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Paul [167]
3 years ago
11

What is the weight of an object (mass = 60 kilograms) on Mars, where the acceleration due to gravity is 3.75 meters/second2?. Se

lect one of the options below as your answer:. A. 1.6 newtons. B. 16 newtons. C. 22.5 newtons. D. 225 newtons. E. 2250 newtons.
Physics
1 answer:
juin [17]3 years ago
5 0
Weight = mass * gravity = 60 kg * 3.75 m/s² = 225 N

<span>Option D.</span>
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Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.​
Tems11 [23]

Answer:

<em>The range is 35.35 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and g=9.8m/s^2 the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

\displaystyle d={\frac  {v_o^{2}\sin(2\theta )}{g}}

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

\displaystyle d={\frac  {20^{2}\sin(2\cdot 30^\circ )}{9.8}}

\displaystyle d={\frac  {400\sin(60^\circ )}{9.8}}

d=35.35\ m

The range is 35.35 m

7 0
3 years ago
If an object absorbs all colors but red, we see
julia-pushkina [17]
[I researched for you, since I am not in that particular level to know that knowledge yet. I assure this is accurate info :)]

The answer is A, red.
"Remember, the color you see is light REFLECTING off the surface of that object. If all colors are absorbed in to the surface EXCEPT red, red must be reflected, and you'll see red." - Yahoo User @Chap
8 0
3 years ago
A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

7 0
3 years ago
A garden hose has a radius of 0.0120 m, and water initially comes out at a speed of 2.88m/s. Dasha puts her thumb over the end ,
creativ13 [48]

Answer:

v = 12.4 [m/s]

Explanation:

With the speed and Area information, we can determine the volumetric flow.

V=v*A\\A=\pi *r^{2}

where:

r = radius = 0.0120 [m]

v = 2.88 [m/s]

A=\pi *(0.0120)^{2} \\A=4.523*10^{-4} [m]\\

Therefore the flow is:

V=2.88*4.523*10^{-4} \\V=1.302*10^{-3} [m^{3}/s ]

Despite the fact that you cover the inlet with the finger, the volumetric flow rate is the same.

v=V/A\\v=1.302*10^{-3} /1.05*10^{-4} \\v=12.4[m/s]

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3 years ago
When approaching an intersection, bridge, or railroad crossing, you should never drive (pass on the left half of the roadway whe
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You should not go into the left side of the roadway when within 100 feet of the crossing. Moreover, you should also turn on your turn signal when within 100 of a turn. These precautions prevent accidents as it makes clear to other drivers what your intentions are and drivers making turns are not endangered. 
8 0
3 years ago
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