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GrogVix [38]
4 years ago
9

How fast must a 350 gram softball be traveling in order to have a momentum of .600 kg

Physics
1 answer:
BartSMP [9]4 years ago
4 0
Momentum is mass times velocity. So here we can just substitute in our givens and solve for velocity.

.600kg*m/s/.350kg=1.71m/s

Hope this helps! Thank you!
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Compare mass and mass number.
devlian [24]

Answer:

Atomic mass is the weighted average mass of an atom of an element based on the relative natural abundance of that element's isotopes. The mass number is a count of the total number of protons and neutrons in an atom's nucleus.

7 0
3 years ago
Calculate the energy of the green light emitted, per photon, by a mercury lamp with a frequency of 5.49 × 1014 hz. calculate the
olga nikolaevna [1]
The energy of a single photon is given by:
E=hf
where
E is the energy
h is the Planck constant
f is the frequency of the light

The light in our problem has a frequency f=5.49 \cdot 10^{14}Hz, so the energy of each photon of that light is:
E=hf=(6.6 \cdot 10^{-34} Js)(5.49 \cdot 10^{14} Hz)=3.64 \cdot 10^{-19} J
5 0
3 years ago
Hunter wants to measure the weight of a dumbbell. he should
pshichka [43]
He should cleverly, using charm and subterfuge, entice the dumbbell slowly in the direction of where he had hidden the scale. As long as Hunter isn't obvious about it, the dumbbell will swallow his faux charm, and will remain naively oblivious to his true purpose.
7 0
4 years ago
A horizontal circular platform (m = 119.1 kg, r = 3.23m) rotates about a frictionless vertical axle. A student (m = 54.3kg) walk
Murrr4er [49]

Answer:

\omega_2=5.1rad/s

Explanation:

Since there is no friction angular momentum is conserved. The formula for angular momentum thet will be useful in this case is L=I\omega. If we call 1 the situation when the student is at the rim and 2 the situation when the student is at r_2=1.39m from the center, then we have:

L_1=L_2

Or:

I_1\omega_1=I_2\omega_2

And we want to calculate:

\omega_2=\frac{I_1\omega_1}{I_2}

The total moment of inertia will be the sum of the moment of intertia of the disk of mass m_D=119.1 kg and radius r_D=3.23m, which is I_D=\frac{m_Dr_D^2}{2}, and the moment of intertia of the student of mass m_S=54.3kg at position r (which will be r_1=r=3.23m or r_2=1.39m) will be I_{S}=m_Sr_S^2, so we will have:

\omega_2=\frac{(I_D+I_{S1})\omega_1}{(I_D+I_{S2})}

or:

\omega_2=\frac{(\frac{m_Dr_D^2}{2}+m_Sr_{S1}^2)\omega_1}{(\frac{m_Dr_D^2}{2}+m_Sr_{S2}^2)}

which for our values is:

\omega_2=\frac{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(3.23m)^2)(3.1rad/s)}{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(1.39m)^2)}=5.1rad/s

6 0
3 years ago
A normal human heart, beating about once per second, creates a maximum 4.00 mV 4.00 mV potential across 0.250 m 0.250 m of a per
ElenaW [278]

Answer:

1) maximum electric field strength = 0.016 V/m

2)maximum magnetic field strength = 5.33 x 10^(-11) T

3) wavelength = 3 x 10^(8)m

Explanation:

A) We are given;

Frequency (f) = 1 Hz

Maximum Potential ΔV = 4 mV = 4 x 10^(-3) V

Length; d = 0.25 m

Now, maximum potential formula is given as;

ΔV = E_o•d

Where E_o is maximum electric field strength.

Thus,

E_o= ΔV/d = (4 x 10^(-3))/0.25 = 0.016 V/m

B) The maximum magnetic field strength is given by;

B_o = E_o/c

Where;

c is speed of light = 3 x 10^(8)

B_o is maximum magnetic field strength

Thus,

B_o = 0.016/(3 x 10^(8)) = 5.33 x 10^(-11) T

C) Speed of any electromagnetic wave is given by;

c = fλ

Where;

f is frequency

λ is wavelength

c is speed of light = 3 x 10^(8)

Thus, making λ the subject,

λ = c/f

λ = (3 x 10^(8))/1 = 3 x 10^(8)m

8 0
4 years ago
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