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Illusion [34]
3 years ago
14

Measure the length of the arrow in centimeters using correct significant figures.

Physics
2 answers:
UkoKoshka [18]3 years ago
8 0

Answer:

The significant figures is 3.3 cm.

Explanation:

Significant :

Significant figures of a number are numbers that have significance to contribute to its resolution of measurements.

Centimeter is unit of length.

According to figure,

The significant figures is

l = 3.3\ cm

Hence, The significant figures is 3.3 cm.

Zinaida [17]3 years ago
6 0

Based on the very tip of the arrow the best answer would be; 3.3cm

But i could very well be wrong and it may be 3.35, but i would say 3.3 if it just wants to the nearest 10th

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A 3.91 kg cart is moving at 5.7 m/s when it collides with a 4 kg cart which was at rest. They collide and stick together.
Nesterboy [21]

Answer:

<em>The velocity after the collision is 2.82 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of two bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The common velocity after this situation is:

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.

After the collision, both cars stick together. Let's compute the common speed after that:

\displaystyle v'=\frac{3.91*5.7+4*0}{3.91+4}

\displaystyle v'=\frac{22.287}{7.91}

\boxed{v' = 2.82\ m/s}

The velocity after the collision is 2.82 m/s

6 0
3 years ago
Which measurement is equivalent to 350 milliliters?
DENIUS [597]
That is 0.35 liters. i hope that is the correct answer? 
5 0
4 years ago
Read 2 more answers
If a marathon runner averages 9.51 mi/hr, how long does it take him to run a 26.220-mile marathon. Express your answers in hours
ASHA 777 [7]

Explanation:

Speed of the marathon runner, v = 9.51 mi/hr

Distance covered by the runner, d = 26.220 mile

Let t is the time taken by the marathon runner. We know that the speed of the runner is given by total distance divided by total time taken. Mathematically, it is given by :

v=\dfrac{d}{t}

t=\dfrac{d}{v}

t=\dfrac{26.220\ mi}{9.51\ mi/hr}

t = 2.75 hours

Since, 1 hour = 60 minutes

t = 165 minutes

Since, 1 minute = 60 seconds

t = 9900 seconds

Hence, this is the required solution.

7 0
4 years ago
The x coordinate of an electron is measured with an uncertainty of 0.200 mm . What is vx, the x component of the electron's velo
Vikentia [17]

Answer:

Velocity of electron along x direction is 57.9 m/s

Explanation:

The uncertainty in x coordinate of electron, Δx = 0.200 mm = 0.2 x 10⁻³ m

Let vₓ be the x component of electrons velocity.

The uncertainty in x component of electrons momentum is:

Δpₓ = mΔvₓ

Here m is mass of the electron.

The uncertainty in velocity x component is 1% i.e. 0.01.

So, the above equation can be written as :

Δpₓ = 0.01mvₓ    ....(1)

The minimum uncertainty principle is:

\Delta x\Delta p_{x} = \frac{h}{2\pi }    ....(2)

Here h is Planck's constant.

From equation (1) and (2),

\Delta x\times0.01m v_{x} = \frac{h}{2\pi }

Substitute 0.2 x 10⁻³ m for Δx, 9.1 x 10⁻³¹ kg for m and 6.626 x 10⁻³⁴ m²kg/s in the above equation.

0.2\times10^{-3} \times0.01\times9.1\times10^{-31}\times v_{x} = \frac{6.626\times10^{-34} }{2\pi }

vₓ = 57.9 m/s

8 0
3 years ago
What is dark energy?
olga2289 [7]

Explanation:

Dark Energy. Dark Energy is a hypothetical form of energy that exerts a negative, repulsive pressure, behaving like the opposite of gravity. It has been hypothesised to account for the observational properties of distant type Ia supernovae, which show the universe going through an accelerated period of expansion

6 0
3 years ago
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