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guajiro [1.7K]
3 years ago
8

When a space shuttle was launched, the astronauts onboard experienced an acceleration of 32.0 m/s2 . If one of the astronauts ha

d a mass of 40.0 kg, what net force in newtons did the astronaut experience?
Physics
1 answer:
jarptica [38.1K]3 years ago
8 0

Answer:

F = 1280 N

Explanation:

Given that,

Acceleration experienced by a space shuttle, a = 32 m/s²

Mass of the astronauts, m = 40 kg

We need to find the force experienced by the astronaut.

We know that the net force is equal to the product of acceleration and its mass. So,

F = ma

F = 40 kg × 32 m/s²

So,

F = 1280 N

So, 1280 N of force is experienced by the Astronaut.

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Alex

Answer:

The amplitude  is  A =  90.2 \ m

Explanation:

From the question we are told that

    The frequency of when sound is approaching observer is   f = 392 Hz

     The frequency as the move away from observer  is  f_ a =  330 \ Hz

    The time between the pitch are t =  10 \ s

Here you are the observer and your friends are the source of the sound

The period is mathematically evaluated as

       T =  2 t

as it is the time to complete one oscillation which from on highest pitch to the next highest pitch

Now T can also be mathematically represented as

          T = \frac{2 \pi}{w}

Where  w is the angular velocity

=>   \frac{2 \pi}{w}  =  2 * 10

=>   w =  0.314 \ rad/sec

Now using Doppler Effect,

   The source of the sound is approaching the observer

The

          f = f_o (\frac{v}{v- wA} )

         392  = f_o (\frac{v}{v- wA} )

Where A is the amplitude

    So when the source is moving away from the observer

         f_a =  f_o (\frac{v}{v+ wA} )  

        330  =  f_o (\frac{v}{v+ wA} )  

Here  f_o is the fundamental frequency

Dividing the both equation  we have

           \frac{392}{330}  =  \frac{f_o(\frac{v}{v-wA} )}{f_o(\frac{v}{v+wA}}

           1.1878  = \frac{v+wA}{v-wA}

         1.1878 v -  1.1878 wA = v+wA

        1.1878 v = 2.1878 wA

=>     A =  \frac{(0.1878 * (330))}{(2.1878)* (0.314)}

         A =  90.2 \ m

7 0
3 years ago
A toy cannon uses a spring to project a 5.24-g soft rubber ball. The spring is originally compressed by 5.01 cm and has a force
salantis [7]

Answer:

Speed will be equal to 1.40 m/sec

Explanation:

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Spring is compressed by 5.01 cm

So x = 5.01 cm = 0.0501 m

Spring constant k = 8.08 N/m

Frictional force f = 0.031 N

Distance moved by ball d = 15.8 cm = 0.158 m

Energy gained by spring

KE=\frac{1}{2}kx^2=\frac{1}{2}\times 8.08\times 0.0501^2=0.0101J

Energy lost due to friction

W=Fd=0.031\times 0.158=0.0048J

So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J

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\frac{1}{2}\times 0.00524\times v^2=0.0052

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3 years ago
A bus travels north on some busy city streets for 2.5 km, and a trip
d1i1m1o1n [39]

Answer:

V = 4.63 m/s

V = 11.31 m/s

Explanation:

Given,

The distance traveled by the bus, towards north, d = 2.5 km

                                                                                     = 2500 m

The time taken by the trip is, t = 9 min

                                                  = 540 s

The velocity of the bus,

                                       V = d / t

                                           = 2500 / 540

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Therefore the velocity becomes

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<em>Convert 1nanosecond in to its SI init</em>

<em>In SI units, nano is 1000th part of micro which in turn is 1000th part of mini which in turn is 1000th part of main unit. Now, when you affix nano to any unit, here in case, second, it means that you are referring to 1000th part of 1000th part of 1000th part of second or in short, 1000000000th(10^9) part of a second.</em>

<em>In SI units, nano is 1000th part of micro which in turn is 1000th part of mini which in turn is 1000th part of main unit. Now, when you affix nano to any unit, here in case, second, it means that you are referring to 1000th part of 1000th part of 1000th part of second or in short, 1000000000th(10^9) part of a second.So to convert nanosecond into second, just multiply the nanosecond with 0.000000001 or (10^-9)</em>

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