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3 years ago

Convert the unit of 0.00023 kilograms into grams. (Answer in scientific notation)

Physics

2 answers:

Rus_ich [418]3 years ago

4 0

Answer:

2.3 × $10^{-1}$

Explanation:

1 kg = 1000 g.

0.00023 kg x 1000 g = 0.23 grams

Irina-Kira [14]3 years ago

3 0

Answer:

0.23×10⁴

Explanation:

kilogram to gram ÷ 1000

0.00023kg ÷ 1000

=0.23g

scientific notation=0.23×10⁴

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Running at 2 m/s, Bruce the 45 kg quarterback, collides with Biff, the 90 kg tackle, who is traveling at 7 m/s in the other dire

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Given data

*The mass of Bruce is m_1 = 45 kg

*The initial velocity of the Bruce is u_1 = 2 m/s

*The mass of the biff is m_2 = 90 kg

*The initial velocity of the Biff is u_2 = -7 m/s

*The final velocity of the first glider is v_2 = -1 m/s

According to the law of conservation of linear momentum, the total linear momentum of a system remains constant

Applying the law of conservation of momentum as

$\begin{gathered} p_i=p_f \\ m_1u_1+m_2u_2=m_1v_1+m_2v_2 \\ v_1=\frac{m_1u_1+m_2u_2-m_2v_2_{}_{}_{}_{}}{m_1} \end{gathered}$

Substitute the known values in the above expression as

$\begin{gathered} v_1=\frac{(45)(2)+(90)(-7)-(90)(-1)}{45} \\ =-10\text{ m/s} \end{gathered}$

Hence, the speed of the bruce knock backwards is v_1 = -10 m/s

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2 years ago

Describe the conditions of an continental tropical air mass

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3 years ago

A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find

sukhopar [10]

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge $q_{1}=9\times10^{-6}\ C$ at origin

Second charge $q_{2}=6\times10^{-6}\ C$

Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

$E=\dfrac{kq}{r^2}$

$0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}$

$\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}$

$\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1$

$\dfrac{1}{d}=1.82$

$d=\dfrac{1}{1.82}$

$d=0.55\ m$

Hence, The coordinates of the point is (0,0.55).

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3 years ago

Water flows through a water hose at a rate of Q1 = 860 cm3/s, the diameter of the hose is d1 = 1.85 cm. A nozzle is attached to

Snowcat [4.5K]

Answer:

a) A1 = $\frac{\pi (d1)^{2} }{4}$

b) A1 = 2.688 cm $^{2}$

c) Q1 = A1 x v1

d) v1 = 3.1994 m/s

e) A2 = $\frac{A1 X v1}{v2}$

f) A2 = 0.7963cm $^{2}$

Explanation:

a) Area = $\pi r^{2}$

r = $\frac{d}{2}$

thus,

area = $\pi (\frac{d}{2})^{2}$

A1 = $\frac{\pi (d1)^{2} }{4}[/tex]b) d1 = 1.85 cmsubstituting in the above equation,A1 = [tex]\frac{\pi (d1)^{2} }{4}$

A1 = $\frac{\pi (1.85)^{2} }{4}$

A1 = 2.688 cm $^{2}$

c) Flow rate = Area x velocity ( refer brainly.com/question/13997998)

Q1 = A1 x v1

d) From the above equation,

v1 = $\frac{Q1}{A1}$ = $\frac{860}{2.688}$ = 319.94 cm/s = 3.1994 m/s

e) Since the flow rate Q1 is constant throughout the hose, Av is a constant.

i.e. A1 x v1 = A2 x v2

thus,

A2 = $\frac{A1 X v1}{v2}$

f) v2 = 10.8 m/s.

substituting the values in the above equation,

A2 = $\frac{2.688 X 3.1994}{10.8}$ = 0.7963cm $^{2}$

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4 years ago