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3 years ago

A car speeds up from 0 m/s to 14.0 m/s in 325 s. What is the acceleration?

Physics

1 answer:

Anton [14]3 years ago

8 0

Answer: $0.04308 m/s^2$

Explanation: Plug the variables into the equation: $a=\frac{v-v_{0} }{t}$ | $v$ equals initial speed, $v_{0}$ equals final speed and $t$ equals time.

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S Problem Set<br> 2.) 6.4 x 109 nm to cm

anyanavicka [17]

Answer:

$6.4\cdot 10^2 cm$

Explanation:

First of all, let's convert from nanometres to metres, keeping in mind that

$1 nm = 10^{-9} m$

So we have:

$6.4\cdot 10^9 nm \cdot 10^{-9} m/nm = 6.4 m$

Now we can convert from metres to centimetres, keeping in mind that

$1 m = 10^2 cm$

So, we find:

$6.4 m \cdot 10^2 cm/m = 6.4\cdot 10^2 cm$

8 0

2 years ago

When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a

sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = $\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}$

= $4.59\times 10^{-19} \ J$

or,

= $\frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}$

= $2.87 \ eV$

(b)

As we know,

⇒ $Vq=\frac{hc}{\lambda}-\Phi_0$

By substituting the values, we get

⇒ $1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0$

⇒ $\Phi_0=2.3\times 10^{-19} \ J$

or,

⇒ $=\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}$

⇒ $=1.4375 \ eV$

5 0

2 years ago

A2 kg object moves at a constant 5 m/s across a level surface. Determine its kinetic energy.

AURORKA [14]

KE = 1/2mv^2
KE= 1/2(2)(5)^2
KE= 25 J

6 0

2 years ago

In which of these situations would there be the least kinetic energy?

Alexandra [31]

C. a truck rolling down hill

7 0

2 years ago

The current theory of the structure of the

Elza [17]

Answer: 91.4 J

Explanation:

Kinetic energy is the energy possessed by a body due to virtue of its motion.

K.E. = 0.5 m v²

Mass of the continent is given, m = 1.819 × 10²¹ kg

Side of the block of continent, s = 4150 km = 4150000 m

Depth of the block of continent, d = 38 km = 38000 m

(Mass = density × volume

m = 2780 kg/m³× (4150 × 10³ m)²× 38 × 10³ m = 1.819 × 10²¹ kg)

The continent is moving at the rate of, v = 1 cm /year = 0.01 m / 31556926 s = 3.17 × 10⁻¹⁰ m/s

⇒ K.E. = 0.5 × 1.819 × 10²¹ kg × (3.17 × 10⁻¹⁰ m/s)²= 91.4 J

Hence, mass of the continent has 91.4 J of kinetic energy.

5 0

3 years ago