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iren2701 [21]
2 years ago
10

Two hockey pucks with mass 0.1 kg slide across the ice and collide. Before

Physics
1 answer:
svp [43]2 years ago
3 0

Answer:

Puck 1 = 12 m/s

Explanation:

Conservation of energy

KE_1a + KE_2a = KE_1b + KE_2b

(1/2)(0.1kg)(15m/s)^2 + (1/2)(0.1kg)(12m/s)^2 = (1/2)(0.1kg)(V)^2 + (1/2)(0.1kg)(15m/s)^2

V = 12m/s

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A 1500-W electric heater is plugged into the outlet of a 120-V circuit that has a 20-A circuit breaker. You plug an electric hai
Korolek [52]

Answer:

b) 900 W

Explanation:

The breaker trips when the current is equal to 20 A. The power (P) is the ddp (V) multiplied by the current. So, for the electric heater, the current is:

P = V*i

1500 = 120*i

i = 12.5 A

So, to become in 20 A, it's needed 7.5 A, which must come from the hairdryer. Its power must be:

P = 120*7.5

P = 900 W

8 0
3 years ago
A rope exerts a 280 N force while pulling an 80 Kg skier upward along a hill inclined at 12o. The rope pulls parallel to the hil
user100 [1]

Answer:

The speed of the skier after moving 100 m up the slope are of V= 25.23 m/s.

Explanation:

F= 280 N

m= 80 kg

α= 12º

μ= 0.15

d= 100m

g= 9,8 m/s²

N= m*g*sin(α)

N= 163 Newtons

Fr= μ * N

Fr= 24.45 Newtons

∑F= m*a

a= (280N - 24.5N) / 80kg

a= 3.19 m/s²

d= a * t² / 2

t=√(2*d/a)

t= 7.91 sec

V= a* t

V= 3.19 m/s² * 7.91 s

V= 25.23 m/s

4 0
2 years ago
A 91.0-kg hockey player is skating on ice at 5.50 m/s. another hockey player of equal mass, moving at 8.1 m/s in the
never [62]

The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2=  91.0-kg

(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.

(u₂) is the velocity before the collision of hockey player 2=?

a)

Momentum before the collision;

\rm  m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

1237.6 kg m/s² =(2 ×91.0-kg )V

V=6.8 m/sec.

Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

8 0
2 years ago
Suppose a straight 1.00-mm-diameter copper wire could just "float" horizontally in air because of the force due to the Earth’s m
insens350 [35]

To solve this problem it is necessary to apply the concepts related to the Force since Newton's second law, as well as the concept of Electromagnetic Force. The relationship of the two equations will allow us to find the magnetic field through the geometric relations of density and volume.

F_{mag}= BIL

Where,

B = Magnetic Field

I = Current

L = Length

<em>Note: F_{mag}  is a direct adaptation of the vector relation F=q \times V \times B</em>

From Newton's second law we know that the relation of Strength and weight is determined as

F_g = mg

Where,

m = Mass

g = Gravitational Acceleration

For there to be balance the two forces must be equal therefore

F_{mag} = F_g

BIL = mg

Our values are given as,

Diameter (d) = 1.0mm = 1*10^{-3}m

Radius (r) = \frac{d}{2} = \frac{1*10^{-3}}{2} = 0.5*10^{-3}m

Magnetic Field (B) = 5.0*10^{-5} T

From the relationship of density another way of expressing mass would be

\rho = \frac{m}{V} \rightarrow m = \rho V

At the same time the volume ratio for a cylinder (the shape of the wire) would be

V = \pi r^2 L \rightarrow L =Length, r= Radius

Replacing this two expression at our first equation we have that:

BIL = mg

BIL = ( \rho V)g

BIL = ( \rho \pi r^2 L)g

Re-arrange to find I

I = \frac{( \rho \pi r^2 L)g}{BL}

I = \frac{( \rho \pi r^2 )g}{B}

We have for definition that the Density of copper is 8.9*10^3 Kg/m^3, gravity acceleration is 9.8m/s^2 and the values of magnetic field (B) and the radius were previously given, then:

I = \frac{( (8.9*10^3 ) \pi (0.5*10^{-3})^2 )(9.8)}{5.0*10^{-5}}

I = 1370.05A

The current is too high to be transported which would make the case not feasible.

8 0
2 years ago
How do you do this?Or what are the formulus
LuckyWell [14K]
You got the formulas on the sheet on the top :) So just use those, exchanging v (as in velocity, expressed in m/s) and the d (in meters) and t (in seconds). Hope you will manage it.
4 0
3 years ago
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