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3 years ago

You are asked to determine the density of a liquid. What lab equipment is most helpful to find density of the liquid?

Physics

2 answers:

crimeas [40]3 years ago

7 0

The answer is D, the graduated cylinder and a triple beam-balance

AnnyKZ [126]3 years ago

6 0

D. The graduated cylinder is used to find the volume, and triple beam balance is used to find the mass.

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The brakes of a 125 kg sled are applied while it is moving at 8.1 m/s, which exerts a force of 261 N to slow the sled down. How

sammy [17]

Answer:

15.7 m

Explanation:

m = mass of the sled = 125 kg

v₀ = initial speed of the sled = 8.1 m/s

v = final speed of sled = 0 m/s

F = force applied by the brakes in opposite direction of motion = 261

d = stopping distance for the sled

Using work-change in kinetic energy theorem

- F d = (0.5) m (v² - v₀²)

- (261) d = (0.5) (125) (0² - 8.1²)

d = 15.7 m

6 0

3 years ago

A roller coaster car rapidly picks up speed as it rolls down a slope as it starts down the slope its speed is 4m/s but 3 seconds

Gwar [14]

Answer:

The acceleration is 6 [m/s^2]

Explanation:

We can find the acceleration of the roller coaster using the kinematic equation for uniformly accelerated motion.

$v_{f} =v_{i} + a*t\\where:\\v_{f} = final velocity = 22 [m/s]\\v_{i} = initial velocity = 4 [m/s]\\t = time = 3 [s]\\$

Now replacing the values we have:

$a=\frac{v_{f} - v_{i} }{t} \\a=\frac{22 - 4 }{3}\\a = 6 [m/s^{2} ]$

3 0

3 years ago

Which pair of equations can describe the path of a particle moving with an acceleration that is perpendicular to the velocity of

antoniya [11.8K]

Answer:

The question clearly describes the circular motion.

The circular motion equation is

$a_{radial} = \frac{v^2}{r}$

The path of the particle is circular.

Explanation:

In circular motion, the radial acceleration is always towards the center and constant in magnitude. Furthermore, the velocity of the circular motion is always tangential to the circle, that is it is always perpendicular to the radius, hence the acceleration.

6 0

3 years ago

The average distance from the sun to Pluto is approximately 6.10 × 109 km. How long does it take light from

Scorpion4ik [409]

$V= \frac{S}{t}$
$t= \frac{S}{V}$ <u />
$t= \frac{S}{c}$
$t= \frac{6.1*10^{12}}{299792458}$
$t=20347.4098071s$

It takes 20347.4098071s for light from the sun to reach Pluto.
The 6.1*10^9 is replaced by 6.1*10^12 on line 4 because we convert the distance from km to m.
c = speed of light. If a different value was given in the previous question then use that instead of the value I used to do the final calculation.

3 0

4 years ago

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$