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MrRissso [65]
2 years ago
5

A glider moving with a speed of 200 kilometers/hour experiences a cross wind of 30 kilometers/hour. What is the resultant speed

of the glider relative to the ground?
Physics
1 answer:
liraira [26]2 years ago
5 0
You didn't say so, but we must assume that the "200 km/hr" is
the glider's air-speed, that is, speed relative to the air. 

If the air itself is moving at 30 km/hr relative to the ground and
across the glider's direction, then the glider's speed relative to
the ground is

                 √(200² + 30²)

             =  √(40,000 + 900)

             =  √(40,900)  =  202.24... km/hr   (rounded) 
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if you apply a Force of F1 to area A1 on one side of a hydraulic jack, and the second side of the jack has an area that is twice
Firlakuza [10]

Answer:

2F_{1}

Explanation:

F₁ = Force on one side of the jack

A₁ = Area of cross-section of one side of the jack

F₂ = Force on second side of the jack

A₂ = Area of cross-section of second side of the jack = 2 A₁

Using pascal's law

\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{A_{_{2}}}

\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{2A_{_{1}}}

F_{1}= \frac{F_{2}}{2}\\

F_{2}= 2F_{1}

7 0
2 years ago
A helicopter travels west at 80 mph. It is moving above a car traveling on a highway at 80 mph. Given this information, you can
gavmur [86]

Answer:

d. at the same velocity

Explanation:

I will assume the car is also travelling westward because it was stated that the helicopter was moving above the car. In that case, it depends where the observer is. If the observer is in the car, the helicopter would look like it is standing still ( because both objects have the same velocity). If the observer is on the side of the road, both objects would be travelling at the same velocity. Also recall that, velocity is a vector quantity; it is direction-aware. Velocity is the rate at which the position changes but speed is the rate at which object covers distance and it is not direction wise. Hence velocity is the best option.

5 0
2 years ago
The force on a wire is a maximum of6.71 10-2 N when placed between the pole faces of a magnet.The current flows horizontally to
Taya2010 [7]

Answer:

B.   i=2.79A

C.   F=0.066N

Explanation:

A) By the right hand rule we have that

F=iL x B

F=iLBsin(α)

If the wire jump toward the observer the top pole face is the magnetic southpole.

B) The diameter of the pole face is 15cm. We can take this value as L (the length in which the wire perceives the magnetic field). Hence, we have

F=iLBsin(\alpha)\\\alpha=90°\\F=iLB\\i=\frac{F}{LB}=\frac{6.71*10^{-2}N}{(0.15m)(0.16T)}=2.79A

C) Now the length of the wire that feels B is

L=\frac{0.15m}{cos(10\°)}=0.152m

and the force will be (by taking the degrees between the magnetic field vector and current vector as 80°)

F=iLBsin(\alpha)\\F=(2.79A)(0.152m)(0.16T)(sin(80\°))=0.066N

I hope this is useful for you

regards

8 0
2 years ago
A man runs at 7.5 m s-1 due east across the deck of a ship which moves at 10.0 m s-1 due north. Determine the resultant velocity
german

Answer:

12.5 m/s North/east

Explanation:

I thought of it as a graph. I basically just found the distance between the points (0,0) and (10, 7.5)!

7 0
2 years ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
alexandr1967 [171]

Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

m_e is the mass of the electron

Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

r=5.3\cdot 10^{-11}m

while the electron mass is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
2 years ago
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