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Naily [24]
2 years ago
12

A particle moving along a straight line with constant acceleration has a velocity of 2.35 m/s at t=3.42 s, and a velocity of -8.

72 m/s at t=5.59s what is the particle's acceleration in m/s?
Physics
1 answer:
mrs_skeptik [129]2 years ago
4 0

The particle's acceleration is 5.1 m/s²

<h3>What is Acceleration ?</h3>

Acceleration can be defined as the rate at which velocity is changing. It is a vector quantity and it is measured in m/s²

Given that a particle is moving along a straight line with constant acceleration has a velocity of 2.35 m/s at t=3.42 s, and a velocity of -8.72 m/s at t=5.59s

The given parameters are;

  • V1 = 2.35 m/s
  • V2 = - 8.72 m/s
  • T1 = 3.42s
  • T2 = 5.59s

Acceleration a = ΔV ÷ ΔT

a = (2.35 + 8.72) / (5.59 - 3.42)

a = 11.07 / 2.17

a = 5.1 m/s²

Therefore, the particle's acceleration is 5.1 m/s²

Learn more about Acceleration here: brainly.com/question/9069726

#SPJ1

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How many minutes of daylight do we gain after winter solstice
serg [7]

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Explanation:

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A truck travels down the highway at a speed of 110 km/hr. How long does the trip last if the truck covered 2200km?
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8 0
3 years ago
A projectile is fired at v 0 = 381.0 m/s at an angle of θ = 73.5 ∘ , with respect to the horizontal. Assume that air friction wi
Ipatiy [6.2K]

Answer:

Explanation:

velocity of projection, vo = 381 m/s

angle of projection, θ = 73.5°

The formula for the range is

R=\frac{u^{2}Sin2\theta }{g}

R=\frac{381^{2}Sin147 }{9.8}

R = 8067.4 m

Range in shorten by 34.1 %

So, the new range is

R' = 8067.4 - 34.1 x 8067.4/100

R' = 5316.4 m

5 0
3 years ago
A 1900 kg car rounds a curve of 55 m banked at an angle of 11 degrees? . If the car is traveling at 98 km/h, How much friction f
Nat2105 [25]

Answer:

22000 N

Explanation:

Convert velocity to SI units:

98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s

Draw a free body diagram.  There are three forces acting on the car.  Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.

I'm going to assume the friction force is pointed down the bank.  If I get a negative answer, that'll just mean it's actually pointed up the bank.

Sum of the forces in the radial direction (+x):

∑F = ma

N sin θ + F cos θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ - F sin θ - W = 0

To solve the system of equations for F, first solve for N and substitute.

N = (W + F sin θ) / cos θ

Substituting:

((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r

(W + F sin θ) tan θ + F cos θ = m v² / r

W tan θ + F sin θ tan θ + F cos θ = m v² / r

W tan θ + F (sin θ tan θ + cos θ) = m v² / r

W tan θ + F sec θ = m v² / r

F sec θ = m v² / r - W tan θ

F = m v² cos θ / r - W sin θ

F = m (v² cos θ / r - g sin θ)

Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:

F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)

F = 21577 N

Rounding to two sig-figs, you need at least 22000 N of friction force.

4 0
3 years ago
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