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3 years ago

Suppose an ice hockey puck strikes a hockey stick that lies flat on the ice and is free to move in any direction. Which

Physics

1 answer:

Rudik [331]3 years ago

5 0

Answer:

Linear momentum

Explanation:

The most likely conservation candidate is the linear momentum. The law of momentum conservation states that the sum of momenta before and after an (elastic or inelastic) collision will remain constant.

The kinetic energy is another possible, but less likely suspect. It is conserved in elastic collisions (i.e., those with no kinetic energy loss), but we are not told this collision is assumed elastic. In fact the real setup would be nowhere close to an elastic collision, as the stick lies on ice, which hasn't be zambonied for an entire period of rough skating, there's rough surface and the stick's shaft is also slightly stuck to the surface through frost. So when the puck hits the stick, a portion of its kinetic energy is spent to unstick the stick and get it moving. And so, kinetic energy is not conserved.

Angular momentum is not applicable with the puck-stick scenario.

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A 35 kg boy is riding a 65 kg go-cart. He pushes on the gas pedal, causing the cart to accelerate at 5 m/s2. Use the equation F

trasher [3.6K]

Answer:

$\boxed {\boxed {\sf 500 \ Newtons }}$

Explanation:

The equation for force is given:

$F=m*a$

First, we must find the total mass, which is the sum of the boy's mass and the go-cart's mass.

total mass= boy's mass + go cart's mass

The boy's mass is 35 kilograms and the go cart's is 65 kilograms.

total mass= 35 kg+ 65 kg=100 kg

Now we know the total mass and the acceleration.

$m= 100 \ kg \\a= 5 \ m/s^2$

Substitute the values into the formula.

$F=100 \ kg * 5 \ m/s^2$

Multiply.

$F= 500 \ kg*m/s^2$

1 kilograms meter per square second is equal to 1 Newton.
Our answer of 500 kg*m/s² is equal to 500 Newtons.

$F= 500 \ N$

The force exerted by the go cart engine is 500 Newtons.

4 0

3 years ago

Read 2 more answers

If 2.0 j of work Is done raising a 180 g Apple how far was it lifted

andrew11 [14]

The apple is lifted by 1.1 m

Explanation:

Given that, mgh = 2 J and mass of apple = 180g
Step 1: convert grams to kg as work is given in SI units.

180 g = 0.18 kg

Step 2: substitute in the formula for work done.

0.18 * 10 * h = 2

Step 3: calculate the value of h from this formula.

h = 2/1.8

h = 1.11 m

6 0

3 years ago

If the absolute temperature of a gas is 600 K, the temperature in degrees Celsius is: A. 705Â°C. B. 873Â°C. C. 273Â°C. D. 327Â°C

zlopas [31]

Answer:

D). $327 ^0 C$

Explanation:

As we know that temperature scale is linear so we will have

$\frac{^0C - 0}{100 - 0} = \frac{K - 273}{373 - 273}$

now we have

$\frac{^0 C - 0}{100} = \frac{K - 273}{100}$

so the relation between two scales is given as

$^0 C = K - 273$

now we know that in kelvin scale the absolute temperature is 600 K

so now we have

$T = 600 - 273 = 327 ^0 C$

so correct answer is

D). $327 ^0 C$

4 0

3 years ago

A 1,100 kg car is traveling initially 20 m/s when the brakes are applied. The brakes apply a constant force while bringing the c

Natalka [10]

Answer:

Work done = -220,000 Joules.

Explanation:

Given the following data;

Mass = 1100kg

Initial velocity = 20m/s

To find workdone, we would calculate the kinetic energy possessed by the car.

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

$K.E = \frac{1}{2}MV^{2}$

Where,

K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

Substituting into the equation, we have;

$K.E = \frac{1}{2}*1100*20^{2}$

$K.E = 550*400$

K.E = 220,000J

Therefore, the workdone to bring the car to rest would be -220,000 Joules because the braking force is working to oppose the motion of the car.

4 0

3 years ago

You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the oars) toward you, it moves a dis

tangare [24]

Answer:

56.86153 N

Explanation:

t =Time taken

F = Force

Power

$P=\frac{W}{t}\\\Rightarrow W=P\times t\\\Rightarrow W=77\times 0.96\\\Rightarrow W=73.92\ Joules$

Work done

$W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow F=\frac{73.92}{1.3}\\\Rightarrow F=56.86153\ N$

The magnitude of the force that is exerted on the handle is 56.86153 N

5 0

3 years ago