force can be obtained using newton's second law as

$Force=\frac{Weight}{g}\times a$

Applying values we get

$Force=\frac{1.0\times 10^4}{9.81}\times -2.78\\\\\therefore F=-2833.84Newtons$

The negative direction indicates that the force is opposite to the motion of the object.

Part b)

The time required to stop the car can be found using the first equation of kinematics as

$v=u+at$ with symbols having the same meanings

Applying values we get

$0=10.56-2.78\times t\\\\\therefore t=\frac{10.56}{2.78}=3.8seconds$

Part c)

From the developed relation of stopping distance we can see that the for same force( Same acceleration) the stopping distance is proportional to the square of the initial speed thus doubling the initial speed increases the stopping distance 4 times.

Part d)

From the relation of stopping time and the initial speed we can see that the stopping distance is proportional initial speed thus if we double the initial speed the stopping time also doubles.

8 0

3 years ago