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Gnoma [55]
3 years ago
13

A train travels 600 km in 1 hour what i the traina velocity

Physics
2 answers:
bulgar [2K]3 years ago
6 0
The answer is 166.66m/s
oksian1 [2.3K]3 years ago
3 0

Answer:

v = 166,7 m/seg

Explanation:

To obtain the velocity of the train we must use the velocity formula for a uniform line movement:

v = x/t

Where x is the space and t is time.

replacing given values:

v = 600 km / 1 h

v = 600 km/h

to pass this value to international units:

v = 600 km/h * ( 1 h / 3600 seg) * ( 1000 m / 1 km)

v = 600 km/h * 0,28 h.m/seg.km

canceling the units we obtain:

v = 166,7 m/seg

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A hopper jumps straight up to a height of 1.3 m. With what velocity did he leave the floor
xxMikexx [17]

The velocity with which the jumper leaves the floor is 5.1 m/s.

<h3>What is the initial velocity of the jumper?</h3>

The initial velocity of the jumper or the velocity with which the jumper leaves the floor is calculated by applying the principle of conservation of energy as shown below.

Kinetic energy of the jumper at the floor = Potential energy of the jumper at the maximum height

¹/₂mv² = mgh

v² = 2gh

v = √2gh

where;

  • v is the initial velocity of the jumper on the floor
  • h is the maximum height reached by the jumper
  • g is acceleration due to gravity

v = √(2 x 9.8 x 1.3)

v = 5.1 m/s

Learn more about initial velocity here: brainly.com/question/19365526
#SPJ1

3 0
1 year ago
A double-slit experiment uses light of wavelength 650 nm with a slit separation of 0.100 mm and a screen placed 4.0 m away. a) W
dezoksy [38]

Answer:

Explanation:

a ) Slit separation d = .1 x 10⁻³ m

Screen distance D = 4 m

wave length of light  λ = 650 x 10⁻⁹ m

Width of central fringe = λ D / d

= \frac{650\times10^{-9}\times4}{.1\times10^{-3}}

= 26 mm

b ) Distance between 1 st and 2 nd bright fringe will be equal to width of dark fringe which will also be equal to 26 mm

c ) Angular separation between the central maximum and 1 st order maximum will be equal to angular width of fringe which is equal to

λ  / d

= \frac{650\times10^{-9}}{.1\times10^{-3}}

= 6.5 x 10⁻³ radian.

8 0
3 years ago
A 50.0 N box sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at on
Vikki [24]

Answer:

-4.0 N

Explanation:

Since the force of friction is the only force acting on the box, according to Newton's second law its magnitude must be equal to the product between mass (m) and acceleration (a):

F_f = ma (1)

We can find the mass of the box from its weight: in fact, since the weight is W = 50.0 N, its mass will be

m=\frac{W}{g}=\frac{50.0 N}{9.8 m/s^2}=5.1 kg

And we can fidn the acceleration by using the formula:

a=\frac{v-u}{t}

where

v = 0 is the final velocity

u = 1.75 m/s is the initial velocity

t = 2.25 s is the time the box needs to stop

Substituting, we find

a=\frac{0-1.75 m/s}{2.25 s}=-0.78 m/s^2

(the acceleration is negative since it is opposite to the motion, so it is a deceleration)

Therefore, substituting into eq.(1) we find the force of friction:

F_f = (5.1 kg)(-0.78 m/s^2)=-4.0 N

Where the negative sign means the direction of the force is opposite to the motion of the box.

6 0
3 years ago
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