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-Dominant- [34]

3 years ago

15

Two football players are attempting to tackle each other. If one football player has a mass of 100 kg and pushes with a force of

100N onto another football player whose mass is 80kg, how much force does the other player push back with?

1 answer:

Vaselesa [24]3 years ago

8 0

Answer:

is it 20kg. Two opposing forces pushing onto each other

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A student buys a plastic dart gun and tries to find the maximum horizontal range. The student shoots the gun straight up and it

ryzh [129]

Answer:

The value is $R_{max} = 33.54 \ m$

Explanation:

From the question we are told that

The total time of flight is $t = 3.7 \ s$

Generally from kinematic equation

$v = u - g * \frac{t}{2}$

So v is the velocity at maximum height and the value is v = 0 m/s

So

$0 = u - 9.8 * \frac{ 3.7}{2}$

=> $u = 18.13 \ m/s$

Here u is the initial velocity of the dart as it leaves that gun

Gnerally the horizontal range of the dart is mathematically represented as

$R = \frac{u ^2 sin 2\theta }{g}$

For maximum horizontal range the value of $\theta = 45^o$

So

$R_{max} = \frac{ 18.13 ^2 sin 2(45) }{9.8}$

=> $R_{max} = 33.54 \ m$

6 0

3 years ago

What term refers to the part of a spacecraft that is occupied by the crew for takeoff and landing?

Semenov [28]

Command module ✅

service module

lunar module

annum module

7 0

2 years ago

A nuclear particle with no charge

pishuonlain [190]

A nuclear particle with no charge is a neutron

7 0

3 years ago

A car accelerates uniformly at 2m/s2 for 3minutes. what is the velocity of the car

grigory [225]

Answer:

56.7 nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn

4 0

4 years ago

What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Eart

Wittaler [7]

Answer:

The centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ is $1.020x10^{-3}m/s^{2}$

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

$\cos \theta = \frac{adjacent}{hypotenuse}$

$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

$r_{71.9^{\circ}}$ can be isolated from equation 3:

$r_{71.9^{\circ}} = r_{e} \cos \theta$ (4)

$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

$r_{71.9^{\circ}} = 1.97x10^{6} m$

Then, equation 2 can be used

$v = \frac{2 \pi (1.97x10^{6} m)}{24h}$

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

$T = 24h . \frac{3600s}{1h}$ ⇒ $84600s$

$v = \frac{2 \pi (1.97x10^{6} m)}{84600s}$

$v = 146.31m/s$

Finally, equation 1 can be used:

$a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}$

$a = 0.010m/s^{2}$

Hence, the centripetal acceleration is $0.010m/s^{2}$

To given the centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ it is gotten:

$\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}$

6 0

3 years ago