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Studentka2010 [4]

2 years ago

5

A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes

and the car comes to a rest uniformly in a distance of 200 m. What are the magnitude and direction of the net force applied to the car to bring it to rest?

1 answer:

Fofino [41]2 years ago

6 0

Answer:

Force will be acting southward and the magnitude of force will be 1000 N

Explanation:

given,

mass of car = 1000 Kg

initial speed of the car (u) = 20 m/s

final speed of the car (v) = 0 m/s

distance to stop the car = 200 m

using equation of motion

v² = u² + 2 a s

0 = 20² + 2 x a x 200

400 a = -400

a = -1 m/s²

Now, we know

Force = mass x acceleration

F = 1000 x -1

F = -1000 N

- ve sign of force represent force will be acting in the opposite direction of motion.

Force will be acting southward and the magnitude of force will be 1000 N

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Answer:

The magnitude of buoyancy force is equal to that of ball's weight.

Explanation:

Ball 1 is floating on water. Weight of ball 1 is Fg=m1g is acting vertically downward

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Net force acting on the ball is zero, FB=Fg

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2 years ago

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A single slit, which is 0.050 mm wide, is illuminated by light of 550 nm wavelength. What is the angular separation between the

likoan [24]

Answer:

The separation between the first two minima on either side is 0.63 degrees.

Explanation:

A diffraction experiment consists on passing monochromatic light trough a small single slit, at some distance a light diffraction pattern is projected on a screen. The diffraction pattern consists on intercalated dark and bright fringes that are symmetric respect the center of the screen, the angular positions of the dark fringes θn can be find using the equation:

$a\sin \theta_n=n\lambda$

with a the width of the slit, n the number of the minimum and λ the wavelength of the incident light. We should find the position of the n=1 and n=2 minima above the central maximum because symmetry the angular positions of n=-1 and n=-2 that are the angular position of the minima below the central maximum, then:

for the first minimum

$a\sin \theta_1=(1)\lambda$

solving for θ1:

$\theta_1=\arcsin (\frac{\lambda}{a})=\arcsin (\frac{550\times10^{-9}}{0.05\times10^{-3}})$

$\theta_1=0.63 degrees$

for the second minimum:

$a\sin \theta_2=(2)\lambda$

$\theta_2=\arcsin (\frac{2\lambda}{a})=\arcsin (\frac{2*550\times10^{-9}}{0.05\times10^{-3}})$

$\theta_2=1.26 degrees$

So, the angular separation between them is the rest:

$\Delta \theta =1.26-0.63$

$\Delta \theta=0.63$

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2 years ago

An electric winch is used to raise a 40-kg package and a 10-kg package vertically up the side of a building as pictured in the d

just olya [345]

Answer:

Explanation:

40 divided by 10 then which would equal 4. Add the 1.0 , 2 ,and 15 together. Then multply the 60 by 18.0 after you are done dividing the answer is 3 with a remainder of 6.

3 0

2 years ago

Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The

Eddi Din [679]

Answer:

(a) Steel rod: $1.1 * 10^{-4}$

Copper rod: $1.88 * 10^{-4}$

(b) Steel rod: $8.3 * 10^{-5} m$

Copper rod: $1.41 * 10^{-4} m$

Explanation:

Length of each rod = 0.75 m

Diameter of each rod = 1.50 cm = 0.015 m

Tensile force exerted = 4000 N

(a) Strain is given as the ratio of change in length to the original length of a body. Mathematically, it is given as

Strain = $\frac{1}{Y} * \frac{F}{A}$

where Y = Young modulus

F = Fore applied

A = Cross sectional area

For the steel rod:

Y = 200 000 000 000 $N/m^{2}$

F = 4000N

A = $\pi r^{2}$ (r = d/2 = 0.015/2 = 0.0075 m)

=> A = $\pi * (0.0075)^{2}$

=> A = 0.000177 $m^{2}$

∴ $Strain = \frac{4000}{200000000000 * 0.000177} \\\\Strain = \frac{4000}{35400000}\\ \\Strain = 0.000113 = 1.13 * 10^{-4}$

For the copper rod:

Y = 120 000 000 000 N/m²

F = 4000N

A = $\pi r^{2}$ (r = d/2 = 0.015/2 = 0.0075 m)

=> A = $\pi * (0.0075)^{2}$

=> A = 0.000177 $m^{2}$

$Strain = \frac{4000}{120 000 000 000 * 0.000177} \\\\Strain = \frac{4000}{21240000}\\ \\Strain = = 1.88 * 10^{-4}$

(b) We can find the elongation by multiplying the Strain by the original length of the rods:

Elongation = Strain * Length

For the steel rod:

Elongation = $1.1 * 10^{-4} * 0.75 = 8.3 * 10^{-5} m$

For the copper rod:

Elongation = $1.88 * 10^{-4} * 0.75 = 1.41 * 10^{-4} m$

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2 years ago

Airplane flight recorders must be able to survive catastrophic crashes. Therefore, they are typically encased in crash-resistant

chubhunter [2.5K]

Answer:

3.2 MN

Explanation:

Given that

Mass of the box, m = 52 kg

Initial velocity of the box, u = 400 m/s

Final velocity of the box, v = 0 m/s

Time taken for the collision, t = 0.0065 s

Using the equation of motion

V = u + at, we turn around and make acceleration, a the subject of the formula. Now we have,

a = (v - u) / t

a = (0 - 400) / 0.0065

a = 61538.5 m/s²

The acceleration(which is negative acceleration or retar.dation actually) is 61538.5 m/s². We then proceed to is this acceleration in the basic Force equation, to get the magnitude of force needed.

Remember,

Force = mass * acceleration

F = ma, we already have our mass and acceleration, all we do is multiply

F = 52 * 61538.5

F = 3200002 N or 3.2 MN

Therefore, the magnitude of the force that acts on the box during collision is 3.2 MN

8 0

2 years ago