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Brrunno [24]
3 years ago
15

I need help with question 26this is from my physics textbook

Physics
1 answer:
Neko [114]3 years ago
6 0

Suppose a,b,c are measured in meters. Then the semiperimeter s=\dfrac{a+b+c}2 is also given in meters, so s-a,s-b,s-c are all also given in meters. When multiplied together, (s-a)(s-b)(s-c) is measured in cubic meters. Divide this by s, measured in meters, and this reduces to square meters. Take the square root and you end up back with meters once again. So the formula is indeed dimensionally consistent.

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Help please<br> It’s kinda urgent
user100 [1]

Answer:

a = - 50 [m/s²]

Explanation:

To solve this problem we simply have to replace the values supplied in the given equation.

Vf = final velocity = 0.5 [m/s]

Vi = initial velocity = 10 [m/s]

s = distance = 100 [m]

a = acceleration [m/s²]

Now replacing we have:

(0.5)^{2}-(10)^{2} = 2*a*(100)\\0.25-10000=200*a\\200*a=-9999.75\\a =-50 [m/s^{2} ]

The negative sign of acceleration means that the ship slows down its velocity in order to land.

4 0
2 years ago
Use this formula to solve this problem:
Maslowich
Well, you gave us the formula to calculate power from work and time,
but you didn't give us the formula for work.  We have to know that.

             Work = (force) x (distance)

The work to raise Sara to the top of the hill is

               Work = (300 N) x (15 meters)

                         =   4,500 newton-meters  =  4,500 joules .

Now we're ready to use the formula that you gave us.  (Thank you.)

                Power = (work) / (time)

                            = (4,500 joules) / (10 seconds)

                            450 joules/second  =  450 watts.       
6 0
3 years ago
Read 2 more answers
Can anyone explain how tides work
Mila [183]
The position of the sun and the moon affect how high the tide is 
6 0
3 years ago
Read 2 more answers
You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics
spayn [35]

Answer:

v_0 = 3.53~{\rm m/s}

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}

For the y-direction gives

v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t

Combining both equation yields the y_component of the final velocity

v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}

6 0
3 years ago
A furlong is an old british unit of length equal to 0.125 mi, derived from the length of a furrow in an acre of ploughed land. a
castortr0y [4]

The speed of light is: c = 3x10^8 m/s <span>

or 

c = 186,000,000 miles/sec = 1.86x10^8 mi/s 

1 furlong = 0.125 mile 

1 fortnight = 2 weeks(7d/wk)(24h/d)(3600s/h) 
= 1209600s = 1.2096x10^6 s 

Therefore, 

c =1.86x10^8 mi/s(1furl/0.125mi)(1.2096x10^6s/fort) 

<span>c = 18x10^14 furlong/fortnight = 18x10^8 Mfurlong/fortnight</span></span>

4 0
3 years ago
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