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faust18 [17]
3 years ago
14

A cannon ball launched horizontally with a speed of 20m/s and a baseball dropped off a cliff and it accelerates at a rate of 10m

/s^2. Which will hit the ground first?
Physics
1 answer:
miskamm [114]3 years ago
8 0
If they both start from the same height, then they both hit the ground at the
same time.  It makes no difference if their horizontal speeds aren't equal.
The cannon ball still accelerates downward at the same rate as the baseball.
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An object is at rest in front of a compressed spring. It travels over a surface that exerts a kinetic frictional force on it and
PolarNik [594]

Answer:

the object will travel 0.66 meters before to stop.

Explanation:

Using the energy conservation theorem:

E_i+K_i+W_f=K_f+U_f

The work done by the friction force is given by:

W_f=F_f*d\\W_f=\µ*m*g*d\\W_f=0.35*4*9.81*d\\W_f=13.7d[J]

so:

\frac{1}{2}1800*(10*10^{-2})+0-13.7d=0+0\\d=0.66m

3 0
3 years ago
The table represents the speed of a car in a northern direction over several seconds. Column 1 would be on the x-axis, and Colum
TEA [102]

"The table represents the speed of a car in a northern direction over several seconds. Column 1 would be on the x-axis, and Column 2 would be on the y-axis."


typical plot is speed or velocity on the y-axis n time on the x-axis so the ans is Column 1 should be titled “Time,” and Column 2 should be titled “Velocity.”

7 0
3 years ago
Read 2 more answers
7. A block of copper of unknown mass has an initial temperature of 65.4oC. The copper is immersed in a beaker containing 95.7g o
dolphi86 [110]

Answer:

37.34372 kg

Explanation:

m = Mass

\Delta T = Change in temperature

1 denotes water

2 denotes copper

c = Heat capacity

Heat is given by

Q=mc\Delta T

In this case the heat transfer will be equal

m_1c_1\Delta T_1=m_2c_2\Delta T_2\\\Rightarrow m_2=\frac{m_1c_1\Delta T_1}{c_2\Delta T_2}\\\Rightarrow m_2=\frac{95.7\times 4.18(24.2-22.7)}{0.39(65.4-24.2)}\\\Rightarrow m_2=37.34372\ kg

Mass of copper block is 37.34372 kg

5 0
3 years ago
A motorcycle and its driver has a mass of 145 kg. Answer the following questions about it, using correct units.
inysia [295]

a) KE=0.5*mv^2==0.5*145*25^2=45312.5 J

b) PE=mgh=145*9.8*3.5=4973.5 J

c) ME=KE+PE=m(0.5v^2+gh)=62524 J

4 0
3 years ago
Read 2 more answers
Ultraviolet light of wavelength 270 nm strikes a metal whose work function is 2.3 eV.What is the shortest de Broglie wavelength
soldier1979 [14.2K]

Answer:

The shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

Explanation:

Given;

wavelength of ultraviolet light, λ = 270 nm

work function of the metal, φ = 2.3 eV = 2.3 x 1.602 x 10⁻¹⁹ J = 3.685 x 10⁻¹⁹ J

The energy of the ultraviolet light is given by;

E = \frac{hc}{\lambda}\\\\E = \frac{(6.626*10^{-34} )(3*10^{8}) }{270*10^{-9} }\\\\E = 7.362 * 10^{-19} \ J

The energy of the incident light is related to kinetic energy of the electron and work function of the metal by the following equation;

E = φ  + K.E

K.E = E - φ

K.E = (7.362 x 10⁻¹⁹ J) - (3.685 x 10⁻¹⁹ J )

K.E = 3.677 x 10⁻¹⁹ J

K.E = ¹/₂mv²

mv² = 2K.E

velocity of the electron is given by;

V = \sqrt{\frac{2K.E}{m} }\\\\V =  \sqrt{\frac{2(3.677*10^{-19}) }{9.1*10^{-31} } }\\\\V = 8.99*10^{5}  \ m/s

the shortest de Broglie wavelength for the electrons is given by;

\lambda = \frac{h}{mv}\\ \\\lambda = \frac{6.626*10^{-34} }{(9.1*10^{-31})( 8.99*10^{5} )}\\\\\lambda = 8.10*10^{-10} \ m\\\\\lambda = 0.81 \ nm

Therefore, the shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

7 0
3 years ago
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