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2 years ago

What is the momentum of a 50-kilogram ice skater gliding across the ice at a speed of 5 m/s? (1 point)

Physics

2 answers:

LenaWriter [7]2 years ago

4 0

P=mv
p=50kgx5ms-1
p=250

serious [3.7K]2 years ago

3 0

Momentum is a term used to quantify the motion of an object has. It is calculated as the the product of the object's mass and the velocity. It is expressed as:

Momentum = m x v
Momentum = 50 kg x 5 m/s
Momentum = 250 kg m/s

Therefore, the correct answer is the last option.

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The electric potential in a particular region of space varies only as a function of y-position and is given by the function V(y)

nikdorinn [45]

Answer:

$E = 55.9583\ Volts/meter$

Explanation:

First let's find the electric potential using y = 22.5:

$V(y) = 1.69y^2 +15.6y+52.5$

$V(22.5) = 1.69(22.5)^2 + 15.6*22.5 + 52.5$

$V(22.5) = 1259.0625\ Volts$

Then, to find the magnitude of the electric field, we just need to divide the electric potential by the distance y:

$E = V/d$

$E = 1259.0625/22.5$

$E = 55.9583\ Volts/meter$

3 0

2 years ago

An air-track cart with mass m1=0.28kg and initial speed v0=0.75m/s collides with and sticks to a second cart that is at rest ini

arsen [322]

Kinetic energy is calculated through the equation,

KE = 0.5mv²

At initial conditions,

m₁: KE = 0.5(0.28 kg)(0.75 m/s)² = 0.07875 J

m₂ : KE = 0.5(0.45 kg)(0 m/s)² = 0 J

Due to the momentum balance,

m₁v₁ + m₂v₂ = (m₁ + m₂)(V)

Substituting the known values,

(0.29 kg)(0.75 m/s) + (0.43 kg)(0 m/s) = (0.28 kg + 0.43 kg)(V)

V = 0.2977 m/s

The kinetic energy is,
KE = (0.5)(0.28 kg + 0.43 kg)(0.2977 m/s)²
KE = 0.03146 J

The difference between the kinetic energies is 0.0473 J.

7 0

2 years ago

The air surrounding an airplane in flight exerts a drag force that acts opposite to the airplane's motion. When an Airbus A380 i

Rainbow [258]

Answer:

$4.32\cdot 10^5 hp$ , $3.22\cdot 10^8 W$

Explanation:

The jet is flying at constant velocity: this means that its acceleration is zero, so the net force acting on the jet is also zero.

Therefore, we can write:

$4F_T - F_d = 0$

where

$F_T = 322,000 N$ is the thrust force generated by each engine of the jet

$F_d$ is the drag force

Solving for Fd,

$F_d = 4 F_T = 4(322,000)=1.288\cdot 10^6 N$

The velocity of the jet is

$v=250 m/s$

So, the rate at which the drag force does work (which is the power) is

$P=F_d v$

and substituting

$F_d = 1.288\cdot 10^6 N\\v = 250 m/s$

we find

$P=(1.288\cdot 10^6)(250)=3.22\cdot 10^8 W$

Converting into horsepower,

$P=\frac{3.22\cdot 10^8}{746}=4.32\cdot 10^5 hp$

4 0

2 years ago

A tuna fish body is streamlined to reduce friction. Which describes the type of friction the fish must overcome? kinetic frictio

Wittaler [7]

I do believe it is fluid friction.
Hope this helps~

8 0

2 years ago

Read 2 more answers

Steam enters a counterflow heat exchanger operating at steady state at 0.07 MPa with a specific enthalpy of 2431.6 kJ/kg and exi

PIT_PIT [208]

Answer:

Explanation:

Given:

Steam Mass rate, ms = 1.5 kg/min

= 1.5 kg/min × 1 min/60 sec

= 0.025 kg/s

Air Mass rate, ma = 100 kg/min

= 100 kg/min × 1 min/60 sec

= 1.67 kg/s

Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.

xf, quality = 0.9.

Tsat = 89.9°C

hf = 376.57 kJ/kg

hfg = 2283.38 kJ/kg

Using the equation for specific enthalpy,

hi = hf + (hfg × xf)

= 376.57 + (2283.38 × 0.9)

= 2431.552 kJ/kg

The specific enthalpy of the outlet, h2 = hf

= 376.57 kJ/kg

Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy

= ms × (hi - h2)

= 0.025 × (2431.552 - 376.57)

= 0.025 × 2055.042

= 51.37455 kW

= 51.38 kW.

5 0

2 years ago