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Anastasy [175]
3 years ago
9

A particular field is 111 yards long. Express this length in meters.

Physics
2 answers:
romanna [79]3 years ago
8 0

Answer:

101.498 m

Explanation:

STALIN [3.7K]3 years ago
7 0

1 yard = 0.9144 meter

111 yards = <em>101.5 meters</em>

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20.0 -kg cannonball is fired from a cannon with muzzle speed of 1000m/s at an angle of 37.0° with the horizontal. A second ball
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.

To know about energy, refer to the below link:

brainly.com/question/1932868

#SPJ4

5 0
1 year ago
A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha
USPshnik [31]

Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

\Phi_E=\frac{Q}{\epsilon_o}

ФE: electric flux trough the Gaussian surface

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C

Finally, you obtain for E:

E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

3 0
3 years ago
How much heat is needed to warm 0.072kg of gold from 20 celsius and 90 celsius if the specific heat of gold 136 joules
dybincka [34]

Heat supplied to the gold will raise the temperature of the gold from 20 degree Celsius to 90 degree Celsius.

Mass of the gold (m) = 0.072 kg

Temperature change (ΔT) = 90 - 20 = 70 degree Celsius

Specific heat capacity of the gold (c) = 136 J/kg C

Heat supplied = m × c × ΔT

Heat supplied = 0.072 × 136 × 70

Heat supplied = 685.44 Joules

Hence, the heat supplied to the gold to raise the temperature from 20 degree Celsius to 90 degree Celsius = 685.44 Joules

5 0
3 years ago
The acquisition of electric charge without contact between charged and/or uncharged substances.
Lera25 [3.4K]

Answer:

induction

Explanation:

I just did usatestprep

4 0
2 years ago
Read 2 more answers
How much heat is needed to change the temperature of 3 grams of gold (c = 0.129 ) from 21°C to 363°C? The answer is expressed to
Temka [501]

Q= mcΔT

Where Q is heat or energy

M is mass, c is heat capacitance and t is temperature

You have to convert Celsius into kelvin in order to use this formula I believe

Celsius + 273 = Kelvin

21 + 273 = 294K

363 + 273 = 636K

Now...

Q= (0.003)(0.129)(636-294)

Q= 0.132 J if you are using kilograms, in terms of grams which seems more appropriate the answer would be 132J of energy.  

3 0
2 years ago
Read 2 more answers
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