All categories

3 years ago

How many significant figures are there in : (a) 0.000054 (b) 3.001 x 10^5 (c) 5.600

Physics

1 answer:

melomori [17]3 years ago

4 0

Answer:

(a) 2 (b) 4 (c) 4

Explanation:

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.
All non-zero numbers are always significant. For example: 654, 6.54 and 65.4 all have three significant figures.
All zero’s between integers are always significant. For example: 5005, 5.005 and 50.05 all have four significant figures.
All zero’s preceding the first integers are never significant. For example: 0.0078 has two significant figures.
All zero’s after the decimal point are always significant. For example: 4.500, 45.00 and 450.0 all have four significant figures.
All zeroes used solely for spacing the decimal point are not significant. For example : 8000 has one significant figure.

As per question,

0.000054 has 2 significant figures.

3.001 x 10⁵ has 4 significant figures.

5.600 has 4 significant figures.

You might be interested in

Find V. in the circuit of the following figure:

Angelina_Jolie [31]

Answer:

where is the figure

Explanation:

8 0

3 years ago

NaOH + FeCl3* Na Cl + Fe 10H)3<br> balanced

zvonat [6]

3NaOH + FeCl3 → 3NaCl + Fe(OH)3

8 0

3 years ago

A bowling ball with a momentum of 18kg-m/s strikes a stationary bowling pin. After the collision, the ball has a momentum of 13k

Veronika [31]

Answer:

$14.98\ \text{kg m/s}$

$45.26^{\circ}$

Explanation:

$P_1$ = Initial momentum of the pin = 13 kg m/s

$P_i$ = Initial momentum of the ball = 18 kg m/s

$P_2$ = Momentum of the ball after hit

$55^{\circ}$ = Angle ball makes with the horizontal after hitting the pin

$\theta$ = Angle the pin makes with the horizotal after getting hit by the ball

Momentum in the x direction

$P_i=P_1\cos55^{\circ}+P_2\cos\theta\\\Rightarrow P_2\cos\theta=P_i-P_1\cos55^{\circ}\\\Rightarrow P_2\cos\theta=18-13\cos55^{\circ}\\\Rightarrow P_2\cos\theta=10.54\ \text{kg m/s}$

Momentum in the y direction

$P_1\sin55=P_2\sin\theta\\\Rightarrow P_2\sin\theta=13\sin55^{\circ}\\\Rightarrow P_2\sin\theta=10.64\ \text{kg m/s}$

$(P_2\cos\theta)^2+(P_2\sin\theta)^2=P_2^2\\\Rightarrow P_2=\sqrt{10.54^2+10.64^2}\\\Rightarrow P_2=14.98\ \text{kg m/s}$

The pin's resultant velocity is $14.98\ \text{kg m/s}$

$P_2\sin\theta=10.64\\\Rightarrow \theta=sin^{-1}\dfrac{10.64}{14.98}\\\Rightarrow \theta=45.26^{\circ}$

The pin's resultant direction is $45.26^{\circ}$ below the horizontal or to the right.

4 0

3 years ago

An alpha particle (α), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an elect

beks73 [17]

Answer:

Speed of the alpha particle is $v=1.8180\times 10^3m/sec$

Explanation:

We have given charge on alpha particle $q=3.2\times 10^{-19}C$

Mass of the alpha particle $m=6.68\times 10^{-27}kg$

Potential difference $V=-3.45\times 10^{-3}volt$

We have to find the speed of the alpha particle

From energy conservation we know that

$\frac{1}{2}mv^2=qV$

$\frac{1}{2}\times 6.68\times 10^{-27}\times v^2=3.2\times 10^{-19}\times 3.45\times 10^{-3}$

$v=1.8180\times 10^3m/sec$

4 0

3 years ago

An electron is accelerated by a 3.6 kv potential difference. the charge on an electron is 1.60218 × 10−19 c and its mass is 9.10

katen-ka-za [31]

By definition, the potential energy is:
U = qV
Where,
q: load
V: voltage.
Then, the kinetic energy is:
K = mv ^ 2/2
Where,
m: mass
v: speed.
As the power energy is converted into kinetic energy, we have then:
U = K
Equating equations:
qV = mv ^ 2/2
From here, we clear the speed:
v = root (2qV / m)
Substituting values we have:
v = root ((2 * (1.60218 × 10 ^ -19) * 3600) /9.10939×10^-31))
v = 3.56 × 10 ^ 7 m / s
Then, the centripetal force is:
Fc = Fm
mv ^ 2 / r = qvB
By clearing the magnetic field we have:
B = mv / qr
Substituting values:
B = (9.10939 × 10 ^ -31) * (3.56 × 10 ^ 7) / (1.60218 × 10 ^ -19) * 0.059
B = 3.43 × 10 ^ -3 T
Answer:
A magnetic field that must be experienced by the electron is:
B = 3.43 × 10 ^ -3 T

6 0

3 years ago