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3 years ago

If a car starts from rest and attains a velocity of 10m/square in 5 secondscalculate the accerleration

Physics

2 answers:

stepan [7]3 years ago

8 0

Explanation:

Acceleration = Velocity / Time

Acceleration = 10 / 5

Acceleration = 2m/s ²

bogdanovich [222]3 years ago

7 0

Answer:

acceleration may be 34 beciase if calcuation maybr

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g The magnetic force on a charged particle A. depends on the sign of the charge on the particle. B. depends on the velocity of t

Leona [35]

Answer:

E) is described by all of these

Explanation:

The magnetic force on a charged particle is expressed as:

F = qv * B = qvBsinθ

Where,

q = charge on particle

θ = angle between the magnetic field and the particle velocity.

v = velocity of the particle

B = magnitude of field vector

From here, we could denote that magnetic force, F depends on charge on particle, velocity of particle, magnitude of field vector.

The magnetic force on a charged particle is at right angles to both the velocity of the particle. The magnetic force and magnetic field in a charged particle are perpendicular to each other, the right hand rule is used to determine the direction of force.

The correct option is E.

3 0

2 years ago

Read 2 more answers

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

A person walks first at a constant speed of 5.50 m/s along a straight line from point A to point B and then back along the line

Sav [38]

Answer:

4.25 m/s

Explanation:

They walked the first distance at 5.50 m/s, then the same distance at 3 m/s.

Since the distances are equal, the average speed is simply the average of 5.50 and 3.

(5.50 + 3) / 2 = 4.25

Her average speed over the entire trip is 4.25 m/s.

8 0

3 years ago

Breathing on a cold window can cause it to fog up. which change of state causes this

Dominik [7]

When you breathe on a cold window and it fogs up, that change of state is called: Condensation.

Condensation is: The process in which molecules of a gas slow down, come together, and form a liquid water that collects as droplets on a cold surface when humid air is in contact with it.

I know this because I learned it in science class in school.

Hope I helped!

- Debbie

3 0

3 years ago

A 100kg crate slides along a floor with a starting velocity of 21ms If the force due to friction is 8N how long will it take for

GaryK [48]

A 100kg crate slides along a floor with a starting velocity of 21 m/s. If the force due to friction is 8N, then, it will take 262.5 s for the box to come to rest.

We'll begin by calculating the declaration of the box. This can be obtained as follow:

Force (F) = –8 N (opposition)

Mass (m) = 100 Kg

<h3>Deceleration (a) =? </h3>

–8 = 100 × a

Divide both side by 1000

$a = \frac{-8}{100}$

Therefore, the deceleration of the box is –0.08 ms¯²

Finally, we shall determine the time taken for the box to come to rest. This can be obtained as follow:

Deceleration (a) = –0.08 ms¯²

Initial velocity (u) = 21 ms¯¹

Final velocity (v) = 0 ms¯¹

0 = 21 + (–0.08×t)

0 = 21 – 0.08t

Collect like terms

0 – 21 = –0.08t

–21 = –0.08t

Divide both side by –0.08

$t = \frac{-21}{-0.08}\\\\$

Therefore, it will take 262.5 s for the box to come to rest.

Learn more: brainly.com/question/14446351

7 0

2 years ago

If a car starts from rest and attains a velocity of 10m/square in 5 secondscalculate the accerleration​

If a car starts from rest and attains a velocity of 10m/square in 5 secondscalculate the accerleration