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Lady bird [3.3K]
3 years ago
6

A 0.5 kg football is thrown so that it has a momentum of 16 kg*m/s. What is its speed? ​

Physics
1 answer:
anygoal [31]3 years ago
3 0
The answer is v= 32 m/s
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<span>Radio waves are the type of wave that most likely transmit cell phones messages. Radio waves can be used to transmit data and information at various frequencies. Gamma rays are high frequency and are dangerous to our health. Ultraviolet rays come from the sun. Infrared waves are ones we experience every day, even though it is invisible to the naked eye. This is the heat we feel from the sun. </span>
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3 years ago
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A transformer has two sets of coils, the primary with N1 = 160 turns and the secondary with N2 = 1400 turns. The input rms volta
vovikov84 [41]

To solve the problem it is necessary to apply the concepts related to the voltage in a coil, through the percentage relationship that exists between the voltage and the number of turns it has.

So things our data are given by

N_1 = 160

N_2 = 1400

\Delta V_{1rms} = 62V

PART A) Since it is a system in equilibrium the relationship between the two transformers would be given by

\frac{N_1}{N_2} = \frac{\Delta V_{1rms}}{\Delta V_{2rms}}

So the voltage for transformer 2 would be given by,

\Delta V_{2rms} = \frac{N_2}{N_1} \Delta V_{1rms}

PART B) To express the number value we proceed to replace with the previously given values, that is to say

\Delta V_{2rms} = \frac{N_1}{N_2} \Delta V_{1rms}

\Delta V_{2rms} = \frac{1400}{160} 62V

\Delta V_{2rms} = 1446.66V

7 0
3 years ago
Two children hang by their hands from the same tree branch. the branch is straight, and grows out from the tree trunk at an angl
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The net torque exerted by the children on the branch of the tree is 1382 N-m.

The torque exert by the kids is calculated as

T= 45.6*9.8*1.28*cos27.5°+36*9.8*(2.25-1.28)cos27.5°

T=1382 N-m

Hence, The net torque exerted by the children on the branch of the tree is 1382 N-m.

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4 years ago
How does the vertical acceleration at point a compare to the vertical acceleration at point c?
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4 0
3 years ago
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The coefficients of friction between the 20-kg crate and the inclined surface are µ,8 = 0.24 and J.lk = 0.22. If the crate start
Yanka [14]

Answer:5.60 m/s

Explanation:

Given

Coefficient of static friction \mu _s=0.24

Coefficient of kinetic friction \mu _k=0.22

mass of crate m=20\ kg

Force applied F=200\ N

maximum static Friction F_s=\mu _sN

N=mg

F_s=0.24\times 20\times 9.8

F_s=47.04\ N

thus applied force is greater than Static friction therefore kinetic friction will come into play

F_k=\mu _kN

F_k=0.22\times 20\times 9.8=43.12\ N

net Force on crate F-F_k=ma

a=\frac{200-43.12}{20}=7.84\ m/s^2

Magnitude of velocity can be obtained by using

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

here initial velocity is zero as crate start from rest

v^2-0=2\times 7.84\times 2

v=\sqrt{31.37}

v=5.60\ m/s                                          

7 0
4 years ago
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