The mass of the earth is 6x10^24 kg and its radius is 6400 km.What is the mass of a man weighing 977 N in a spring balance.

You are designing an airport for small planes. One kind of airplane that might use this airfield must reach a speed before takeo

ladessa [460]

Explanation:

Given that,

Initial speed of the airfield, u = 0

Final speed, v = 27.8 m/s

Acceleration of the airfield, $a=2\ m/s^2$

Length of the runway, d = 150 m

Let v' is the speed of the airplane to reach the required speed for takeoff. Finding v' using third equation of motion as :

$v'^2-u^2=2ad\\\\v'=\sqrt{2ad} \\\\v'=\sqrt{2\times 2\times 150} \\\\v'=24.49\ m/s$

This speed is not enough as the airfield must reach a speed before takeoff of at least 27.8 m/s. Now, the required length of the runways is :

$v^2=2ax\\\\x=\dfrac{v^2}{2a}\\\\x=\dfrac{(27.8)^2}{2\times 2}\\\\x=193.21\ m$

So, the minimum length of the runways is 193.21 meters.

8 0

3 years ago

A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the

Mariana [72]

Answer:

17.72° or 72.28°

Explanation:

u = 6.5 m/s

R = 2.5 m

Let the angle of projection is θ.

Use the formula for the horizontal range

$R=\frac{u^{2}Sin2\theta }{g}$

$2.5=\frac{6.5^{2}Sin2\theta }{9.8}$

Sin 2θ = 0.58

2θ = 35.5°

θ = 17.72°

As we know that the range is same for the two angles which are complementary to each other.

So, the other angle is 90° - 17.72° = 72.28°

Thus, the two angles of projection are 17.72° or 72.28°.

8 0

3 years ago

•Would a moving fan have energy? Why or why not.

Pepsi [2]

Moving fan has rotational kinetic energy
Non moving fan has no energy since it is in rest

7 0

3 years ago

The tires on your truck have 0.35 m radius. In a straight line, you drive 2600 m. What is the angular displacement of the tire,

Travka [436]

1) In a circular motion, the angular displacement $\theta$ is given by
$\theta = \frac{S}{r}$
where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire: $S=2600 m$ . Using the information about the radius, $r=0.35 m$ , we find the total angular displacement:
$\theta = \frac{2600 m}{0.35 m} =7428 rad$

2) If we put larger tires, with radius $r=0.60 m$ , the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
$\theta = \frac{2600 m}{0.60 m}=4333 rad$

8 0

3 years ago

X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr

mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

$\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)$

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

$\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m$

b) The change in photon energy is given by:

$\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm$

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

$P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\$

$E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV$

5 0

3 years ago

The mass of the earth is 6x10^24 kg and its radius is 6400 km.What is the mass of a man weighing 977 N in a spring balance.​

The mass of the earth is 6x10^24 kg and its radius is 6400 km.What is the mass of a man weighing 977 N in a spring balance.