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JulijaS [17]
2 years ago
14

A car starting from rest (i.e. initial velocity = 0.0 m/s), moves in the positive X-direction with a constant average accelerati

on = 3.1 m/s2 for 7.9 s. What will be its velocity at that time.
Physics
1 answer:
olga_2 [115]2 years ago
8 0

Answer:

Final speed of the car, v = 24.49 m/s

Explanation:

It is given that,

Initial velocity of the car, u = 0

Acceleration, a=3.1\ m/s^2

Time taken, t = 7.9 s

We need to find the final velocity of the car. Let it is given by v. It can be calculated using first equation of motion as :

v=u+at

v=0+3.1\times 7.9

v = 24.49 m/s

So, the final speed of the car is 24.49 m/s. Hence, this is the required solution.

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Now suppose the initial velocity of the train is 4 m/s and the hill is 4 meters tall. If the train has a mass of 30000 kg, what
hoa [83]

Answer:

<h2>187,500N/m</h2>

Explanation:

From the question, the kinectic energy of the train will be equal to the energy stored in the spring.

Kinetic energy = 1/2 mv² and energy stored in a spring E = 1/2 ke².

Equating both we will have;

1/2 mv² = 1/2ke²

mv² = ke²

m is the mass of the train

v is the velocity of then train

k is the spring constant

e is the extension caused by the spring.

Given m = 30000kg, v = 4 m/s, e = 4 - 2.4 = 1.6m

Substituting this values into the formula will give;

30000*4² =  k*1.6²

k = \frac{30,000*16}{1.6^2}\\ \\k = \frac{480,000}{2.56}\\ \\k = 187,500Nm^{-1}

The value of the spring constant is 187,500N/m

7 0
2 years ago
Gas a bG1 5.22 0.0289G2 1.05 0.0388G3 2.31 0.0467G4 4.05 0.0310Based on the given van der Waals constants for four hypothetical
inysia [295]

Answer:

Gas 2, Gas 3, Gas 4, Gas 5 is the order of decreasing strength of inter-molecular forces.

Explanation:

The strength increases as there is a decrease in the vanderwaals constant and vice versa.

3 0
1 year ago
A 1700kg rhino charges at a speed of 50.0km/h. what is the magnitude of the average force needed to bring the rhino to a stop in
stepladder [879]
Impulse equals Change in Momentum
F = average applied force = to be determined
Δt = time during which the force is applied = 0.50 s
m = mass = 1,700 kg
Δp = change in momentum = to be determined
Δv = change in velocity = to be determined
v1 = initial velocity = 50.0 km/h = 50,000 m/h = 13.9 m/s
v2 = final velocity = 0.00 km/h = 0.00 m/s

F∙Δt = Δp
F∙Δt = m∙Δv
F∙Δt = m∙(v2 - v1)

F = m∙(v2 - v1) / Δt
F = 1,700 kg∙(0.00 m/s - 13.9 m/s) / 0.50 s
<span>F = -47,222 N The negative sign means that the force vector is </span>
<span>applied AGAINST the momentum vector of the rhinoceros.</span>
7 0
2 years ago
Some hydrogen gas is enclosed within a chamber being held at 200∘c with a volume of 0.0250 m3. the chamber is fitted with a mova
Mrac [35]

Answer: The final volume V₂ of the container is  0.039 m³.

Explanation:

Since the temperature is constant, the gas would expand isothermally.

For isothermal expansion,

P₁V₁=P₂V₂

Where, P₁ and P₂ are the initial and final pressure and V₁ and V₂ are initial and final volume.

It is given that:

V₁ = 0.0250 m³

P₁ = 1.5 × 10⁶ Pa

P₂ = 0.950 × 10⁶ Pa

V₂ = ?

⇒ 1.5 × 10⁶ Pa × 0.0250 m³ = 0.950 × 10⁶ Pa × V₂

⇒V₂ = 0.039 m³

Hence, the final volume V₂ of the container is  0.039 m³.

4 0
1 year ago
A cyclotron particle accelerator (sometimes called an “atom smasher” in the popular press) is a device for accelerating charged
Iteru [2.4K]

Answer:

v=1.54\times 10^{7}}\ \textup{m/s}

Explanation:

Given:

The accelerated energy, U = 1.25 MeV = 1.25 × 10⁶ eV

we know,

1 eV = 1.6 × 10⁻¹⁹ J

thus,

1.25 eV = (1.6 × 10⁻¹⁹) × (1.25) J = 2 × 10⁻¹³ J

Now, Applying the law of conservation of energy, the energy due to acceleration will be equal to the kinetic energy

mathematically,

K.E = U

\frac{1}{2}mv^2=2\times 10^{-13} \ \textup{J}

where,

m = mass of the particle = 1.67 × 10⁻²⁷ kg

v = velocity of the particle

on substituting the values we get

\frac{1}{2}\times 1.67\times 10^{-27}\times v^2=2\times 10^{-13} \ \textup{J}

or

v^2=\frac{2\times 10^{-13}}{1.67\times 10^{-27}}

or

v=\sqrt{2.39\times 10^{14}}

or

v=1.54\times 10^{7}}\ \textup{m/s}

7 0
2 years ago
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