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3 years ago

Which mass is the same as 0.031 g?

Physics

2 answers:

geniusboy [140]3 years ago

8 0

Answer:

0.000031 kg

Explanation:

There are 1000 g in 1 kg, so

0.031 g / 1000 g = 0.000031 kg

LenaWriter [7]3 years ago

6 0

0.031 g is equal to 31 grams

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Answer:

Explanation:

from the question we are told that

Load $L=15kg$

Force $F=70N$

Angle of inclination $=20.0 degres$

Displacement $m=5 meters$

coefficient of kinetic friction $\alpha =0.300$

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3 years ago

Find the components to write this vector in unit vector notation: 63.5 A please help

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j component is 63.5

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3 years ago

A particle moves along the x axis. It is initially at the position 0.270 m, moving with velocity 0.140 m/s and acceleration 20.3

mixas84 [53]

Answer:

a) -2.34 m

b) -1.3 m/s

c) 0.056 m

d) 0.320 m/s

Explanation:

part a

Given:

s(0) = 0.27 m

v(0) = 0.14 m/s

a = -0.320 m/s^2

t = 4.50s

Using kinematic equation of motion for constant acceleration:

s (t) = s(0) + v(0)*t + 0.5*a*t^2

s ( 4.5 s ) = 0.27 + 0.14*4.5 + 0.5*-0.32*4.5^2

s(4.5) = -2.34 m

part b

Using kinematic equation of motion for constant acceleration:

v(t) = v(0) + a*t

v(4.5s) = (0.14) + (-0.32)(4.5) = -1.3 m/s

part c

Use equation for simple harmonic motion:

s(t) = A*cos(w*t)

v(t) = -A*w*sin (w*t)

a(t) = -A*(w)^2 * cos (w*t)

0.27 m = A*cos(w*t) .... Eq 1

0.14 m/s = - A*w*sin (w*t) .....Eq2

-0.320 = -A*(w)^2 * cos (w*t) .... Eq3

Solve the three equations above for A, w, and t

Divide 1 and 3:

w^2 = 0.32 / 0.27

w = 1.0887 rad / s

Divide 2 and 1:

w*tan(wt) = 0.14 / 0.27

tan(1.0887*t) = 0.476289

t = 0.4083 s

A = 0.27 / cos (1.0887*0.4083) = 0.3 m

Hence, the SHM is s(t) = 0.3*cos(1.0887*t)

s(4.5) = 0.3*cos(1.0887*4.5) = 0.056 m

part d

v(t) = - 0.32661 * sin (1.0887*t)

v(4.5) = - 0.32661 * sin (1.0887*4.5) = 0.320 m/s

3 0

3 years ago

Marvin uses a long copper wire with resistivity 1.68 x 10^-8 Ω⋅m and diameter 1.00 x 10^−3 m to create a solenoid that has a 3

Mazyrski [523]

Answer with Explanation:

We are given that

Resistivity of copper wire= $\rh0=1.68\times 10^{-8}\Omega m$

Diameter=d= $1.00\times 10^{-3} m$

Radius of copper wire= $r=\frac{d}{2}=\frac{1}{2}\times 10^{-3} m$

Radius of solenoid=r' $=3 cm=3\times 10^{-2} m$

1 m=100 cm

a.Length of wire=l=11.3 m

Area of wire=A= $\pi r^2$

Where $\pi=3.14$

A= $3.14\times (\frac{1}{2}\times 10^{-3})^2$

Resistance, R= $\rho \frac{l}{A}$

Using the formula

$R=1.68\times 10^{-8}\times\frac{11.3}{3.14\times (\frac{1}{2}\times 10^{-3})^2}$

$R=0.24\Omega$

B.Length of solenoid= $2\pi r'=2\times 3.14\times 3\times 10^{-2}=0.188$ m

Number of turns= $n_0=\frac{l}{2\pi r'}=\frac{11.3}{0.188}$

$n_0$ =60

C.Potential difference,V=3 V

Current,I= $\frac{V}{R}$

I= $\frac{3}{0.24}=12.5 A$

D.Total length =0.1 m

Number of turns per unit length,n= $\frac{60}{0.1}=600$

Magnetic field along central axis inside of the solenoid,B= $\mu_0 nI$

$B=4\pi\times 10^{-7}\times 12.5\times 600=9.42\times 10^{-3} T$

4 0

3 years ago

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