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DanielleElmas [232]

3 years ago

7

which of the following laboratory tool would be most appropiate for measuring the approximate volume of a liquid?

1 answer:

SIZIF [17.4K]3 years ago

7 0

Beaker would be most appropriate for measuring the approximate volume of a liquid.

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Find the resistance of an electric light bulb if a current of 0.08 A flows when the potential difference across the bulb is 120

lora16 [44]

U=RI Ohm's law
then R=U/I
=120/0.08
=2250Ω
hope this helps you

8 0

3 years ago

A dump truck and a little toy matchbox car have the same velocity. which one has the greater momentum?

Zina [86]

That would be the dump truck. Momentum depends on how heavy a certain object is in motion. The more weight it has the harder it is to stop.

4 0

3 years ago

A 1,100 kg car is traveling initially 20 m/s when the brakes are applied. The brakes apply a constant force while bringing the c

Natalka [10]

Answer:

Work done = -220,000 Joules.

Explanation:

<u>Given the following data;</u>

Mass = 1100kg

Initial velocity = 20m/s

To find workdone, we would calculate the kinetic energy possessed by the car.

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

$K.E = \frac{1}{2}MV^{2}$

Where,

K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

Substituting into the equation, we have;

$K.E = \frac{1}{2}*1100*20^{2}$

$K.E = 550*400$

K.E = 220,000J

Therefore, the workdone to bring the car to rest would be -220,000 Joules because the braking force is working to oppose the motion of the car.

4 0

2 years ago

A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block

Usimov [2.4K]

Answer:

a) $v'=-0.227\ m.s^{-1}$

b) $v=1.36\ m.s^{-1}$

Explanation:

Given:

mass of the lighter block, $m'=0.02\kg$

velocity of the lighter block, $u'=0.68\ m.s^{-1}$

mass of the heavier block, $m=0.04\ kg$

velocity of the heavier block, $u=0\ m.s^{-1}$

a)

Using conservation of linear momentum:

$m'.u'+m.u=m'.v'+m.v$

where:

$v'=$ final velocity of the lighter block

$v=$ final velocity of the heavier block

$m'.u'=m'.v'+m.v$

$m'(u'-v')=m.v$ ........................(1)

Since kinetic energy is conserved in elastic collision:

$\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2$

$m'(u'^2-v'^2)=m.v^2$

$m'(u'-v')(u'+v')= m.v^2$

divide the above equation by eq. (1)

$v=u'+v'$ .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

$m'(u'-v')=m(u'+v')$

$\frac{m'+m}{m'-m} =\frac{u'}{v'}$

$\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}$

$v'=-0.227\ m.s^{-1}$ (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

$m'(u'-v+u')=m.v$

$2m'.u'=(m-m')v$

$2\times 0.02\times 0.68=(0.04-0.02)\times v$

$v=1.36\ m.s^{-1}$

6 0

3 years ago

Select the correct answer.

spin [16.1K]

Answer:

There isnt enough in your question to answer the question bro, like we need a picture or something bro.

Explanation:

7 0

3 years ago