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SSSSS [86.1K]
3 years ago
12

If anyone knows how to do any of these PLEASE help me....im am so confused rn and our teacher sucks at explaining this stuff....

..

Physics
2 answers:
Taya2010 [7]3 years ago
7 0
Take 68.2/60 = 1.137 hr
take 56.9/1.137 = 50.043 mi/hr

take 189/211 = 0.896

24.8/2 = 12.4 m
12.4/82.3 = 0.15s

Mama L [17]3 years ago
4 0
Well, five answers for only five points is pretty slim.  But I've already got
a lot of points, and your plea for help sounds so desperate, that I think
I ought to jump in here and relieve some of your pain.

These aren't difficult problems.  The bummer is that whoever wrote them
got so interested in his dinosaur stories that he kind of hid the actual Math
and Physics under all kinds of stuff that doesn't matter.

1).     Speed  =  (distance covered) / (time to cover the distance)

                       =    (56.9 miles)  /  (68.2 minutes)

                       =      (56.9 / 68.2)  miles/minute

                       =          0.834  mile per minute.

BUT ... the question wants to know the speed in 'mph' ... miles per hour.

Well ... there are 60 minutes in an hour.  If 0.834 mile in covered in 1 minute,
then 60 times as much would be covered in 1 hour.

                    (0.834 mile/minute)  x  (60 minute/hour)  =  50.06 mile/hour<span>

                      </span>            Rounded to no decimal places  =  50 mph


2).  Use the same formula:

         Speed  =  (distance covered) / (time to cover the distance)

                       =    (189 meters)  /  (211 seconds)

                       =      (189 / 211)  meters/seconds

                       =          0.8957 meters/second

     Round to 3 decimal places:      0.896  m/s


3).  The attacker jumped from 24.8 meters away.
       He already covered half of the distance.
       How far away is he now ?      1/2 of 24.8 m  =  12.4 m .
       He still has 12.4 meters left to go.

       He's moving at 82.3 meters per second.
       When will he arrive here ?

       Time  =  (distance)  /  (speed)

                  =      (12.4 meters)  /  (82.3 m/s)

                  =          (12.4 / 82.3)  seconds

                  =              0.1506 second

       Two decimal places:    0.15 second


4).    Distance  =  (speed) x (time)

                          =  ( 237 km/hr)  x (16.8 seconds) .

Do you see the problem here ?  We have the distance covered in an hour,
but we need to know how much distance is covered in only 16.8 seconds.
We really need to work out the speed in (meters per second).

OK.  237 km = 237,000 meters, so the speed is  237,000 meters/hour.

1 hour is 3,600 seconds. 
So the speed is only one 3600th of 237,000 meters in 1 second.

           (237,000 meters/hour) x (1 hour / 3600 seconds)  =  65.8333 m/s

           Distance  =  (speed)  x  (time)

                            =  (65.8333 meter/second)  x  (16.8 seconds)

                            =   (65.8333 x 16.8)  meters

                            =      1106 meters.     


5).  This is a nasty trick question.  The whole story is irrelevant,
and we don't even care how far Vinny dropped.

Anything that drops on Earth falls with the acceleration of gravity.
It doesn't matter how big or heavy or small or light it is.
Everything falls with the acceleration of gravity on Earth ...  about   9.8 m/s²  .
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In 2017, the company SpaceX became the first private company to send supplies to the International Space Station with a reusable
pav-90 [236]

Answer:

Approximately 3.98\; \rm m \cdot s^{-2}.

Assumption: air resistance on the rocket is negligible. Take g = \rm 9.81\; m \cdot s^{-2}.

Explanation:

By Newton's Second Law of Motion, the acceleration of the rocket is proportional to the net force on it.

\displaystyle \text{Acceleration} = \frac{\text{Net Force}}{\text{Mass}}.

Note that in this case, the uppercase letter \rm M in the units stands for "mega-", which is the same as 10^6 times the unit that follows. For example, \rm 1\; Mg = 10^6\; g, while \rm 1\; MN = 10^6\; N.

Convert the mass of the rocket and the thrust of its engines to SI standard units:

  • The standard unit for mass is kilograms: \displaystyle m = \rm 552\; Mg = 552 \times 10^6\; g \times \frac{1\; \rm kg}{10^3\; g}  = 552 \times 10^3 \; kg.
  • The standard for forces (including thrust) is Newtons: \text{Thrust} = \rm 7.61 \; MN = 7.61 \times 10^6\; N.

At launch, the velocity of the rocket would be pretty low. Hence, compared to thrust and weight, the air resistance on the rocket would be pretty negligible. The two main forces that contribute to the net force of the rocket would be:

  • Thrust (which is supposed to go upwards), and
  • Weight (downwards due to gravity.)

The thrust on the rocket is already known to be \rm 7.61 \times 10^6\; N. Since the rocket is quite close to the ground, the gravitational acceleration on it should be approximately 9.81\; \rm m \cdot s^{-2} = 9.81 \; N \cdot kg^{-1}. Hence, the weight on the rocket would be approximately 9.81\; \rm N \cdot kg^{-1} \times 552 \times 10^3\; kg = 5.41412\times 10^6\; N.

The magnitude of the net force on the rocket would be

\begin{aligned}&\text{Thrust} - \text{Weight} \\ &= 7.61 \times 10^6\; \rm N - 5.41412\times 10^6\; N \\ &\approx 2.19 \times 10^6\; \rm N\end{aligned}.

Apply the formula \displaystyle \text{Acceleration} = \frac{\text{Net Force}}{\text{Mass}} to find the net force on the rocket. To make sure that the output (acceleration) is in SI units (meters-per-second,) make sure that the inputs (net force and mass) are also in SI units (Newtons for net force and kilograms for mass.)

\begin{aligned}\displaystyle &\text{Acceleration} \\ &= \frac{\text{Net Force}}{\text{Mass}} \\ &= \frac{2.19 \times 10^6\; \rm N}{552 \times 10^3\; \rm kg}  \\ &\approx \rm 3.98\; \rm m \cdot s^{-2}\end{aligned}.

6 0
3 years ago
Three objects lie in the x, y plane. Each rotates about the z axis with an angular speed of 5.58 rad/s. The mass m of each objec
QveST [7]

Answer:

a) V1=11.05m/s    V2=92.07m/s     V3=17.24m/s

b) KE = 16238.26J

Explanation:

For tangential speeds:

V1 = \omega*R1=5.58*1.98=11.05m/s

V2 = \omega*R2=5.58*16.5=92.07m/s

V3 = \omega*R3=5.58*3.09=17.24m/s

For the kinetic energy, it can be calculated as:

KE=1/2*\omega^2*(I1+I2+I3)

Where:

I1 = m1*R1^2=5.46*1.98^2=21.4kg.m^2

I2 = m2*R2^2=3.64*16.5^2=990.99kg.m^2

I3 = m3*R3^2=3.21*3.09^2=30.65kg.m^2

So,

KE=1/2*5.58^2*(21.4+990.99+30.65)

KE=16238.26J

4 0
3 years ago
When ship running in the sea<br> so why ship does not drowned in the deep of see?
katen-ka-za [31]

Answer:

It floats

Explanation:

Boats don't drown they row.

4 0
4 years ago
Kat is creating a model to show how plate movements are linked to the formation of new rocks, and she is trying to place the ste
Jobisdone [24]
1. Tectonic plates move
2. Magma forms and rises through the cracks
3. Lava flows onto the Earth’s surface
4. Lava cools and hardens to form igneous rock
8 0
3 years ago
Let's say you have a plot for Pendulum experiment. Let's assume g for this experiment was measured (from the slope of the plot)
Gemiola [76]

Answer:

0.0052 m

Explanation:

T^2=0.021\ s^2

g = Acceleration due to gravity = 9.78 m/s²

Time period

T=2\pi \sqrt{\frac{L}{g}}\\\Rightarrow T^2=4\pi^2 \frac{L}{g}

\Rightarrow L=\frac{T^2g}{4\pi^2}

\Rightarrow L=\frac{0.021\times 9.78}{4\pi^2}\\\Rightarrow L=0.0052\ m

The length of the pendulum is 0.0052 m

3 0
3 years ago
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