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3 years ago

A plane traveled west for 4.0 hours and covered a distance of 4,400 kilometers. What was its velocity? 18,000 km/hr

Physics

2 answers:

Leto [7]3 years ago

7 0

The correct answer is 1100, I know this answer is correct cause i got it right when i turned it in.

olga55 [171]3 years ago

5 0

Velocity is the same as the formula for speed, the only difference is velocity has direction. Velocity is distance over time. Given is 4,400 kilometers travelled west in 4 hours. Applying the given equation, we wil have 4,400/4 = 1,100 km/hr west

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An object of mass 4kg is moving along a horizontal plane. If the coefficient of kinetic friction is 0.2 find the friction force

lana66690 [7]

Answer:

The friction force acting on the object is 7.84 N

Explanation:

Given;

mass of object, m = 4 kg

coefficient of kinetic friction, μk = 0.2

The friction force acting on the object is calculated as;

F = μkN

F = μkmg

where;

F is the frictional force

m is the mass of the object

g is the acceleration due to gravity

F = 0.2 x 4 x 9.8

F = 7.84 N

Therefore, the friction force acting on the object is 7.84 N

5 0

3 years ago

6th grade science !!!!!!!!!!!!!!!

posledela

Answer:

Explanation:

5 0

3 years ago

The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9

kirill [66]

Answer:

Speed of the car 1 = $V_1=8.98m/s$

Speed of the car 2 = $V_2=17.96m/s$

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

$KE =\frac{1}{2}mv^2$

where,

m = mass of the body

v = velocity of the body

thus,

$\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

$\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

$2M_2V_1^2=0.5\times M_2V_2^2$

$2V_1^2=0.5\times V_2^2$

$4V_1^2= V_2^2$

$2V_1= V_2$ .................(1)

also,

$\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2$

$\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2$

$2(V_1+9.0)^2=(2V_1+9.0)^2$

$\sqrt{2}(V_1+9.0)=(2V_1+9.0)$

$(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)$

$(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)$

$(2V_1-\sqrt{2}V_1)=(12.72-9.0)$

$(0.404V_1)=(3.72)$

$V_1=8.98m/s$

and, from equation (1)

$V_2=2\times 8.98m/s = 17.96m/s$

Hence,

Speed of car 1 = $V_1=8.98m/s$

Speed of car 2 = $V_2=17.96m/s$

8 0

2 years ago

Read 2 more answers

At a certain location, wind is blowing steadily at 9 m/s. Determine the mechanical energy of air per unit mass and the power gen

Misha Larkins [42]

Answer:

The specific mechanical energy of the air in the specific location is 40.5 J/kg.
The power generation potential of the wind turbine at such place is of 2290 kW
The actual electric power generation is 687 kW

Explanation:

The mechanical energy of the air per unit mass is the specific kinetic energy of the air that is calculated using: $\frac{1}{2} V^2$ where V is the velocity of the air.
The specific kinetic energy would be: $\frac{1}{2}(9\frac{m}{s})^2=40.5\frac{m^2}{s^2}=40.5\frac{m^2 }{s^2}\frac{kg}{kg}=40.5\frac{N*m }{kg}=40.5\frac{J}{kg}$ .
The power generation of the wind turbine would be obtained from the product of the mechanical energy of the air times the mass flow that moves the turbine.
To calculate mass flow it is required first to calculate the volumetric flow. To calculate the volumetric flow the next expression would be: $\frac{V\pi D_{blade}^2}{4} =\frac{9\frac{m}{s}\pi(80m)^2}{4} =45238.9\frac{m^3}{s}$
Then the mass flow is obtain from the volumetric flow times the density of the air: $m_{flow}=1.25\frac{kg}{m^3}45238.9\frac{m^3}{s}=56548.7\frac{kg}{s}$
Then, the Power generation potential is: $40.5\frac{J}{kg} 56548.7\frac{kg}{s} =2290221W=2290.2kW$
The actual electric power generation is calculated using the definition of efficiency: $\eta=\frac{E_P}{E_I}}$ , where η is the efficiency, $E_P$ is the energy actually produced and, $E_I$ is the energy input. Then solving for the energy produced: $E_P=\eta*E_I=0.30*2290kW=687kW$

6 0

3 years ago

On an unknown planet in the outer-reaches of the solar system, a pendulum with a 12 g bob and a string length of 4 m oscillates

11Alexandr11 [23.1K]

The acceleration due to gravity (g) on this planet is 39.44 m/s²

<h3>What is solar system?</h3>

Solar system consists of all the planets and the most importantly the center of the solar system is Sun.

Given is an unknown planet in the outer-reaches of the solar system, a pendulum with a 12 g bob and a string length of 4 m oscillates with a period of 2 seconds.

The time period of the pendulum is

T = 2π √l/g

Squaring both sides, we get

l/g = T² / 4π²

g = 4π²l/ T²

Substitute Time period T = 2s and length l = 4m, we get

g = 4π²x 4/ 2²

g =39.44 m/s²

Thus, the acceleration due to gravity on this planet is 39.44 m/s²

Learn more about solar system.

brainly.com/question/12075871

#SPJ1

3 0

2 years ago