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g100num [7]
3 years ago
12

A ball rolls down an incline with an acceleration of 10 cm/s^2. If it starts with an initial velocity of 0 cm/s and has a velocy

of 50 cm/s when it reaches the bottom of the ramp, how long is the ramp?
Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
8 0
Given a = 10 cm/s²
          u = 0 cm/s
          v = 50 cm/s
we know that 
         v²=u²+2aS
        2500=2×10×S
        2500÷20 = S
        S= 125 cm
The ramp is 125 cm
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Which two properties of light relate to the amount of energy carried by the light? A. speed and amplitude
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6 0
3 years ago
A 2.0-mm-diameter copper ball is charged to 40 nC . What fraction of its electrons have been removed? The density of copper is 8
murzikaleks [220]

Answer:

0.02442 × 10⁻⁹

Explanation:

Given:

Diameter of copper ball = 2.00 mm = 0.002 m

Charge on ball = 40 nC = 40 × 10⁻⁹ C

Density of copper = 8900 Kg/m³

Now,

The number of electrons removed, n = \frac{\textup{Charge on ball}}{\textup{Charge of an electron}}

also, charge on electron = 1.6 × 10⁻¹⁹ C

Thus,

n = \frac{40\times10^{-9}}{1.6\times10^{-19}}

or

n = 25 × 10¹⁰ Electrons

Now,

Mass of copper ball = volume × density

Or

Mass of copper ball =  \frac{4}{3}\pi(\frac{d}{2})^3  × 8900

or

Mass of copper ball =  \frac{4}{3}\pi(\frac{0.002}{2})^3  × 8900

or

Mass of copper ball = 0.03726 grams

Also,

molar mass of copper = 63.546 g/mol

Therefore,

Number of mol of copper in  0.03726 grams = \frac{ 0.03726}{63.546}

or

Number of mol of copper in  0.03726 grams = 5.86 × 10⁻⁴ mol

and,

1 mol of a substance contains = 6.022 × 10²³ atoms

Therefore,

5.86 × 10⁻⁴ mol of copper contains = 5.86 × 10⁻⁴ × 6.022 × 10²³ atoms.

or

5.86 × 10⁻⁴ mol of copper contains = 35.88 × 10¹⁹ atoms

Now,

A neutral copper atom has 29 electrons.  

Therefore,

Number of electrons in ball = 29 × 35.88 × 10¹⁹ = 1023.37 × 10¹⁹ electrons.

Hence,

The fraction of electrons removed = \frac{25\times10^{10}}{1023.37\times10^{19}}

or

The fraction of electrons removed = 0.02442 × 10⁻⁹

3 0
3 years ago
a spring gun initially compressed 2cm fires a 0.01kg dart straight up into the air. if the dart reaches a height it 5.5m determi
Vikki [24]

Answer:

2697.75N/m

Explanation:

Step one

This problem bothers on energy stored in a spring.

Step two

Given data

Compression x= 2cm

To meter = 2/100= 0.02m

Mass m= 0.01kg

Height h= 5.5m

K=?

Let us assume g= 9.81m/s²

Step three

According to the principle of conservation of energy

We know that the the energy stored in a spring is

E= 1/2kx²

1/2kx²= mgh

Making k subject of formula we have

kx²= 2mgh

k= 2mgh/x²

k= (2*0.01*9.81*5.5)/0.02²

k= 1.0791/0.0004

k= 2697.75N/m

Hence the spring constant k is 2697.75N/m

7 0
3 years ago
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