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Neko [114]
2 years ago
7

If it were possible to remove gravity and friction, think about what would happen to a football if it were tossed into the air.

Which statement best describes the force(s) acting on the football? A force must be applied to it when it slows to a position of rest. A force must be applied to it before it slows to a position of rest. No force is needed; an object in motion stays in motion. Many forces must act upon it to keep it in motion.
Physics
2 answers:
elena-14-01-66 [18.8K]2 years ago
6 0
Ignoring fluid resistance, football will <span>maintain a constant speed until other forces accelerate the football.</span>
Naily [24]2 years ago
6 0

Answer:

No force is needed; an object in motion stays in motion.

Explanation:

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Why is the transverse spatial extent of a photon proportional to its wavelength which is a longitudinal quantity?
otez555 [7]
In quantum mechanics, particularly the wave-particle theory, it states that light behaves like a wave or a particle. For the wave behavior, its movement is measured in wavelengths while the time for each wavelength is the frequency. For the particle behavior, according to Planck, the energy of the photon (light particle) is determined as

E = hc/wavelength, where h is the Planck's constant (<span>6.626 x 10-34 J-s per particle) and c is the speed of light ( 3 x 10^8m/s)

As you can see, the energy of the photon is INVERSELY PROPORTIONAL to the wavelength with the Planck's constant as the constant of proportionality.</span>
7 0
2 years ago
A two-resistor voltage divider employing a 2-k? and a 3-k? resistor is connected to a 5-V ground-referenced power supply to prov
vesna_86 [32]

Answer:

circuit sketched in first attached image.

Second attached image is for calculating the equivalent output resistance

Explanation:

For calculating the output voltage with regarding the first image.

Vout = Vin \frac{R_{2}}{R_{2}+R_{1}}

Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V

For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.

so.

R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.  

if the -5% is applied to both resistors the Voltage is still 5V because the quotient  has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:

Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V

Vout = 5 \frac{2100}{5100}\\Vout = 5 \frac{21}{51} = 2.05 V

R_{out} = R_{2} || R_{1}\\R_{out} = 1900||2850= \frac{1900*2850}{1900+2850} = 1140

R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260

so.

V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}

4 0
2 years ago
A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7
maxonik [38]
First, let's find the speed v_i of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
p_i = m_1 v_1
After the collision, the two blocks stick together and so now they have mass m_1 +m_2 and they are moving with speed v_i:
p_f = (m_1 + m_2)v_i
For conservation of momentum
p_i=p_f
So we can write
m_1 v_1 = (m_1 +m_2)v_i
From which we find
v_i =  \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force \mu (m_1+m_2)g. The work done by the frictional force to stop the two blocks is
\mu (m_1+m_2)g  d
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
\frac{1}{2} (m_1+m_2)v_i^2
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g  d
From which we can find the value of the coefficient of kinetic friction:
\mu =  \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49
3 0
2 years ago
If in a vernier callipers 10 VSD coincides with 8 MSD then the least count of varnier calliper is
ratelena [41]

Answer:0.2mm

Explanation:

The length of one VSD=8/10=0.8mm

The least count of the instrument is the difference between the length of one MSD and length of one VSD

The length of inebriated MSD=1mm

Therefore,

The least count=1-0.8=0.2mm

3 0
2 years ago
Read 2 more answers
Briana swings a ball on the end of a rope in a circle. The rope is 1.5 m long. The ball completes a full circle every 2.2 s. Wha
schepotkina [342]
The radius of the circular path is 1.5 m.

The circumference is then
1.5\ m*2\pi=3\pi\ m

The ball moves 3π m every 2.2 s, so the speed is
\frac{3\pi\ m}{2.2\ s}\approx 4.3\ m/s
9 0
2 years ago
Read 2 more answers
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