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mojhsa [17]
3 years ago
6

An empty glass beaker has a mass of 103 g. When filled with water, it has a total mass of 361g.

Physics
1 answer:
aivan3 [116]3 years ago
7 0

Answer:

0.96 gcm¯³

Explanation:

From the question given above, the following data were obtained:

Mass of empty beaker = 103 g

Mass of beaker + water = 361 g

Mass of beaker + oil = 351 g

Density of water = 1 gcm¯³

Density of cooking oil =?

Next, we shall determine the mass of water. This can be obtained as follow:

Mass of empty beaker = 103 g

Mass of beaker + water = 361 g

Mass of water =?

Mass of water = (Mass of beaker + water) – (Mass of empty beaker)

Mass of water = 361 – 103

Mass of water = 258 g

Next, we shall determine the volume of the beaker. This can be obtained by calculating the volume of water in the beaker.

Density of water = 1 gcm¯³

Mass of water = 258 g

Volume of water =?

Density = mass /volume

1 = 258 / volume

Cross multiply

1 × volume = 258

Volume of water = 258 cm³

Thus the volume of the beaker is 258 cm³.

Next, we shall determine the mass of the cooking oil. This can be obtained as follow:

Mass of empty beaker = 103 g

Mass of beaker + oil = 351 g

Mass of cooking oil =?

Mass of cooking oil = (Mass of beaker + oil) – (Mass of empty beaker)

Mass of cooking oil = 351 – 103

Mass of cooking oil = 248 g

Finally, we shall determine the density of the cooking oil. This can be obtained as follow:

Mass of cooking oil = 248 g

Volume of the beaker = 258 cm³

Density of cooking oil =?

Density = mass / volume

Density = 248 / 258

Density of cooking oil = 0.96 gcm¯³

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Dovator [93]

The mechanical energy for the first and the second ball is

10 ^{7}  \: joules.

Mass of the first ball = 20 kg

The initial speed at which a cannonball is fired from a cannon =1000 m/s

The angle made by the cannonball while being fired from the cannon = 37°

The maximum height reached by the first ball is,

=   \frac{ u {}^{2} _{1}sin {}^{2} θ}{2g}

=    \frac{ {1000}^{2} sin {}^{2}37°}{2 \times 9.8}

= 18478.69 \: m

The maximum height of the first cannonball is 17478.69 m.

The initial speed at which a cannonball is fired from a cannon =1000 m/s

The angle made by the cannonball while being fired from the cannon = 90 °

=   \frac{ u {}^{2} _{2}sin {}^{2} θ}{2g}

=   \frac{ 1000{}^{2}sin^{2} 90°}{2 \times 9.8}[tex] = 51020.41 \: m

For the first ball, total mechanical energy= Potential energy at maximum height + kinetic energy at the maximum height

So, the total mechanical energy is,

= mgh \: + \frac{1}{2}mv {}^{2} _{x}[/tex]

= 20 \times 9.8 \times 18478.64  \times  \frac{20}{2} (1000 \: cos37 °)

= 10 ^{7}  The potential energy at the maximum height, = m _{2}gh

= 20 \times 9.8  \times 51020.41

= 10 ^{7} \:J

Therefore, the total mechanical energy for the first and the

\:second \:  cannonball \:  is  \: 10 ^{7}  \:joules.

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2 years ago
A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and a
Zolol [24]
     Considering the unknown resistence as R and using the Ohm's First Law, we have:

i= \frac{V}{R_{eq}}  \\ 0.5= \frac{12}{R+10}  \\ R+10=24 \\ R=14-Ohm
 
     The equivalent resistence is given by the resistor series with the lamp resistence.

R_{eq}=R+10 \\ R_{eq}=14+10 \\ \boxed {R_{eq}=24-Ohm}

If you notice any mistake in my english, please let me know, because i am not native.

6 0
2 years ago
| A 1.0 kg stone is dropped from a bridge 100 m above a canyon. What will be the kinetic energy of the stone after it
Mnenie [13.5K]

Answer:

Option D

490 J

Explanation:

When at a height of 100 am above and released, the ball initially posses only potential energy. When it falls, some potential energy is converted to kinetic energy.

Initial potential energy= mgh where m is the mass, g is the acceleration due to gravity and h is height. Substituting 1 Kg for m, 9.81 for g and 100 m for h then

PE initial = 1*9.81*100= 981 J

At 50 m, PE will be 1*9.81*50=490.5 J

Subtracting PE at 50 m from initial PE we get the energy that has been converted to kinetic energy hence

981-490.5= 490.5 J

Approximately, 490 J

8 0
3 years ago
A 9.0-kg bowling ball on a horizontal, frictionless surface experiences a net force of 6.0 n. what will be its acceleration?
Vladimir [108]

This question involves the concepts of Newton's Second Law of Motion.

The acceleration of the bowling ball will be "0.67 m/s²".

<h3>Newton's Second Law of Motion</h3>

According to Newton's Second Law of Motion, when an unbalanced force is applied on an object, it produces an acceleration in it, in the direction of the applied force. This acceleration is directly proportional to the force applied and inversely proportional to the mass of the object. Mathematically,

F=ma\\\\a=\frac{F}{m}

where,

  • a = acceleration = ?
  • F = Magnitude of the applied force = 6 N
  • m = Mass of the ball = 9 kg

Therefore,

a=\frac{6\ N}{9\ kg}

a = 0.67 m/s²

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2 years ago
The energy flow per unit time per unit area (S) of an electromagnetic wave has an average value of 310 mW/m2. The maximum value
Aleksandr-060686 [28]

Answer:

5.096*10^-8

Explanation:

Given that

The average value of the electromagnetic wave is 310 mW/m²

To find the maximum value of the magnetic field the wave is closest to, we say

Emax = √Erms

Emax = √[(2 * 0.310 * 3*10^8 * 4π*10^-7)]

Emax = √233.7648

Emax = 15.289

Now, with our value of maximum electromagnetic wave gotten, we divide it by speed of light to get our final answer

15.289 / (3*10^8) = 5.096*10^-8 T

Suffice to say, The maximum value of the magnetic field in the wave is closest to 5.096*10^-8

6 0
2 years ago
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