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4 years ago

Estimate the water volume in the graduated cylinder to the nearest 0.1 mL.

Physics

1 answer:

earnstyle [38]4 years ago

5 0

Answer:

With a 50-mL graduated cylinder, read and record the volume to the nearest 0.1 mL. The 10-mL graduated cylinder scale is read to the nearest 0.01 mL and the 500-mL graduated cylinder scale is read to the nearest milliliter (1 mL). A buret is a scaled cylindrical tube attached to a stopcock, or valve.

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a track coach with a meter stick and a stopwatch is trying to determine if a student is walking with constant speed.he should...

Aleksandr [31]

Give you a device that tracks your speed?

7 0

3 years ago

Please help me it's URGENT (#10 btw and also how do I solve for time

Afina-wow [57]

Answer:

t = 12s

Explanation:

Given:

v-initial = 0 m/s

x = 360 m

a = 5.0 m/s^2

Solve:

x = (v-initial)t + 1/2(a*t^2)

360 = 0t + 1/2 (5.0t^2)

360 = 2.5 t^2

144 = t^2

t = sqrt(144) = 12

Therefore, it takes 12 seconds.

5 0

3 years ago

Am arrow of mass 1000kg is shot into a wooden block of mass 5000lg lying at rest in a smooth surface.If the arrow travels 15m/s

Ber [7]

Answer:

Vf=3

Explanation:

you must first write your data

data before impact

M1=1000 M2=5000

V1=0 m/s V2 =0m/s

data after impact

M1=1000 M2=5000

V1=15m/s V2=?

M1V1 +M2V2=M1V1 +M2V2f

(1000)(0)+(5000)(0)=(1000)(15)+(5000)Vf

0=15000+5000Vf

- 15000÷5000=5000Vf÷5000

Vf= -3

Vf =3

6 0

3 years ago

There are Z protons in the nucleus of an atom, where Z is the atomic number of the element. An α particle carries a charge of +2

qaws [65]

Answer:

$r = 2.84 \times 10^{-14} m$

Explanation:

As per energy conservation we know that the electrostatic potential energy of the charge system is equal to the initial kinetic energy of the alpha particle

So here we can write it as

$\frac{1}{2}mv^2 = \frac{k(2e)(ze)}{r}$

now we know that

$m = 1.67 \times 10^{-27} kg$

$e = 1.6 \times 10^{-19} C$

z = 79

here kinetic energy of the incident alpha particle is given as

$KE = 6.4 \times 10^{-13} J$

now we have

$6.4 \times 10^{-13} = \frac{(9\times 10^9)(1.6 \times 10^{-19})^2(79)}{r}$

now we have

$r = 2.84 \times 10^{-14} m$

7 0

3 years ago

particle of mass 59 g and charge 51 µC is released from rest when it is 32 cm from a second particle of charge −14 µC. Determine

AleksAgata [21]

Answer:

The initial acceleration of the 59g particle is $1062.7\frac{m}{s^{2}}$

Explanation:

Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:

$F=ma$ (1)

We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:

$F=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}$

with q1 and q2 the charge of the particles, r the distance between them and k the constant $k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}$ . So:

$F=(9.0\times10^{9})\frac{\mid (51\times10^{-6})(-14\times10^{-6})\mid}{0.32^{2}}$

$F=62.7 N$

Using that value on (1) and solving for a

$a=\frac{F}{m}=\frac{62.7}{0.059}=1062.7\frac{m}{s^{2}}$

4 0

4 years ago