Ask question

Ask question

All categories

2 years ago

15

Multiply the following numbers, using scientific notation and the correct amount of significant digits. 1.003 m⋅3.09 = _____ 3.0

99 m 3.10 m 3.0993 m 3.1 m

2 answers:

Misha Larkins [42]2 years ago

6 0

Answer:

3.10 m

Explanation:

When we multiply or divide there is a rule for significant digits given below:

“The answer will be rounded off to the shortest number of significant digits among the numbers which are being multiplied or divided”.

When we multiply 1.003 m and 3.10, we get 3.1093 m. But we have round it off according to the rule up to 3 significant digits because one of the numbers has 3 significant digits. Hence, answer will be 3.10m.

lozanna [386]2 years ago

3 0

Answer

correct answer is 3.10

Explanation:

in this question we have to multiply two numbers 1.003 and 3.09.

1.003 has 4 significant digits and 3.09 has 3 significant digits so answer must have 3 significant digits.

$1.003\times 3.09=3.09927$

$1.003\times 3.09=3.10$

Hope it will help you

You might be interested in

A man walks along a straight path at a speed of 4 ft/s. A searchlight is located on the ground 6 ft from the path and is kept fo

BARSIC [14]

We are given that,

$\frac{dx}{dt} = 4ft/s$

We need to find $\frac{d\theta}{dt}$ when $x=8ft$

The equation that relates x and $\theta$ can be written as,

$\frac{x}{6} tan\theta$

$x = 6tan\theta$

Differentiating each side with respect to t, we get,

$\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}$

$\frac{dx}{dt} = (6sec^2\theta)\cdot \frac{d\theta}{dt}$

$\frac{d\theta}{dt} = \frac{1}{6sec^2\theta} \cdot \frac{dx}{dt}$

Replacing the value of the velocity

$\frac{d\theta}{dt} = \frac{1}{6} cos^2\theta (4)^2$

$\frac{d\theta}{dt} = \frac{8}{3} cos^2\theta$

The value of $cos \theta$ could be found if we know the length of the beam. With this value the equation can be approximated to the relationship between the sides of the triangle that is being formed in order to obtain the numerical value. If this relation is known for the value of x = 6ft, the mathematical relation is obtained. I will add a numerical example (although the answer would end in the previous point) If the length of the beam was 10, then we would have to

$cos\theta = \frac{6}{10}$

$\frac{d\theta}{dt} = \frac{8}{3} (\frac{6}{10})^2$

$\frac{d\theta}{dt} = \frac{24}{25}$

Search light is rotating at a rate of 0.96rad/s

4 0

2 years ago

A person carries a mass of 10 kg and walks along the +x-axis for a distance of 100m with a constant velocity of 2 m/s. What is t

gizmo_the_mogwai [7]

Since the direction of the force and the direction of the path is perpendicular, the person is not doing any physical work.

3 0

2 years ago

Read 2 more answers

Which atomic model proposed that electrons move in specific orbits around the nucleus of an atom

BabaBlast [244]

Bohr's atomic model

8 0

2 years ago

On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

irinina [24]

Answer:

$v_a=0.8176 m/s$

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is

$M_1=m_av_a+m_bv_b$

Since b is initially at rest

$M_1=m_av_a$

After the collision and being $v'_a$ and $v'_b$ the respective velocities, the total momentum is

$M_2=m_av'_a+m_bv'_b$

Both momentums are equal, thus

$m_av_a=m_av'_a+m_bv'_b$

Solving for $v_a$

$v_a=\frac{m_av'_a+m_bv'_b}{m_a}$

$v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}$

$v_a=0.8176 m/s$

The initial kinetic energy can be found as (provided puck b is at rest)

$K_1=\frac{1}{2}m_av_a^2$

$K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J$

The final kinetic energy is

$K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2$

$K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J$

The change of kinetic energy is

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

3 0

2 years ago

MATCH THESE ^-^ Match Newton's law with the correct statement.

ikadub [295]

1 and A
2 and B
3 and D
4 and C

4 0

2 years ago

Read 2 more answers