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marta [7]
2 years ago
6

An airplane flies with a constant speed of 840km/h. how far can it travel in 1 hour?

Physics
1 answer:
arlik [135]2 years ago
3 0
Distance = speed X time

In this example, the speed of the airplane = 840km. The time (that the question is asking)is how far can it travel in 1 hour.

So just plug in your numbers.

Distance = 840km X 1 hour = 840km/hour or 840km for short.
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Two tougboats are toeing a ship each exerts a force of 6000N and the angle between the two ropes is 60 calculate the resultant f
Arlecino [84]

Answer:

10392.30N

Explanation:

We proceed by computing the individual force exerted by the boats

For the first boat

The angle is 30 degree to the vertical

Hence

Force = F cos θ

F=6000 cos 30

F=6000*0.866

F=5196.15 N

Since the boats are two and also at the same angle and also exerting the same force

The Net force = 2*5196.15

Net force=10392.30N

7 0
2 years ago
What are the factors that affect the resistance of a wire?
777dan777 [17]

1) Length of the wire.

2) Thickness of the wire.

3) Temperature.

4) Type of metal.

Hope this helps!

-Payshence

6 0
3 years ago
An elevator cable breaks when a 925-kg elevator is 28.5 m above the top of a huge spring Ak = 8.00 * 104 N????mB at the bottom o
rosijanka [135]

Answer:

a) = 258352.5J

b) = 23.63 m/s

c) = 1.8m

Explanation:

Data;

Mass = 925kg

Distance (s) = 28.5m

Force constant (k) = 8.0*10⁴ N/m

g = 9.8 m/s²

a) = work = force * distance

But force = mass * acceleration

Force = 925 * 9.8 = 9065N

Work = F * s = 9065 * 28.5 = 258352.5J

b) acceleration (a) = (v² - u²) / 2s

a = v² / 2s

v² = a * 2s

v² = 9.8 * (2 * 28.5)

v² = 9.8 * 57

v² = 558.6

v = √(558.6)

V = 23.63 m/s

C). The work stops when the work done to raise the spring equals the work done to stop it by the spring

W = ½kx²

258352.5 = ½ * 8.0*10⁴ * x²

(2 * 258352.5) = 8.0*10⁴x²

516705 = 8.0*10⁴x²

X² = 516705 / 8.0*10⁴

X² = 6.46

X = √(6.46)

X = 2.54m

The compression was about 2.54m

3 0
3 years ago
Read 2 more answers
Your friend is bragging about his motorcycle. He claims that it can go from a stopped position to 50 miles per hour in three sec
nalin [4]
If the motorcycle can go from 0 to 50mph in 3 seconds the person is describing the motorcycle's acceleration. 
8 0
3 years ago
Read 2 more answers
A mass is oscillating with amplitude A at the end of a spring.
Dmitry_Shevchenko [17]

A) x=\pm \frac{A}{2\sqrt{2}}

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

E=\frac{1}{2}kA^2 (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2 (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

U=\frac{1}{3}K

Using (2) we can rewrite this as

U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}

And using (1), we find

U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2

Substituting U=\frac{1}{2}kx^2 into the last equation, we find the value of x:

\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}

B) x=\pm \frac{3}{\sqrt{10}}A

In this case, the kinetic energy is 1/10 of the total energy:

K=\frac{1}{10}E

Since we have

K=E-U

we can write

E-U=\frac{1}{10}E\\U=\frac{9}{10}E

And so we find:

\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A

3 0
2 years ago
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