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2 years ago

If the time of impact in a collision is extended by 3 times, by hour much is

Physics

1 answer:

vodomira [7]2 years ago

4 0

Answer:

The force of the impact would be smaller

Explanation: Examples:

If the force is big then the time would be small (2500N of Force = 10 seconds)

If the force is small then the time would be big (250N of Force = 50 seconds)

Impulse/Collision -> [Ft] = [M (vf-vo)] <- Change in momentum

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What is the amount of displacement of a runner who runs exactly 2 laps around a 400 meter track?

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Displacement is the distance and direction from the start point to the end point. Our runner finished exactly where he started. His displacement is zero.

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3 years ago

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What is the magnitude of the force that is exerted on a 20 kg mass to give it an acceleration of 10.0

ANEK [815]

Answer:Mass of the body = 20 kg.

Final Velocity = 5.8 m/s.

Initial velocity = 0

Time = 3 seconds.

Using the Formula,

Acceleration = (v - u)/ t

= (5.8 - 0)/ 3

= 1.6 m /s².

Now, Using the Formula,

Force = mass × acceleration

= 20 × 1.6

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2 years ago

An organization's _______ expresses the need the firm will fill, the operations of the business, its components and functions, a

sergiy2304 [10]

Answer:

Business Model

Explanation:

A business model is basically a plan of how a business will make a profit. It indicates what type of services or goods they will deliver, who they will provided services to, and the expenses to expect.

A good business model can be used to attract investors, motivate staff and management and recruit potential talent into the business.

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3 years ago

What is the wavelength of a monochromatic light beam, where the photon energy is 2.70 × 10^−19 J? (h = 6.63 ×10^−34 J⋅s, c = 3.0

SOVA2 [1]

Answer:

Wavelength = 736.67 nm

Explanation:

Given

Energy of the photon = 2.70 × 10⁻¹⁹ J

Considering:

$Energy=h\times frequency$

where, h is Plank's constant having value as 6.63 x 10⁻³⁴ J.s

The relation between frequency and wavelength is shown below as:

c = frequency × Wavelength

Where, c is the speed of light having value = 3×10⁸ m/s

So, Frequency is:

Frequency = c / Wavelength

So, Formula for energy:

$Energy=h\times \frac {c}{\lambda}$

Energy = 2.70 × 10⁻¹⁹ J

c = 3×10⁸ m/s

h = 6.63 x 10⁻³⁴ J.s

Thus, applying in the formula:

$2.70\times 10^{-19}=6.63\times 10^{-34}\times \frac {3\times 10^8}{\lambda}$

Wavelength = 736.67 × 10⁻⁹ m

1 nm = 10⁻⁹ m

So,

<u>Wavelength = 736.67 nm</u>

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3 years ago

Find the quantity of heat needed

krok68 [10]

Answer:

Approximately $3.99\times 10^{4}\; \rm J$ (assuming that the melting point of ice is $0\; \rm ^\circ C$ .)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

where

$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}$ .

Similarly, for the third part of energy input, $c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}$ .

The second part of energy input requires a different equation. The energy $Q$ required to melt a sample of mass $m$ and latent heat of fusion $L_\text{f}$ is:

$Q = m \cdot L_\text{f}$ .

Apply this equation to find the size of the second part of energy input:

$\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}$ .

Find the sum of these three parts of energy:

$\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}$ .

3 0

2 years ago