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Lesechka [4]
10 months ago
6

If a football quarterback can throw the ball at 20.0 m/s, what is the furthest distance the quarterback could throw the ball neg

lecting air resistance?
Physics
1 answer:
kvasek [131]10 months ago
4 0

The furthest distance the quarterback could throw the ball neglecting air resistance is 81.63 meters.

The situation given above is an example of projectile motion.

The maximum distance up to which a projectile can be thrown is known as its range, which is given by the formula,

Range(R) = 2u^2 sin(2θ)/g

Here u is the maximum velocity of the projectile= 20 m/s

θ is the angle at which the projectile is launched = 45°

g is the acceleration due to gravity = 9.8 m/s^2

Putting the above values in the given equation,

The range comes out to be 81.63 meters.

Hence, the furthest distance the quarterback could throw the ball neglecting air resistance is 81.63 meters.

To know more about "projectile motion", refer to the following link:

brainly.com/question/11422992?referrer=searchResults

#SPJ4

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A car is pushed with a force of 450 N for 19.4 seconds. What impulse was applied to the car?​
harina [27]

Answer:

impulse = 8820 kg·\frac{m}{s} or 8820 N·s

Explanation:

Impulse J is equal to the average force F_{av} multiplied by the elapsed time Δt or in equation form, J = F_{av}Δt

As long as your force of 450 N is constant then that value is your average force F_{av} and your elapsed time is 19.4 seconds.

Multiply these values.

You will get an impulse of 8820 kg·\frac{m}{s} or 8820 N·s.

6 0
2 years ago
I need help with homework! what should i choose as the greatest british invention?
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THE MINI Alec Issigonis, 1959
The Telephone Alexander Graham Bell, 1876

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4 0
3 years ago
A ball hits a wall. What is true about the magnitude of the force experienced by the ball compared with the force experienced by
Ivahew [28]
<span>This is best understood with Newtons Third Law of Motion: for every action there is an equal and opposite reaction. That should allow you to see the answer.</span>
5 0
3 years ago
200-grams of computer chips with a specific heat of 0.3 kJ/kg·K are initially at 25°C. These chips are cooled by placement in 0.
balu736 [363]

Answer:

a. -0.01324 kJ/K,  b.  = 0.03233 kJ/K , c.  = 0.01909, Yes the process is possible

Explanation:

Heat transfer will occur between the chip and the surrounding fluid. Then, finally they will attain a common equilibrium temperature and heat transfer will stop. Now, if we assume that, after heat transfer, chip will attain the temperature of fluid, that is, -34 C,, So , to check whether this is possible

Amount of energy lost by the chip = m . c . (T(i) - T(f))

= 0.2 x 0.3 (25 + 34) = 3.54 KJ

Now, to evaluate the final state of the fluid, after the heat transfer completion,

Energy Gained = m(mew final – mew initial) = m[(μf+ x . μfg) - μf]

Note that heat transfer will change the internal energy of the fluid. Do not consider enthalpy change, as this is not a problem involving fluid flow in and out of the system

M[(μf+ x . μfg) - μf] = m(xμfg)

<u>Energy gained by the fluid will be equal to the energy lost by the chip (No energy loss to the surroundings)</u>

3.54 = 0.1 . X x 203.29

<u>x = 0.1741, which is the dryness fraction of fluid at the final state.</u>

Observe that the total energy lost by the chips is 3.45 kJ and fluid R-134a has got its value of mew fg at -34 C which is = 203.29 kJ/kg

So for 0.1kg of R-134a

0.1 x μfg= <u>20.329 kJ, which is much greater than 3.45 kJ</u>, therefore, it is certain that the state of fluid will be at -34 C only and at the saturation pressure of 69.56 KPa. So the chip will come to attain the temperature of -34 C.  

a. Write the equation for the change of entropy in the chips

ΔSchips = mchips . c . ln(T2/T1), where mc is the mass of chips, c is the specific heat of chips, T2 is the temperature at state 2 and T1 is the temperature at state 1

Substitute mc = 0.2 kg, c = 0.3kJ/kg.K, T1 = 25 + 273, T2 = -34 + 273

delSchips = 0.2 x 0.3 x ln [(-34+273)/ (25+273)]

= -0.01324 kJ/K

There fore the change in entropy of the chips is -0.01324 kJ/K

b. Entropy change of fluid R- 134a

ΔS2 = m[Sfinal – S initial]

= m[Sf + x . Sfg - Sf]

= 0.2 x (0.1741 x 0.92859)

= 0.03233 kJ/K

c. Calculate the total change in the entropy of the entire system

delS = delSchips + delSR -134a

= -0.01324 + 0.03233

= 0.01909

<u>Since the total change in entropy of the entire system is positive that exactly explains that the actual processes are happening in the direction of increase of entropy therefore, the process is possible.</u>

<u />

6 0
3 years ago
Quick please and will give Brainliest!!!
masha68 [24]

25 nC

That is the answer

3 0
3 years ago
Read 2 more answers
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