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3 years ago

A helicopter's speed increases

Physics

1 answer:

sergij07 [2.7K]3 years ago

7 0

Answer:

the helis speed increased its speed by 35 m/s. Unless you are asking me what was the speed per second. if you were asking me for the speed per second then that would be a different answer

Explanation:

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Two wires are parallel and one is directly above the other. Each has a length of 50.0 m and a mass per unit length of 0.020 kg/m

ivann1987 [24]

Answer: The time required for the impluse passing through each other is approximately 0.18seconds

Explanation:

Given:

Length,L = 50m

M/L = 0.020kg/m

FA = 5.7×10^2N

FB = 2.5×10^2N

The sum of distance travelled by each pulse must be 50m since each pulse started from opposite ends.

Ca(t) + CB(t) = 50

Where CA and CB are the velocities of the wire A and B

t = 50/ (CA + CB)

But C = Sqrt(FL/M)

Substituting gives:

t = 50/ (Sqrt( FAL/M) + Sqrt(FBL/M))

t = 50/(Sqrt 5.7×10^2/0.02) + (Sqrt(2.5×10^2/0.02))

t = 50 / (168.62 + 111.83)

t = 50/280.15

t = 0.18 seconds

4 0

3 years ago

Read 2 more answers

A 64 kg swimmer jumps, with a velocity of 4.2 m/s, off the front of a 25 kg kayak when the kayak is moving forward at a velocity

Crank

Answer:

3.88m/s

Explanation:

Using the law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and 2 are the initial velocities

v is the final velocity

Given

m1 = 64kg

u1 = 4.2m/s

m2 = 25kg

u2 = 3.2m/s

Required

Final velocity v

Substitute the given values into the formula

64(4.2)+25(3.2) = (65+25)v

268.8+80 = 90v

348.8 = 90v

v = 348.8/90

v = 3.88m/s

Hence the velocity of the kayak after the swimmer jumps off is 3.88m/s

8 0

3 years ago

An object of mass 6 kg. is resting on a horizontal surface. A horizontal force

son4ous [18]

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force ( $W$ ), measured in joules, is defined by the following expression:

$W = F\cdot \Delta x$ (1)

Where:

$F$ - Force, measured in newtons.

$\Delta x$ - Distance, measured in meters.

If we know that $F = 15\,N$ and $\Delta x = 100\,m$ , then the work done by the force exerted on the object is:

$W = (15\,N)\cdot (100\,m)$

$W = 1500\,J$

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object ( $a$ ), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

$\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}$ (2)

Where $v_{o}$ is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

$\frac{1}{2}\cdot a \cdot t^{2} = \Delta x-v_{o}\cdot t$

$a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}$

If we know that $\Delta x = 100\,m$ , $v_{o} = 0\,\frac{m}{s}$ and $t = 10\,s$ , then the net acceleration experimented by the object is:

$a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}$

$a = 2\,\frac{m}{s^{2}}$

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

$\Sigma F = F - f = m\cdot a$ (3)

Where:

$F$ - External force exerted on the object, measured in newtons.

$f$ - Kinetic friction force, measured in newtons.

If we know that $F = 15\,N$ , $m = 6\,kg$ and $a = 2\,\frac{m}{s^{2}}$ , the kinetic friction force is:

$f = F-m\cdot a$

$f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)$

$f = 3\,N$

The work done by friction ( $W'$ ), measured in joules, is:

$W' = f\cdot \Delta x$ (4)

$W' = (3\,N) \cdot (100\,m)$

$W' = 300\,J$

And the net work experimented by the object is:

$\Delta W = 1500\,J - 300\,J$

$\Delta W = 1200\,J$

By the Work-Energy Theorem we understand that change in translational kinetic energy ( $\Delta K$ ), measured in joules, is equal to the change in net work. That is:

$\Delta K = \Delta W$ (5)

If we know that $\Delta W = 1200\,J$ , then the change in translational kinetic energy is:

$\Delta K = 1200\,J$

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

7 0

3 years ago

I need help with question 6 and 7!!! Also, how to do the back???

lora16 [44]

Reword the questions and you’ll get it right

7 0

2 years ago

Read 2 more answers

calcula la potencia por hora de un radiador, sabiendo que esta conectado a un contacto común 110 v. y requiere 20 Amp.

Ira Lisetskai [31]

calculate the power per hour of a radiator, knowing that it is connected to a common 110 v contact. and requires 20 Amp.

Answer:

2.2kWh

Explanation:

Given parameters:

Potential difference = 110v

Current = 20A

Unknown:

Power = ?

Solution:

To solve this problem, we use the expression below:

Power = IV

Power = 110 x 20 = 2200W

This is therefore 2.2kW

Power per hour = 2.2kWh

8 0

3 years ago