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3 years ago

Whats quantum physics?

Physics

1 answer:

krek1111 [17]3 years ago

8 0

Answer:

How atoms work and how biology and chemistry work.

Explanation:

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Water is falling on the blades of a turbine at a rate of 100 kg/s from a certain spring. If the height of spring be 100m, then t

pashok25 [27]

Answer:

Natae Si Jordan Kaya Sya Napaihe

Explanation:

haha

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3 years ago

A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff

postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

$F_c=\frac{mv^2}{R}$

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

$N=mg$

Therefore, frictional force, $f=\mu mg$

Now, frictional force = centripetal force

$f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}$

Plug in the given values and solve for 'μ'. This gives,

$\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833$

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0

4 years ago

On his way off to college, Russell drags his suitcase 19 m from the door of his house to the car at a constant speed with a hori

Mashcka [7]

Answer:

The work done on the suitcase is, W = 1691 J

Explanation:

Given data,

The force on the suitcase is, F = 89 N

The distance Russell dragged the suitcase, S = 19 m

The work done on the suitcase by Russell is equal to the work done on the suitcase to overcome the friction

The work done on the suitcase by Russell is given by the formula

W = F · S

Substituting the given values,

W = 89 N x 19 m

W = 1691 J

Hence, the work done on the suitcase is, W = 1691 J

8 0

4 years ago

A steel ball and a piece of clay have equal mass. They are dropped from the same height on a horizontal steel platform. The ball

emmasim [6.3K]

Answer: The ball (option A)

Explanation: change in momentum is defined by the formulae m(v - u) where m = mass of object, v = final velocity and u = initial velocity.

For the ball, it hits the ground and bounces back with the same speed, that's final velocity equals initials (v = - u)

Change in momentum = m( -u- u) = m(-2u) = m(-2u) = -2mu

For the clay, it final velocity is zero since it sticks to the floor, hence (v =0)

m(v - u) = m(0 - u) = - mu.

-2mu (change in momentum from the ball) is greater than - mu ( change in momentum of clay)

6 0

3 years ago

How many neutrons are present in an atom of Iron-54

Flura [38]

there are 28 neutrons present

3 0

3 years ago