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Mariana [72]
2 years ago
10

A 12-kg object is moving rightward with a constant velocity of 4 m/s. How much net force is required to keep the object moving w

ith
the same speed and in the same direction?
Physics
1 answer:
Nadusha1986 [10]2 years ago
3 0
C an s and yes ihavetotypemore
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A spaceship whose rest length is 350m has a speed of .82c
igomit [66]

Answer:

t'=1.1897*10^{-6} s

t'=1.1897 μs

Explanation:

First we will calculate the velocity of micrometeorite relative to spaceship.

Formula:

u=\frac{u'+v}{1+\frac{u'*v}{c^{2}}}

where:

v is the velocity of spaceship relative to certain frame of reference =  -0.82c (Negative sign is due to antiparallel track).

u is the velocity of micrometeorite relative to same frame of reference as spaceship = .82c (Negative sign is due to antiparallel track)

u' is the relative velocity of micrometeorite with respect to spaceship.

In order to find u' , we can rewrite the above expression as:

u'=\frac{v-u}{\frac{u*v}{c^{2} }-1 }

u'=\frac{-0.82c-0.82c}{\frac{0.82c*(-0.82c)}{c^{2} }-1 }

u'=0.9806c

Time for micrometeorite to pass spaceship can be calculated as:

t'=\frac{length}{Relatie seed (u')}

t'=\frac{350}{0.9806c}     (c = 3*10^8 m/s)

t'=\frac{350}{0.9806* 3.0*10^{8} }

t'=1.1897*10^{-6} s

t'=1.1897 μs

4 0
3 years ago
A manufacturer selected a metal to use in producing a lightweight button for clothing. A metal that has a density of 2.71 g/cm3
Natali5045456 [20]

Just find the density of every metal and select the one with a density of 2.71 g/cm³ . This is:

Metal 1

ρ = m/V

ρ = 22.1 g / 3 cm³

ρ = 7.367 g / cm³

Metal 2

ρ = m/V

ρ = 42 g / 4 cm³

ρ = 10.5 g / cm³

Metal 3

ρ = m/V

ρ = 9.32 g / 5 cm³

ρ = 1.864 g / cm³

Metal 4

ρ = m/V

ρ = 8.13 g / 3 cm³

ρ = 2.71 g / cm³

<h2>R / Metal 4 was selected.</h2>
4 0
2 years ago
PLEASE HELP ASAP!!!! A huge thanks to anyone who can help me with 14 problems. I'll do anything to return the favor. All true an
snow_lady [41]
Hello, I see you are in a jam. Lemme help.

1.) True
2.) True
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LOL these are all true ;)
4 0
3 years ago
A skateboarder shoots off a ramp with a velocity of 7.1 m/s, directed at an angle of 61° above the horizontal. The end of the ra
dsp73

Answer:

Highest point reached  = 3.37 m

Explanation:

Initial velocity, = 7.1 m/s

Initial vertical velocity = 7.1 sin 61 = 6.21 m/s

Consider the vertical motion of skateboarder,

We have equation of motion, v² = u² + 2as

          Initial velocity, u = 6.21 m/s

          Acceleration, a = -9.81 m/s²

          Final velocity, v = 0 m/s

         Substituting

                       v² = u² + 2as

                       0² = 6.21² + 2 x -9.81 x s

                       s = 1.97 m

So from ramp the position it goes up by 1.97 m

       Highest point reached = 1.97 + 1.4 = 3.37 m    

6 0
2 years ago
Amy counts the wave crest traveling down a stretched string five wave crests pass Amy in two seconds what is the frequency of th
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<span>One end of a uniform meter stick is placed against a vertical wall. The other end is held by a lightweight cord that makes an angle, theta, with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.390. A. what is the maximum value...</span>
8 0
3 years ago
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