Ask question

Ask question

All categories

3 years ago

7

Question 1 of 15

1 answer:

stepan [7]3 years ago

8 0

Answer:

significant

Explanation:

The digits in a measurement that are considered significant are all of those digits that represent marked calibrations on the measuring device plus one additional digit to represent the estimated digit (tenths of the smallest calibration).

You might be interested in

4. How much would an object accelerate if it has a mass of 25 kg and is pushed with a force of 2 Newtons?

slega [8]

Answer:

0.4

Explanation:

f divided by m = a

3 0

3 years ago

A boat that has a speed of 6km / h must cross a 200m wide river perpendicular to the current that carries a speed of 1m / s. Cal

ivanzaharov [21]

Answer:

a) 1.94 m/s

b) 120 m

Explanation:

Convert km/h to m/s:

6 km/h = 1.67 m/s

a) The final speed is found with Pythagorean theorem:

v = √((1.67 m/s)² + (1 m/s)²)

v = 1.94 m/s

b) The time it takes the boat to cross the river is:

t = (200 m) / (1.67 m/s)

t = 120 s

The displacement in the direction of the current is:

x = (1 m/s) (120 s)

x = 120 m

3 0

3 years ago

A block of mass 5.6 kg is attached to a horizontal spring on a rough floor. Initially the spring is compressed 3.5 cm. The sprin

Mariana [72]

Answer:

The velocity of block = 0.188 $\frac{m}{s}$

Explanation:

Mass m = 5.6 kg

k = 1040 $\frac{N}{m}$

$\mu$ = 0.26

$x_{1} =$ 0.035 m , $v_{1}$ = 0

$x_{2} =$ 0.02 m

From work energy theorem

$K_{1} + U_{1} + W_{other} = K_{2} + U_{2}$ --------- (1)

Kinetic energy

$K = \frac{1}{2} k x^{2}$ ------- (1)

Potential energy

$U = \frac{1}{2} k x^{2}$ ------- (2)

Work done

W = F.s ------ (3)

From Newton's second law

$R_{N}$ = mg

$R_{N}$ = 5.6 × 9.81 = 54.9 N

Friction force = 0.4 × 54.9 = 21.9 N

Now the work done by the friction

$W_{f}$ = - 21.9 × 0.015

$W_{f}$ = - 0.329 J

Now kinetic energy

At point 1

$K_{1} = \frac{1}{2} m v^{2} _{1}$

$K_{1} = 0$

$U_{1} = \frac{1}{2} k x^{2}$

$U_{1} = \frac{1}{2} (1040) 0.035^{2}$

$U_{1} =$ 0.637 J

At point 2

$K_{2} = \frac{1}{2} (5.6) v^{2} _{2}$

$K_{2} = 2.8 v_{2} ^{2}$

Potential energy

$U_{2} = \frac{1}{2} k x_2^{2}$

$U_{2} = \frac{1}{2} (1040) 0.02^{2}$

$U_{2} = 0.208$ J

From equation (1) we get

0 + 0.637 - 0.329 = 2.8 $v_{2} ^{2}$ + 0.208

2.8 $v_{2} ^{2}$ = 0.1

$v_{2} =$ 0.188 $\frac{m}{s}$

This is the velocity of block.

6 0

4 years ago

a safety hazard that can occur from having a battery attached for more than a short period of time causes...

My name is Ann [436]

It could shock you since the battery could overheat

4 0

3 years ago

A bus is designed to draw its power from a rotating flywheel that is brought up to 3000 rpm by an electric motor. The flywheel i

Arada [10]

To solve this problem we will apply the concept of rotational kinetic energy. Once this energy is found we will proceed to find the time from the definition of the power, which indicates the change of energy over time. Let's start with the kinetic energy of the rotating flywheel is

$E_r = \frac{1}{2} I\omega^2$

Here

I = moment of inertia

$\omega =$ Angular velocity

Here we have that,

$\omega = 3000\frac{rev}{min}(\frac{2\pi rad}{1rev})(\frac{1min}{60s})$

$\omega = 314.159rad/s$

Replacing the value of the moment of inertia for this object we have,

$E_r = \frac{1}{2} (\frac{MR^2}{2})\omega^2$

$E_r = \frac{1}{2} (\frac{2000(0.5)^2}{2})(314.159)^2$

$E_r = 1.233698*10^7J$

The expression for average power is

$P = \frac{E_r}{\Delta t}$

$\Delta t = \frac{E_r}{P}$

$\Delta t = \frac{1.233698*10^7}{20*10^3}$

$\Delta t = 616.8s \approx 620s$

Therefore the correct answer is 620s.

3 0

4 years ago