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2 years ago

5

A ball is thrown horizontally from a 19 m -high building with a speed of 2.0 m/s . How far from the base of the building does th

e ball hit the ground?

1 answer:

bazaltina [42]2 years ago

3 0

Using the formula: ΔY = V₀y * t + (1/2) * ay * t²

Solve for time and get: 1.968s

Then use: v = d/t in the x-direction and get: d = 3.936

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Is kane brown a true country singer?<br> A.yes<br> B.no

Kryger [21]

Answer: hes not a country singer if thats what your asking

Explanation:

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2 years ago

Read 2 more answers

Relation between wavelength wave velocity and time period

Snezhnost [94]

Answer: wavelength=velocity×period

Explanation:the relation between velocity, wavelength and period is

Wavelength=velocity×period

8 0

2 years ago

A uniform cubical crate is 0.740 m on each side and weighs 600 N. It rests on the floor with one edge against a very small, fixe

Sindrei [870]

Answer:

H = 0.673

Explanation:

given,

side of cubical crate = 0.74

weight of the crate = 600 N

magnitude of force = 330 N

the Horizontal distance of its Center of mass

= 0.74/2

= 0.37

Let the required Height be H

By Balancing the Torques, we get

H x 330 N = 0.37 x 600

330 H = 222

H = 0.673

hence, the height above the floor where force is acting is equal to 0.673 m

6 0

2 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.

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

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

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2 years ago

If a sled has at the top of 10 m hill had 1000 J of potential energy what would happen to the PE if the sled were to moved to a

nataly862011 [7]

Answer:

500J

Explanation:

there are many students who can not get answer step by step and in required time. so

there are a wats group where you can get your answer by trusted experts.

join this wats up group and be smart

7 0

2 years ago