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3 years ago

What is the acceleration of the the object during the first 4 seconds?

Physics

1 answer:

AVprozaik [17]3 years ago

6 0

Answer:

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

Explanation:

What this formula is telling us is that if we know the acceleration of an object, and the ... we can plug in our acceleration of 12.5 m/s2 for a, and 4 seconds for t.

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

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Projectiles that strike objects are good examples of inelastic collisions. A 0.1 kg nail driven by a gas powered nail driver col

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In an inelastic collision, only momentum is conserved, while energy is not conserved.

1) Velocity of the nail and the block after the collision
This can be found by using the total momentum after the collisions:
$p_f=(m+M)v_f=4.8 kg m/s$
where
m=0.1 kg is the mass of the nail
M=10 kg is the mass of the block of wood
Rearranging the formula, we find $v_f$ , the velocity of the nail and the block after the collision:
$v_f= \frac{p_f}{m+M}= \frac{4.8 kg m/s}{0.1 kg+10 kg}= 0.48 m/s$

2) The velocity of the nail before the collision can be found by using the conservation of momentum. In fact, the total momentum before the collision is given only by the nail (since the block is at rest), and it must be equal to the total momentum after the collision:
$p_i = mv_i = p_f$
Rearranging the formula, we can find $v_i$ , the velocity of the nail before the collision:
$v_i = \frac{p_f}{m}= \frac{4.8 kg m/s}{0.1 kg}=48 m/s$

6 0

3 years ago

Read 2 more answers

You are working in cooperation with the Public Health department to design an electrostatic trap for particles from auto emissio

Mademuasel [1]

Answer:

Explanation:

Given that:

Charge (q) on the particle = 3 × 10⁻⁸ C

mass (m) of the particle = 6 × 10⁻⁹ kg

at a distance x = 15 cm , the velocity in the plate = 900 m/s²

For the square plate, the surface charged density σ = -8 × 10⁻⁶ C/m²

To start with calculating the electric field as a result of the square plate; we use the formula;

$E = \dfrac{\sigma }{2 \varepsilon_o}$

$E = \dfrac{8 \times 10^{-6} }{2 \times 8.85 \times 10^{-12}}$

$E = 4.51977 \times 10^5 \ V/m$

On the square plate; The electric force F = Eq

$F = (4.51977 \times 10^5 \ V/m )(3\times 10^{-8} \ C)$

$F = 1.3559 \times 10^{-2} \ N$

The acceleration $a =\dfrac{ F}{m}{$

$a = \dfrac{1.3559\times 10^{-2} \ N}{6 \times 10^{-9} \ Kg}$

$a = 2.25988 \times 10^6 \ m/s^2$

For the particle, the velocity at distance x = 7 m can be calculated by using the formula:

$(\dfrac{1}{2}) mv^2 = \Delta Vq$

$v^2 = \dfrac{2 Eq}{dm}$

$v^2 = \dfrac{2 * 4.51977 \times 10^5 \times 3 \times 10^{-8} }{0.07 \times 6\times 10^{-9} }$

$v^2 = 64568142.86 \ m/s$

$v =\sqrt{ 64568142.86 \ m/s}$

$\mathbf{v = 8.035 \times 10^3 \ m/s}$

From the calculation, we realize that the charge acting between the particle and the plate is said to be "opposite".

Hence, the force is an attractive force.

Similarly, there is a gradual increase exhibited by the velocity of the particle.

Therefore, the particles get to the detector, but the detector failed to get detect due to the velocity which is greater than 1000 m/s.

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3 years ago

A 11 g plastic ball is moving to the left at 29 m/s. How much work must be done on the ball to cause it to move to the right at

Rasek [7]

Answer:

4.25 J

Explanation:

Given that

mass of plastic ball = 11 g

Mass of plastic ball = 0.011 kg

velocity of ball = 29 m/s

We know that from work power energy theorem

$W_{all}=Change\ in\ kinetic\ energy\ of\ system$

We know that kinetic energy of moving mass given as

$KE=\dfrac{1}{2}mv^2$

Now by pitting the values

$KE=\dfrac{1}{2}mv^2$

$KE=\dfrac{1}{2}\times 0.011\times 29^2$

KE= 4.25 J

So the work done on the ball is 4.25 J

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3 years ago

What properties of the wave define why it is found within this area of the spectrum?

MatroZZZ [7]

Answer:

Explanation:

Speed and medium are properties of a wave which is in common within an area of the spectrum of visible light. i.e;

Their medium of propagation

They both travel at the same speed ( speed of light )

The properties mentioned above are properties that define that wave is found within an area spectrum for visible light.

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2 years ago

Which describes Michael Faraday’s work with electricity and magnetism?

Liono4ka [1.6K]

<span>Faraday is best known for his work regarding electricity and magnetism. His first recorded experiment was the construction of a voltaic pile with seven ha'penny coins, stacked together with seven disks of sheet zinc, and six pieces of paper moistened with salt water.</span>

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2 years ago

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