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s344n2d4d5 [400]

2 years ago

12

A car has a weight of 25000 N and its brakes can apply a maximum force of 628 N to stop it. The car is initially moving at a spe

ed of 6.5 m/s and comes to a complete stop.
How much time will it take for the car to stop?

1 answer:

svlad2 [7]2 years ago

5 0

Time required : 25.876 s

<h3>Further explanation</h3>

Given

A car weight = 25000 N

Brake force = 628 N

vo= 6.5 m/s

vt = 0(stop)

Required

time to stop

Solution

W = m . g

25000 = m.10

m=2500 kg

Newton's second law

F = m . a

628 = 2500 . a

$\tt a=\dfrac{628}{2500}=0.2512~m/s^2$

vt=vo-at(deceleration a=-)

0=6.5-0.2512.t

0.2512.t=6.5

t=25.876 s

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Answer:

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4. <u>F = ma</u> <em>F = 50Kg * 4m/s² =</em> 200N

5. <u>F = ma</u> <em>F = 70Kg * 4m/s² =</em> 280N

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8 0

2 years ago

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent

azamat

The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

<em>Length of the string, L = 100 cm</em>

<em />

The wavelengths of the constituent travelling waves is calculated as follows;

$L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}$

for first mode: n = 1

$\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm$

for second mode: n = 2

$\lambda = \frac{2L}{2} = L = 100 \ cm$

For the third mode: n = 3

$\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm$

For fourth mode: n = 4

$\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50 \ cm$

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

Learn more about wavelengths of travelling waves here: brainly.com/question/19249186

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2 years ago

7. You start to walk toward the east towards home at a constant speed of 4 km/hr. At the same time, Someone else leaves your hom

amm1812

A) position time graph for both is shown

here one of the graph is of lesser slope which means it is moving with less speed while other have larger slope which shows larger speed

At one point they intersects which is the point where they both will meet

B) Let the two will meet after time "t"

now we can say that

if they both will meet after time "t"

then the total distance moved by you and other person will be same as the distance between you and home

so it is given as

$v_1t + v_2t = d$

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$t = \frac{3.2}{32} = 0.1 hr$

so they will meet after t = 6 min

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4 0

3 years ago

List the parts of the electromagnetic spectrum from longest wavelength to shortest wavelength

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The rays of the electromagnetic spectrum from shortest to longest wavelength are: radio waves, microwaves, infrared rays, optical rays, ultraviolet rays, X-rays, and gamma-rays.

EDIT: He has these backwards, the shortest wavelength is created by Gamma-Rays and the longest is Radiowaves.

<span> Remember- high energy = short wavelength. </span>

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3 years ago

The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge

lesya [120]

Answer:

Loss, $\Delta E=-10.63\ J$

Explanation:

Given that,

Mass of particle 1, $m_1=m =0.66\ kg$

Mass of particle 2, $m_2=7.4m =4.884\ kg$

Speed of particle 1, $v_1=2v_o=2\times 6=12\ m/s$

Speed of particle 2, $v_2=v_o=6\ m/s$

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

$m_1v_1+m_2v_2=(m_1+m_2)V$

$V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}$

$V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}$

V = 6.71 m/s

Initial kinetic energy before the collision,

$K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)$

$K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)$

$K_i=135.43\ J$

Final kinetic energy after the collision,

$K_f=\dfrac{1}{2}(m_1+m_2)V^2$

$K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2$

$K_f=124.80\ J$

Lost in kinetic energy,

$\Delta K=K_f-K_i$

$\Delta K=124.80-135.43$

$\Delta E=-10.63\ J$

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

7 0

3 years ago