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Volgvan
3 years ago
9

The results from Rutherford's alpha particle scattering experiment led to

Physics
1 answer:
Brut [27]3 years ago
7 0

Answer:

Explanation:

Can you help me in mathematics

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a coin press creates a pressure of 3.20*10^8 Pa on a nickel of radius 0.0106 m. how much force does the press exert on the coin?
Naya [18.7K]

Answer:

1.13 x 10⁵N

Explanation:

Given parameters:

Pressure of the coin press = 3.2 x 10⁸ Pa

radius of the nickel coin = 0.0106m

Unknown:

Force of the press on coil = ?

Solution:

Our knowledge of pressure will help us solve this problem.

Pressure is defined as the force applied per unit area on a body.

              Pressure  = \frac{force}{area}

  Force = Pressure  x  Area

 Since the pressure is known;

Area of the coin = Area of a circle = π r²

 where r is the radius of the coin;

Area of the coin = π x 0.0106²  = 3.53 x 10⁻⁴m²

  Force = 3.2 x 10⁸ x  3.53 x 10⁻⁴  = 1.13 x 10⁵N

3 0
3 years ago
10. Solve the following numerical problems
frosja888 [35]

Answer:

\boxed {\boxed {\sf 120 \ Joules}}

Explanation:

Work is equal to the product of force and distance.

W=F*d

The force is 8 Newtons and the distance is 15 meters.

F= 8 \ N \\d= 15 \ m

Substitute the values into the formula.

W= 8 \ N * 15 \ m

Multiply.

W= 120 \ N*m

  • 1 Newton meter is equal to 1 Joule
  • Our answer of 120 N*m equals 120 J

W= 120 \ J

The work done is <u>120 Joules</u>

3 0
3 years ago
An object is located 10.0 cm from a convex mirror. The magnitude of the
Sunny_sXe [5.5K]

Answer:

B. 6.00 cm

Explanation:

6 0
3 years ago
If your body were a tall building, your skeleton would be
n200080 [17]
The Beams And Joints That Hold It .
4 0
3 years ago
Read 2 more answers
A dime is placed in front of a concave mirror that has a radius of curvature R = 0.40 m. The image of the dime is inverted and t
andrew11 [14]

Answer:

distance between the dime and the mirror, u = 0.30 m

Given:

Radius of curvature, r = 0.40 m

magnification, m = - 2 (since,inverted image)

Solution:

Focal length is half the radius of curvature, f = \frac{r}{2}

f = \frac{0.40}{2} = 0.20 m

Now,

m = - \frac{v}{u}

- 2 = -\frac{v}{u}

\frac{v}{u} = 2                  (2)

Now, by lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

\frac{1}{v} = \frac{1}{f} - \frac{1}{u}

v = \frac{uf}{u - f}            (3)

From eqn (2):

v = 2u

put v = 2u in eqn (3):

2u = \frac{uf}{u - f}

2 = \frac{f}{u - f}

2(u - 0.20) = 0.20

u = 0.30 m

6 0
3 years ago
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